NCERT Exemplar for Class 11 Maths Chapter 9 - Sequences and Series (Book Solutions)

Solved examples

Short Answer Type

Example 1: The first term of an A.P. is a, the second term is b and the last term is c. Show that the sum of the A.P. is $\dfrac{{\left( {b + c - 2a} \right)\left( {c + a} \right)}}{2{b - a}}$

Ans: Let d be the common difference and n be the number of terms of the A.P.

Since the first term is a and the second term is b.

Therefore, d = b – a

Also, the last term is c, so

c = a + (n – 1) (b – a) (since d = b – a)

So, n – 1 = \[\dfrac{{c - a}}{{b - a}}\] 

   $\Rightarrow n = 1 + \dfrac{{c - a}}{{b - a}}$

   $= \dfrac{{b - a + c - a}}{{b - a}}$

   $= \dfrac{{b + c - 2a}}{{b - a}}$

Thus,

${S_n} = \dfrac{n}{2}\left( {a + l} \right) = \dfrac{{\left( {b + c - 2a} \right)}}{{2\left( {b - a} \right)}}\left( {a + c} \right)$

Example 2: The $p^{\text {th }}$ term of an A.P. is a and $q^{\text {th }}$ term is b. Prove that the sum of its (p + q) terms is $\dfrac{{p + q}}{2}\left( {a + b + \dfrac{{a - b}}{{p - q}}} \right).$

Ans: Let A be the first term and D be the common difference of the A.P. It is given that

$\mathrm{t}_{\mathrm{p}}$= a $\Rightarrow$ A + (p – 1) D = a ... (1)

$\mathrm{t}_{\mathrm{q}}$= b $\Rightarrow$ A + (q – 1) D = b ... (2)

Subtracting (2) from (1), we get

(p – 1 – q + 1) D = a – b

$\Rightarrow$ D = $\dfrac{{a - b}}{{p - q}}$... (3)

Adding (1) and (2),

2A + (p + q – 2) D = a + b

$\Rightarrow$ 2A + (p + q – 1) D = a + b + D

$\Rightarrow$ 2A + (p + q – 1) D = a + b + $\dfrac{{a - b}}{{p - q}}$

Now,

  ${S_{p + q}} = \dfrac{{p + q}}{2}\left[ {2A + \left( {p + q - 1} \right)D} \right]$

  $= \dfrac{{p + q}}{2}\left( {a + b + \dfrac{{a - b}}{{p - q}}} \right){\text{   }}\left( {{\text{Using equation }}\left( 3 \right){\text{and}}\left( 4 \right)} \right)$ 

Example 3: If there are (2n + 1) terms in an A.P., then prove that the ratio of the sum of odd terms and the sum of even terms is (n + 1) : n.

Ans: Let a be the first term and d the common difference of the A.P. Also let $\mathrm{S}_{1}$ be the sum of odd terms of A.P. having (2n + 1) terms. Then

  ${S_1} = {a_1} + {a_3} + {a_5} + ... + {a_{2n + 1}}$

  ${S_1} = \dfrac{{n + 1}}{2}\left( {{a_1} + {a_{2n + 1}}} \right)$

  ${S_1} = \dfrac{{n + 1}}{2}\left[ {a + a + \left( {2n + 1 - 1} \right)d} \right]$

  ${S_1} = \left( {n + 1} \right)\left( {a + nd} \right)$ 

Similarly, if $\mathrm{S}_{2}$ denotes the sum of even terms, then

  ${S_2} = \dfrac{n}{2}\left( {2a + 2nd} \right) = n\left( {a + nd} \right)$

  ${\text{Hence,}}$

  $\dfrac{{{S_1}}}{{{S_2}}} = \dfrac{{\left( {n + 1} \right)\left( {a + nd} \right)}}{{n\left( {a + nd} \right)}} = \dfrac{{n + 1}}{n}$ 

Example 4: At the end of each year the value of a certain machine has depreciated by 20% of its value at the beginning of that year. If its initial value was Rs 1250, find the value at the end of 5 years.

Ans: After each year the value of the machine is 80% of its value the previous

year so at the end of 5 years the machine will depreciate as many times as 5.

Hence, we have to find the 6th term of the G.P. whose first term a1 is 1250 and common

ratio r is .8.

Hence, value at the end 5 years $=\mathrm{t}_{6}=\mathrm{a}_{1} \mathrm{r}^{5}=1250(.8)^{5}=409.6$

Example 5: Find the sum of first 24 terms of the A.P. $a_{1}, a_{2}, a_{3}, \ldots$ if it is known that $a_{1}+a_{5}+a_{10}+$ $a_{15}+a_{20}+a_{24}=225$.

Ans: We know that in an A.P., the sum of the terms equidistant from the beginning

and end is always the same and is equal to the sum of first and last term.

Therefore d = b – a

i.e., \[{a_1} + {a_{24}} = {a_5} + {a_{20}} = {a_{10}} + {a_{15}}\]

It is given that $\left(a_{1}+a_{24}\right)+\left(a_{5}+a_{20}\right)+\left(a_{10}+a_{15}\right)=225$

Thus, $\left(a_{1}+a_{24}\right)+\left(a_{1}+a_{24}\right)+\left(a_{1}+a_{24}\right)=225$

$3\left(a_{1}+a_{24}\right)=225$

Thus, $a_{1}+a_{24}=75$

We know that \[{S_n} = \dfrac{n}{2}\left( {a + l} \right)\], where a is the first term and l is the last term of an A.P.

Thus,

 ${S_{24}} = \dfrac{{24}}{2}\left( {{a_1} + {a_{24}}} \right)$

$= 12 \times 75 = 900$ 

Example 6: The product of three numbers in A.P. is 224, and the largest number is 7 times the smallest. Find the numbers.

Ans: Let the three numbers in A.P. be a – d, a, a + d (d > 0)

Now (a – d) a (a + d) = 224

$\Rightarrow$a (a2 – d2) = 224 ... (1)

Now, since the largest number is 7 times the smallest, i.e., a + d = 7 (a – d)

Therefore, $d = \dfrac{{3a}}{4}$

Substituting this value of d in equation (1), we get

  $a\left( {{a^2} - \dfrac{{9{a^2}}}{{16}}} \right) = 224$

  $a = 8$ 

And

$d = \dfrac{{3a}}{4} = \dfrac{3}{4} \times 8 = 6$

Example 7: Show that $\left( {{x^2} + xy + {y^2}} \right),\left( {{z^2} + xz + {x^2}} \right){\text{and }}\left( {{y^2} + yz + {z^2}} \right)$ are consecutive terms of an A.P., if x, y and z are in A.P.

Ans: The terms $\left(x^{2}+x y+y^{2}\right),\left(z^{2}+x z+x^{2}\right)$ and $\left(y^{2}+y z+z^{2}\right)$ will be in A.P. if

$\left(z^{2}+x z+x^{2}\right)-\left(x^{2}+x y+y^{2}\right)=\left(y^{2}+y z+z^{2}\right)-\left(z^{2}+x z+x^{2}\right)$

Thus,

$z^{2}+x z-x y-y^{2}=y^{2}+y z-x z-x^{2}$

i.e., $x^{2}+z^{2}+2 x z-y^{2}=y^{2}+y z+x y$

i.e., $(x+z)^{2}-y^{2}=y(x+y+z)$

i.e., $x+z-y=y$

i.e., $x+z=2 y$

which is true, since $x, y, z$ are in A.P. Hence $x^{2}+x y+y^{2}, z^{2}+x z+x^{2}, y^{2}+y z+z^{2}$ are in A.P.

Example 8: If $a, b, c, d$ are in G.P., prove that $a^{2}-b^{2}, b^{2}-c^{2}, c^{2}-d^{2}$ are also in G.P.

Ans: Let r be the common ratio of the given G.P. 

Then

$\dfrac{b}{a} = \dfrac{c}{d} = \dfrac{d}{c} = r$

$\Rightarrow$ $b=a r, c=b r=a r^{2}, d=c r=a r^{3}$

Now, $a^{2}-b^{2}=a^{2}-a^{2} r^{2}=a^{2}\left(1-r^{2}\right)$

$b^{2}-c^{2}=a^{2} r^{2}-a^{2} r^{4}=a^{2} r^{2}\left(1-r^{2}\right)$

And $c^{2}-d^{2}=a^{2} r^{4}-a^{2} r^{6}=a^{2} r^{4}\left(1-r^{2}\right)$

Therefore,

$\dfrac{{{b^2} - {c^2}}}{{{a^2} - {b^2}}} = \dfrac{{{c^2} - {d^2}}}{{{b^2} - {c^2}}} = {r^2}$

Hence, $a^{2}-b^{2}, b^{2}-c^{2}, c^{2}-d^{2}$ are in G.P.

Example 9: If the sum of m terms of an A.P. is equal to the sum of either the next n terms or the next p terms, then prove that $\left( {m + n} \right)\left( {\dfrac{1}{m} - \dfrac{1}{p}} \right) = \left( {m + p} \right)\left( {\dfrac{1}{m} - \dfrac{1}{n}} \right)$

Ans: Let the A.P. be $a, a+d, a+2 d, \ldots$

We are given $a_{1}+a_{2}+\ldots+a_{m}=a_{m+1}+a_{m+2}+\ldots+a_{m+n} \ldots$ (1)

Adding $a_{1}+a_{2}+\ldots+a_{m}$ on both sides of (1), we get

$2\left[a_{1}+a_{2}+\ldots+a_{m}\right]=a_{1}+a_{2}+\ldots+a_{m}+a_{m+1}+\ldots+a_{m+n}$

$2 S_{m}=S_{m+n}$

Therefore, $2\dfrac{m}{2}\left\{ {3a + \left( {m - 1} \right)d} \right\} = \dfrac{{m + n}}{2}\left\{ {2a + \left( {m + n - 1} \right)d} \right\}$

Putting 2a + (m – 1) d = x in the above equation, we get

  $mx = \dfrac{{m + n}}{2}\left( {x + nd} \right)$

  $\left( {2m - m - n} \right)x = \left( {m + n} \right)nd$

 $\Rightarrow \left( {m - n} \right)x = \left( {m + n} \right)nd{\text{  }}...\left( 2 \right)$ 

Similarly, if $a_{1}+a_{2}+\ldots+a_{m}=a_{m+1}+a_{m+2}+\ldots+a_{m+p}$ Adding $a_{1}+a_{2}+\ldots+a_{m}$ on both sides we get, $2\left(a_{1}+a_{2}+\ldots+a_{m}\right)=a_{1}+a_{2}+\ldots+a_{m+1}+\ldots+a_{m+p}$ or, $2 \mathrm{~S}_{\mathrm{m}}=\mathrm{S}_{\mathrm{m}+\mathrm{p}}$

$2\dfrac{m}{2}\left\{ {3a + \left( {m - 1} \right)d} \right\} = \dfrac{{m + p}}{2}\left\{ {2a + \left( {m + p - 1} \right)d} \right\}$ which gives

i.e., (m – p) x = (m + p)pd ... (3)

Dividing (2) by (3), we get

  $\dfrac{{\left( {m - n} \right)x}}{{\left( {m - p} \right)x}} = \dfrac{{\left( {m + n} \right)nd}}{{\left( {m + p} \right)pd}}$

  $\left( {m - n} \right)\left( {m + p} \right)p = \left( {m - p} \right)\left( {m + n} \right)n$ 

Dividing both sides by mnp, we get

  $\left( {m + p} \right)\left( {\dfrac{1}{n} - \dfrac{1}{m}} \right) = \left( {m + n} \right)\left( {\dfrac{1}{p} - \dfrac{1}{m}} \right)$

$= \left( {m + n} \right)\left( {\dfrac{1}{m} - \dfrac{1}{p}} \right) = \left( {m + p} \right)\left( {\dfrac{1}{m} - \dfrac{1}{n}} \right)$ 

Example 10: If $a_{1}, a_{2}, \ldots, a_{n}$ are in A.P. with common difference $d$ (where $d \neq 0$ ); then the sum of the series $\sin d\left(\operatorname{cosec} a_{2} \operatorname{cosec} a_{2}+\operatorname{cosec} a_{2} \operatorname{cosec} a_{3}+\ldots+\operatorname{cosec} a_{n-1} \operatorname{cosec} a_{n}\right)$ is equal to cot $a_{1}-\cot a_{n}$.

Ans: We have

$\sin d\left(\operatorname{cosec} a_{1} \operatorname{cosec} a_{2}+\operatorname{cosec} a_{2} \operatorname{cosec} a_{3}+\ldots+\operatorname{cosec} a_{n-1} \operatorname{cosec} a_{n}\right)$

$= \sin d\left( {\dfrac{1}{{\sin {a_1}\sin {a_2}}} + \dfrac{1}{{\sin {a_2}\sin {a_3}}} + ... + \dfrac{1}{{\sin {a_{n - 1}}\sin {a_n}}}} \right)$

$= \dfrac{{\sin \left( {{a_2} - {a_1}} \right)}}{{\sin {a_1}\sin {a_2}}} + \dfrac{{\sin \left( {{a_3} - {a_2}} \right)}}{{\sin {a_2}\sin {a_3}}} + ... + \dfrac{{\sin \left( {{a_n} - {a_{n - 1}}} \right)}}{{\sin {a_{n - 1}}\sin {a_n}}}$

$= \dfrac{{\sin {a_2}\cos {a_1} - \cos {a_1}\sin {a_2}}}{{\sin {a_1}\sin {a_2}}} + \dfrac{{\sin {a_3}\cos {a_2} - \cos {a_3}\sin {a_2}}}{{\sin {a_2}\sin {a_3}}} + ... + \dfrac{{\sin {a_n}\cos {a_{n - 1}} - \cos {a_n}\sin {a_{n - 1}}}}{{\sin {a_{n - 1}}\sin {a_n}}}$ 

$=\left(\cot a_{1}-\cot a_{2}\right)+\left(\cot a_{2}-\cot a_{3}\right)+\ldots+\left(\cot a_{n-1}-\cot a_{n}\right)$

$=\cot a_{1}-\cot a_{n}$

Example 11: (i) If a, b, c, d are four distinct positive quantities in A.P., then show that bc > ad

Ans: Since a, b, c, d are in A.P., then A.M. > G.M., for the first three terms.

Thus, $b > \sqrt {ac} {\text{      }}\left( {{\text{Here }}\dfrac{{a + c}}{2} = b} \right)$

Squaring, we get $b^{2}$ > ac ... (1)

Similarly, for the last three terms

AM > GM

$c > \sqrt {bd} {\text{      }}\left( {{\text{Here }}\dfrac{{b + d}}{2} = c} \right)$

$\mathrm{c}^{2}>\mathrm{bd} \ldots(2)$

Multiplying $(1)$ and $(2)$, we get

$\mathrm{b}^{2} \mathrm{c}^{2}>(\mathrm{ac})(\mathrm{bd})$

$\Rightarrow$ bc > ad

(ii) If a, b, c, d are four distinct positive quantities in G.P., then show that a + d > b + c

Ans: Since a, b, c, d are in G.P.

again A.M. > G.M. for the first three terms

$\dfrac{{a + c}}{2} > b{\text{    }}\left( {{\text{Since }}\sqrt {ac}  = b} \right)$

 $a+c>2 b \ldots(3)$

Similarly, for the last three terms

$\dfrac{{b + d}}{2} > c{\text{    }}\left( {{\text{Since }}\sqrt {bd}  = c} \right)$

$\Rightarrow$ b + d > 2c ... (4)

Adding (3) and (4), we get

$(a+c)+(b+d)>2 b+2 c$

$a+d>b+c$

Example 12: If $a, b, c$ are three consecutive terms of an A.P. and $x, y, z$ are three consecutive terms of a G.P. Then prove that $x^{b-c} \cdot y^{c-a} \cdot z^{a-b}=1$

Ans: We have $a, b, c$ as three consecutive terms of A.P. Then

$b-a=c-b=d($ say $)$

$c-a=2 d$

$a-b=-d$

Now $x^{b-c} \cdot y^{c-a} \cdot z^{a-b}=x^{-d} \cdot y^{2 d} \cdot z^{-d}$

  ${ = {x^{-d}}{{(\sqrt {xz} )}^{2d}}{z^{ - d}}({\text{since }}y = {\text{ }}\sqrt {xz} {\text{ }}){\text{ }}as x,y,z are{\text{ }}G.P.)}$ 

  ${ = {x^{-d}}{x^d}{z^d}{z^{-d}}}$

  ${ = {x^{-d}}^{ + d}{z^d}^{-d}}$

$= {x^0}{z^0}{\text{ }}$

$= {\text{ }}1$ 

Example 13: Find the natural number a for which $\sum\limits_{k = 1}^n {f\left( {a + k} \right)}  = 16\left( {{2^n} - 1} \right),$where the function f satisfies f (x + y) = f (x). f (y) for all-natural numbers x, y and further f (1) = 2.

Ans: Given that

$f(x+y)=f(x) \cdot f(y) \text { and } f(1)=2$

$\text { Therefore, }$

$f(2)=f(1+1)=f(1) \cdot f(1)=2^{2}$

$f(3)=f(1+2)=f(1) \cdot f(2)=2^{3}$

$f(4)=f(1+3)=f(1) \cdot f(3)=2^{4}$

$\text { and so on. Continuing the process, we obtain }$

$f(k)=2^{k} \text { and } f(a)=2^{a}$

Hence,

  $\sum\limits_{k = 1}^n {f\left( {a + k} \right)}  = \sum\limits_{k = 1}^n {f\left( a \right)f\left( k \right)}$

$= f\left( a \right)\sum\limits_{k = 1}^n {f\left( k \right)} $

$= {2^a}\left( {{2^1} + {2^2} + {2^3} + ... + {2^n}} \right)$

$= {2^a}\left\{ {\dfrac{{2\left( {{2^n} - 1} \right)}}{{2 - 1}}} \right\} = {2^{a + 1}}\left( {{2^n} - 1} \right){\text{     }}....\left( 1 \right)$ 

But we are given 

 $\sum\limits_{k = 1}^n {f\left( {a + k} \right)}  = 16\left( {{2^n} - 1} \right)$

 ${2^{a + 1}}\left( {{2^n} - 1} \right) = 16\left( {{2^n} - 1} \right)$

 ${2^{a + 1}} = {2^4}$

 $\Rightarrow a + 1 = 4$

 $\Rightarrow a = 3$ 

Example 14: A sequence may be defined as a

(A) relation, whose range $\subseteq$ N (natural numbers)

(B) function whose range $\subseteq$ N

(C) function whose domain $\subseteq$ N

(D) progression having real values

Ans: Option (C) is the correct answer. 

A sequence is a function f : N → X having domain $\subseteq$ N.

Example 15: If x, y, z are positive integers then value of expression (x + y) (y + z) (z + x) is

(A) = 8xyz 

(B) > 8xyz 

(C) < 8xyz 

(D) = 4xyz

Ans: Option (B) is the correct answer. 

Since A.M. > G.M., $\dfrac{{x + y}}{2} > \sqrt {xy} ,\dfrac{{y + z}}{2} > \sqrt {yz} {\text{ and }}\dfrac{{z + x}}{2} > \sqrt {zx}$

Multiplying the three inequalities, we get

  $\dfrac{{x + y}}{2} \cdot \dfrac{{y + z}}{2} \cdot \dfrac{{y + z}}{2} > \sqrt {\left( {xy} \right)\left( {yz} \right)\left( {zx} \right)}$

  $\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right) > 8xyz$ 

Example 16: In a G.P. of positive terms, if any term is equal to the sum of the next two terms. Then the common ratio of the G.P. is

(A) $\sin 18^{\circ}$

(B) $2 \cos 18^{\circ}$

(C) $\cos 18^{\circ}$

(D) $2 \sin 18^{\circ}$

Ans: (D) is the correct answer. 

Since 

  ${t_n} = {t_{n + 1}} + {t_{n + 2}}$

  $a{r^{n - 1}} = a{r^n} + a{r^{n + 1}}$

  $1 = r + {r^2}$

 $\Rightarrow r = \dfrac{{ - 1 \pm \sqrt 5 }}{2},{\text{ since }}r > 0$ 

Therefore,

$r = 2\dfrac{{\sqrt 5  - 1}}{5} = 2\sin 18^\circ$

Example 17: $\text {In an A.P. the } p^{\text {th }} \text { term is } q \text { and the }(p+q)^{\text {th }} \text { term is } 0 \text {. Then the } q^{\text {th }} \text { term is }$

$\text { (A) }-p$

$\text { (B) } p$

$\text { (C) } p+q$

$\text { (D) } p-q$

Ans: Option (B) is the correct answer

$\text { Let a, } d \text { be the first term and common difference respectively. }$

$\text { Therefore, } T_{p}=a+(p-1) d=q \text { and ... (1) }$

$T_{p+q}=a+(p+q-1) d=0 \ldots(2)$

$\text { Subtracting }(1) \text {, from }(2) \text { we get } q d=-q$

$\text { Substituting in (1) we get } a=q-(p-1)(-1)=q+p-1$

$\text { Now, }$

$T_{q}=a+(q-1) d=q+p-1+(q-1)(-1)$

$=q+p-1-q+1=p$

Example 18: Let $S$ be the sum; $P$ be the product and $R$ be the sum of the reciprocals of 3 terms of a G.P. Then $P^{2} R^{3}: S^{3}$ is equal to

(A) $1: 1$

(B) (common ratio) $^{n}: 1$

(C) (first term) $^{2}$:(common ratio)$^{2}$

(D) none of these

Ans: Option (A) is the correct answer

Let us take a G.P. with three terms $\dfrac{a}{r},a,ar$.

Then $S = \dfrac{a}{r} + a + ar = \dfrac{{a\left( {{r^2} + r + 1} \right)}}{r}$

$\dfrac{{{{\text{P}}^2}{{\text{R}}^3}}}{{{{\text{S}}^3}}} = \dfrac{{{a^6} \cdot \dfrac{1}{{{a^3}}}{{\left( {\dfrac{{{r^2} + r + 1}}{r}} \right)}^3}}}{{{a^3}{{\left( {\dfrac{{{r^2} + r + 1}}{r}} \right)}^3}}} = 1$

Therefore, the ratio is 1 : 1.

Example 19: The ${10}^{th}$ common term between the series 3 + 7 + 11 + ... and 1 + 6 + 11 + ... is

(A) 191 

(B) 193 

(C) 211 

(D) None of these

Ans: Option (A) is the correct answer.

The first common term is 11.

Now the next common term is obtained by adding L.C.M. of the common difference 4

and 5, i.e., 20.

Therefore, $10^{\text {th }}$ common term $=T_{10}$ of the AP whose $a=11$ and $d=20$ $\mathrm{T}_{10}=\mathrm{a}+9 \mathrm{~d}=11+9(20)=191$

Example 20: In a G.P. of even number of terms, the sum of all terms is 5 times the sum of the odd terms. The common ratio of the G.P. is

(A) $\dfrac{{ - 4}}{5}$

(B) $\dfrac{1}{5}$

(C) 4 

(D) none the these

Ans: Option (C) is the correct answer.

Let us consider a G.P. a, ar, ar2, ... with 2n terms. We have

$\dfrac{{a\left( {{r^{2n}} - 1} \right)}}{{r - 1}} = \dfrac{{5a\left( {{{\left( {{r^2}} \right)}^n} - 1} \right)}}{{{r^2} - 1}}$

(Since common ratio of odd terms will be r2 and number of terms will be n)

 $\Rightarrow \dfrac{{a\left( {{r^{2n}} - 1} \right)}}{{r - 1}} = 5\dfrac{{a\left( {{r^{2n}} - 1} \right)}}{{\left( {{r^2} - 1} \right)}}$

 $\Rightarrow a\left( {r + 1} \right) = 5a{\text{,  i}}{\text{.e}}{\text{., }}r = 4$ 

Example 21: The minimum value of the expression $3^{x}+3^{1-x}$ $x \in \mathbb{R}$, is

(A) 0 

(B) $\dfrac{1}{3}$

(C) 3 

(D) $\dfrac{2}{3}$

Ans: Option (D) is the correct answer.

We know ${\text{A}}{\text{.M}}{\text{.}} \geqslant {\text{G}}{\text{.M}}{\text{.}}$ for positive numbers.

Therefore,

  $\dfrac{{{3^x} + {3^{1 - x}}}}{2} \geqslant \sqrt {{3^x} \cdot {3^{1 - x}}}$

 $\Rightarrow \dfrac{{{3^x} + {3^{1 - x}}}}{2} \geqslant \sqrt {{3^x} \cdot \dfrac{3}{{{3^x}}}}$

 $\Rightarrow {3^x} + {3^{1 - x}} \geqslant 2\sqrt 3$ 

Exercise

Short Answer Type

1. The first term of an A.P.is a, and the sum of the first p terms is zero, show that the sum of its next q terms is $\dfrac{{ - a\left( {p + q} \right)q}}{{p - 1}}$. $\left[\text {Hint: Required sum } =S_{p+q}-S_{p}\right]$

Ans: Given that $a_{1}=a$ and $S_{p}=0$

Sum of next $q$ terms of the given $A . P .=S_{p+q}-S_{p}$

  ${S_{p + q}} = \dfrac{{p + q}}{2}\left[ {2a + \left( {p + q - 1} \right)d} \right]$

  ${\text{And }}{S_p} = \dfrac{p}{2}\left[ {2a + \left( {p - 1} \right)d} \right] = 0$

 $\Rightarrow 2a + \left( {p - 1} \right)d = 0$

 $\Rightarrow \left( {p - 1} \right)d =  - 2a$

 $\Rightarrow d = \dfrac{{ - 2a}}{{p - 1}}$ 

Sum of next q terms $=\mathrm{S}_{\mathrm{p}+\mathrm{a}}-\mathrm{S}_{\mathrm{p}}$

$= \dfrac{{p + q}}{2}\left[ {2a + \left( {p - 1} \right)d} \right] - 0$

$= \dfrac{{p + q}}{2}\left[ {2a + \left( {p - 1} \right)\left( {\dfrac{{ - 2a}}{{p - 1}}} \right)} \right]$

$= \dfrac{{p + q}}{2}\left[ {2a + \dfrac{{\left( {p - 1} \right)\left( { - 2a} \right)}}{{p - 1}} - \dfrac{{2aq}}{{p - 1}}} \right]$

$= \dfrac{{p + q}}{2}\left[ {2a - 2a - \dfrac{{2aq}}{{p - 1}}} \right]$

$= \dfrac{{\left( {p + q} \right)}}{2}\left( {\dfrac{{ - 2aq}}{{p - 1}}} \right)$

$= \dfrac{{ - a\left( {p + q} \right)q}}{{p - 1}}$ 

Hence, the required sum = $\dfrac{{ - a\left( {p + q} \right)q}}{{p - 1}}$

2. A man saved Rs 66000 in 20 years. In each succeeding year after the first year he saved Rs 200 more than what he saved in the previous year. How much did he saves in the first year?

Ans: Given the total amount saved in 20 years is 66000 hence $S_{20}=66000$

Let the amount saved in first year be '$a$'

Then as he increases 200 Rs every year then the amount in second year will be $a+200$

then in third year '$a+400$' and so on

The sequence will be $a, a+200, a+400 \ldots$

The sequence is AP and the common difference is $d=200$

There are 20 terms in the sequence as he saved money for 20 years

Using the sum formula for AP, 

${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

Given that $S_{20}=66000$

$66000 = \dfrac{{20}}{2}\left[ {2a + \left( {20 - 1} \right)200} \right]$

$\Rightarrow 66000=10(2 a+19 \times 200)$

$\Rightarrow 6600=2 a+3800$

$\Rightarrow 6600-3800=2 a$

$\Rightarrow 2 a=2800$

$\Rightarrow a=1400$

Hence, the amount saved in the first year is Rs. 1400. 

3. A man accepts a position with an initial salary of Rs 5200 per month. It is understood that he will receive an automatic increase of Rs 320 in the very next month and each month thereafter.

(a) Find his salary for the tenth month

Ans: Given the man’s salary in first month is Rs 5200 and then it increases every month by 320rs

Hence the sequence of his salary per month will be 5200, 5200 + 320, 5200 + 640…

The sequence is in AP with first term as a = 5200,

Common difference as d = 320

Now we have to find salary in $10^{\text {th }}$ month that is $10^{\text {th }}$ term of the AP The $n^{\text {th }}$ term of $A P$ is given by $a_{n}=a+(n-1) d$

Where $a$ is the first term and $d$ is the common difference

We have to find $a_{10}$

$\Rightarrow a_{10}=5200+(10-1)(320)$

$\Rightarrow a_{10}=5200+2880$

$\Rightarrow a_{10}=8080$

Hence, salary in $10^{\text {th }}$ month is Rs. 8080 .

(b) What is his total earning during the first year?

Ans: To get the total earnings in first year we have to add first 12 terms of the sequence that is we have to find ${S}_{12}$

The sum of first n terms of AP is given by ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

Where a is the first term and d is common difference

${S_n} = \dfrac{{12}}{2}\left[ {2\left( {5200} \right) + \left( {12 - 1} \right)320} \right]$

$\Rightarrow \mathrm{S}_{12}=6(10400+11(320))$

$\Rightarrow \mathrm{S}_{12}=6(10400+3520)$

$\Rightarrow \mathrm{S}_{12}=6(13920)$

$\Rightarrow \mathrm{S}_{12}=83520$

Hence his total earnings in the first year is Rs 83,520.

4. If the pth and qth terms of a G.P. are q and p respectively, show that its (p + q)th term is ${\left( {\dfrac{{{q^p}}}{{{p^q}}}} \right)^{\dfrac{1}{{p - q}}}}$.

Ans: Let a be the first term and r be the common ratio of a G.P.

Given that $a_{p}=q$

$\Rightarrow a r^{p-1}=q \ldots$ (i)

and $a_{q}=p$

$\Rightarrow a r^{q-1}=p \ldots .$ (ii)

Dividing eq. (i) by eq. (ii),

  $\dfrac{{a{r^{p - 1}}}}{{q{r^{q - 1}}}} = \dfrac{q}{p} \Rightarrow \dfrac{{{q^{p - 1}}}}{{{r^{q - 1}}}} = \dfrac{q}{p}$

 $\Rightarrow {r^{p - q}} = \dfrac{q}{p} \Rightarrow r = {\left( {\dfrac{q}{p}} \right)^{\dfrac{1}{{p - q}}}}$ 

Putting the value of r in eq. (i),

  $a{\left( {\dfrac{q}{p}} \right)^{\dfrac{1}{{p - q}} \times \left( {p - 1} \right)}} = q$

  $a{\left( {\dfrac{q}{p}} \right)^{\dfrac{{p - 1}}{{p - q}}}} = q$

  ${\text{So, }}a = q{\left( {\dfrac{p}{q}} \right)^{\dfrac{{p - 1}}{{p - q}}}}$ 

Now, 

  ${T_{p + q}} = a{r^{p + q - 1}} = q{\left( {\dfrac{p}{q}} \right)^{\dfrac{{p - 1}}{{p - q}}}}{\left( {\dfrac{q}{p}} \right)^{\dfrac{1}{{p - q}} \times \left( {p + q - 1} \right)}}$

$= q{\left( {\dfrac{p}{q}} \right)^{\dfrac{{p - q}}{{p - q}}}}{\left( {\dfrac{p}{q}} \right)^{\dfrac{{p + q - 1}}{{p - q}}}} = q{\left( {\dfrac{p}{q}} \right)^{\dfrac{{p - 1}}{{p - q}}}}{\left( {\dfrac{p}{q}} \right)^{\dfrac{{ - \left( {p + q - 1} \right)}}{{p - q}}}}$

$= q{\left( {\dfrac{p}{q}} \right)^{\dfrac{{p - 1 - p - q + 1}}{{p - q}}}} = q{\left( {\dfrac{p}{q}} \right)^{\dfrac{{ - q}}{{p - q}}}}$

$= \dfrac{{q{p^{\dfrac{{ - q}}{{p - q}}}}}}{{{q^{\dfrac{{ - q}}{{p - q}}}}}} = \dfrac{{{q^{1 + \dfrac{q}{{p - q}}}}}}{{{p^{\dfrac{q}{{p - q}}}}}} = \dfrac{{{q^{\dfrac{{p - q + q}}{{p - q}}}}}}{{{p^{\dfrac{q}{{p - q}}}}}}$

$= \dfrac{{{q^{\dfrac{p}{{p - q}}}}}}{{{p^{\dfrac{q}{{p - q}}}}}} = {\left( {\dfrac{{{q^p}}}{{{p^q}}}} \right)^{\dfrac{1}{{p - q}}}}$ 

5. A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?

Ans: Given first day he made 5 frames, then two frames more than the previous, that is 7 then 9 and so on.

Hence the sequence of making frames each day is 5, 7, 9…

The sequence is AP with first term as a = 5 and common difference d = 2

Total number of frames to be made is 192.

Let carpenter requires n days hence ${S_n}$ = 192

The sum of first n terms of AP is given by ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

Where a is the first term and d is common difference

  $192 = \dfrac{n}{2}\left[ {2 \times 5 + \left( {n - 1} \right)2} \right]$

 $\Rightarrow 192 = \dfrac{n}{2}\left( {10 + 2n - 2} \right)$ 

$\Rightarrow 384=10 n+2 n^{2}-2 n$ 

On computing and simplifying, 

$\Rightarrow 2 n^{2}+8 n-384=0$ 

$\Rightarrow n^{2}+4 n-192=0$ 

$\Rightarrow n^{2}+16 n-12 n-192=0$ 

$\Rightarrow n(n+16)-12(n+16)=0$ 

$\Rightarrow(n-12)(n+16)=0$ 

$\Rightarrow n=12$ and $n=-16$

$\Rightarrow 2 n^{2}+8 n-384=0$

$\Rightarrow \mathrm{n}^{2}+4 \mathrm{n}-192=0$

$\Rightarrow \mathrm{n}^{2}+16 \mathrm{n}-12 \mathrm{n}-192=0$

$\Rightarrow n(n+16)-12(n+16)=0$

$\Rightarrow(n-12)(n+16)=0$

$\Rightarrow n=12$ and $n=-16$

But n represents number of days which cannot be negative hence n = 12

Hence number of days required to finish the job is 12 days.

6. We know the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.

Ans: $\text { Given the sum of interior angles of a polygon having ' } n \text { ' sides is given by }(n-2) \times 180^{\circ}$

$\text { Sum of angles with three sides that is } n=3 \text { is }(3-2) \times 180^{\circ}=180^{\circ}$

$\text { Sum of angles with four sides that is } n=4 \text { is }(4-2) \times 180^{\circ}=360^{\circ}$

$\text { Sum of angles with five sides that is } n=5 \text { is }(5-2) \times 180^{\circ}=540^{\circ}$

$\text { Sum of angles with six sides that is } n=6 \text { is }(6-2) \times 180^{\circ}=720^{\circ}$

$\text { As seen as the number of sides increases by } 1 \text { the sum of interior angles increases by }$ $180^{\circ}$

$\text { Hence the sequence of sum of angles as number of sides' increases is } 180^{\circ}, 360^{\circ}, 540^{\circ} \text {, }$ $720^{\circ} \ldots$

$\text { The sequence is AP with first term as a }=180^{\circ} \text { and common difference as } d=180^{\circ}$

We have to find sum of angles of polygon with 21 sides

Using $(\mathrm{n}-2) \times 180^{\circ}$,

$\Rightarrow$ Sum of angles of polygon having 21 sides $=(21-2) \times 180^{\circ}$

$\Rightarrow$ Sum of angles of polygon having 21 sides $=19 \times 180^{\circ}$

$\Rightarrow$ Sum of angles of polygon having 21 sides $=3420^{\circ}$

7. A side of an equilateral triangle is 20cm long. A second equilateral triangle is inscribed in it by joining the mid points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.

Ans: Let ABC be the triangle with AB = BC = AC = 20 cm

Let D, E and F be midpoints of AC, CB and AB respectively which are joined to form an equilateral triangle DEF

Now we have to find the length of side of $\triangle \mathrm{DEF}$

Consider $\triangle \mathrm{CDE}$

$\mathrm{CD}=\mathrm{CE}=10 \mathrm{~cm} \ldots \mathrm{D}$ and $\mathrm{E}$ are midpoints of $\mathrm{AC}$ and $\mathrm{CB}$

Hence $\triangle \mathrm{CDE}$ is isosceles

$\Rightarrow \angle \mathrm{CDE}=\angle \mathrm{CED}$ (base angles of isosceles triangle)

But $\angle \mathrm{DCE}=60^{\circ}$ ( $\angle \mathrm{ABC}$ is equilateral)

Hence $\angle \mathrm{CDE}=\angle \mathrm{CED}=60^{\circ}$

Hence $\triangle \mathrm{CDE}$ is equilateral

Hence DE = 10 cm

Similarly, we can show that GH = 5 cm

Hence the series of sides of the equilateral triangle will be 20, 10, 5 …

The series is GP with first term a = 20 and common ratio r = $\dfrac{1}{2}$

To find the perimeter of $6^{\text {th }}$ triangle inscribed we first have to find the side of 6th triangle that is the $6^{\text {th }}$ term in the series $\mathrm{n}^{\text {th }}$ term in GP is given by $\mathrm{a}_{\mathrm{n}}=a \mathrm{r}^{\mathrm{n}-1}$

  ${a_6} = a{r^{6 - 1}}$

$= 20 \times {\left( {\dfrac{1}{2}} \right)^5} = 20 \times \dfrac{1}{{32}} = \dfrac{5}{8}{\text{ cm}}$ 

Hence the side of $6^{\text {th }}$ equilateral triangle is $\dfrac{5}{8}$ cm and hence its perimeter would be thrice its side length because it’s an equilateral triangle

Perimeter of $6^{\text {th }}$ equilateral triangle inscribed is 3 × $\dfrac{5}{8}$= $\dfrac{{15}}{8}$cm

8. In a potato race 20 potatoes are placed in a line at intervals of 4 metres with the first potato 24 metres from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

Ans: Given at start he has to run 24m to get the first potato then 28 m as the next potato is 4m away from first and so on

Hence the sequence of his running will be 24, 28, 32 …

There are 20 terms in sequence as there are 20 potatoes

Hence only to get potatoes from starting point he has to run

24 + 28 + 32 + … up to 20 terms

This is only from starting point to potato but he has to get the potato back to starting point hence the total distance will be twice that is

Total distance ran $=2 \times(24+28+32 \ldots) \ldots$ (1)

Let us find the sum using he formula to find sum of n terms of AP

That is ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

There are 20 terms hence n = 20

⇒ ${S_{20}} = \dfrac{{20}}{2}\left[ {2\left( {24} \right) + \left( {20 - 1} \right)4} \right]$

On simplification,

$\Rightarrow \mathrm{S}_{20}=10(48+19(4))$

$\Rightarrow \mathrm{S}_{20}=10(48+76)$

$\Rightarrow \mathrm{S}_{20}=10 \times 124$

$\Rightarrow \mathrm{S}_{20}=1240 \mathrm{~m}$

Using equation $(1)$

$\Rightarrow$ Total distance ran $=2 \times 1240$

$\Rightarrow$ Total distance ran $=2480 \mathrm{~m}$

Hence total distance he has to run is 2480 m.

9. In a cricket tournament 16 school teams participated. A sum of Rs 8000 is to be awarded among themselves as prize money. If the last placed team is awarded Rs 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first-place team receive?

Ans: Let the amount received by first place team be a Rs and d be difference in amount

As the difference is the same hence the second-place team will receive a – d and the third-place a – 2d and so on.

The last team receives 275 Rs.

As there are 16 teams and all teams are given prizes hence the sequence will have 16 terms because there are 16 teams.

Therefore $a, a-d, a-2 d \ldots 275$

The sequence is in AP with first term as a and common difference is '- $d$ '

As the total prize given is of Rs. 8000 .

Hence.

$a+a-d+a-2 d \ldots+275=8000$

The sum of n terms of AP is given by ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ where a is first term and d is common difference.

There are 16 terms $n=16$ and the sum $S_{n}=8000$

$8000 = \dfrac{{16}}{2}\left[ {2a + \left( {16 - 1} \right)\left( { - d} \right)} \right]$

$\Rightarrow 8000=8(2 a-15 d)$

$\Rightarrow 1000=2 a-15 d \ldots(1)$

The last term of AP is 275 and $n^{\text {th }}$ term of AP is $a_{n}=a+(n-1) d$

The last term is $a_{n}=275$

$\Rightarrow 275=a+(16-1)(-d)$

$\Rightarrow 275=a-15 d \ldots(2)$

Subtract equation $(2)$ from equation (1),

$\Rightarrow 1000-275=2 a-15 d-a+15 d$

$\Rightarrow 725=a$

Hence amount received by first place team is Rs.725.

10. If $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are in A.P., where $a_{i}>0$ for all i, show that $\dfrac{1}{{\sqrt {{a_1}}  + \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}}  + \sqrt {{a_3}} }} + ... + \dfrac{1}{{\sqrt {{a_{n - 1}}}  + \sqrt {{a_n}} }} = \dfrac{{n - 1}}{{\sqrt {{a_1}}  + \sqrt {{a_n}} }}$

Ans: Given 

$a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are in A.P., where $a_{i}>0$ for all $i$,

So, $a_{1}-a_{2}=a_{2}-a_{3}=\ldots=a_{n-1}-a_{n}=-d$ (Constant) ... (1)

Now,

  $\dfrac{1}{{\sqrt {{a_1}}  + \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}}  + \sqrt {{a_3}} }} + ... + \dfrac{1}{{\sqrt {{a_{n - 1}}}  + \sqrt {{a_n}} }}$

$= \dfrac{{\sqrt {{a_1}}  - \sqrt {{a_2}} }}{{{a_1} - {a_2}}} + \dfrac{{\sqrt {{a_2}}  - \sqrt {{a_3}} }}{{{a_2} - {a_3}}} + ... + \dfrac{{\sqrt {{a_{n - 1}}}  - \sqrt {{a_n}} }}{{{a_{n - 1}} - {a_n}}}{\text{   }}\left( {{\text{On rationalizing each term}}} \right)$

$= \dfrac{{\sqrt {{a_1}}  - \sqrt {{a_2}} }}{{ - d}} + \dfrac{{\sqrt {{a_2}}  - \sqrt {{a_3}} }}{{ - d}} + ... + \dfrac{{\sqrt {{a_{n - 1}}}  - \sqrt {{a_n}} }}{{ - d}}$

$= \dfrac{1}{{ - d}}\left( {\sqrt {{a_1}}  - \sqrt {{a_n}} } \right)$

$= \dfrac{{{a_1} - {a_n}}}{{ - d\left( {\sqrt {{a_1}}  + \sqrt {{a_n}} } \right)}}{\text{ }}\left( {{\text{On rationalizing}}} \right)$

$= \dfrac{{ - \left( {n - 1} \right)d}}{{ - d\left( {\sqrt {{a_1}}  + \sqrt {{a_2}} } \right)}}{\text{   }}\left( {{\text{As }}{a_n} = {a_1} + \left( {n - 1} \right)d} \right)$

$= \dfrac{{n - 1}}{{\sqrt {{a_1}}  + \sqrt {{a_n}} }}$ 

11.Find the sum of the series $\left(3^{3}-2^{3}\right)+\left(5^{3}-4^{3}\right)+\left(7^{3}-6^{3}\right)+\ldots$ to

(i) $n$ terms

Ans: Given series is: $\left(3^{3}-2^{3}\right)+\left(5^{3}-4^{3}\right)+\left(7^{3}-6^{3}\right)+\ldots n$ terms

  ${T_n} = {\left( {2n + 1} \right)^3} - {\left( {2n} \right)^3}$

$= \left( {2n + 1 - 2n} \right)\left[ {{{\left( {2n + 1} \right)}^2} + \left( {2n + 1} \right)2n + {{\left( {2n} \right)}^2}} \right]$ 

$= 12{n^2} + 6n + 1$

Sum of n terms,

  ${S_n} = \sum\limits_{n = 1}^n {\left( {12{n^2} + 6n + 1} \right)}$

$= 12 \times \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \dfrac{{6n\left( {n + 1} \right)}}{2} + n$

$= 2n\left( {n + 1} \right)\left( {2n + 1} \right) + 3n\left( {n + 1} \right) + n$ 

$= 2n\left( {2{n^2} + 3n + 1} \right) + 3{n^2} + 3n + n$

$= 4{n^3} + 9{n^2} + 6n$

(ii) 10 terms

Ans: Sum of 10 terms = ${S_{10}}$

 ${S_{10}} = 4 \times {10^3} + 9 \times {10^2} + 6 \times 10$

$= 4000 + 900 + 60$ 

$= 4960$

12. Find the $r^{\text {th }}$ term of an A.P. sum of whose first $n$ terms is $2 n+3 n^{2}$. $\left[\text{Hint: } a_{n}=S_{n}-S_{n-1}\right]$

Ans: Sum of first $n$ terms be $S_{n}$ given as $S_{n}=2 n+3 n^{2}$

We have to find the $r^{\text {th }}$ term that is $a_{r}$

Using the given hint $n^{\text {th }}$ term is given as $a_{n}=S_{n}-S_{n-1}$

$\Rightarrow a_{r}=S_{r}-S_{r-1}$

Using $S_{n}=2 n+3 n^{2}$

$\Rightarrow a_{r}=2 r+3 r^{2}-\left(2(r-1)+3(r-1)^{2}\right)$

$\Rightarrow a_{r}=2 r+3 r^{2}-\left(2 r-2+3\left(r^{2}-2 r+1\right)\right)$

$\Rightarrow a_{r}=2 r+3 r^{2}-\left(2 r-2+3 r^{2}-6 r+3\right) \Rightarrow a_{r}=6 r-1$

Hence, the $r^{\text {th }}$ term is $6 r-1$.

Long Answer Type

13. If A is the arithmetic mean and $G_{1}, G_{2}$ be two geometric means between any two numbers, then prove that $2A = \dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}}$

Ans: Let the numbers be a and b.

Then, $A = \dfrac{{a + b}}{2}$ or 2A = a + b

Also, G1 and G2 are geometric means between a and b, then a, $G_{1}, G_{2}$, b are in GP.

Let r be the common ratio.

Then,

  $b = a{r^{4 - 1}} = a{r^3}$

 $\Rightarrow \dfrac{b}{a} = {r^3} \Rightarrow r = {\left( {\dfrac{b}{a}} \right)^{\dfrac{1}{3}}}$ 

So, 

  ${G_1} = ar = a{\left( {\dfrac{b}{a}} \right)^{\dfrac{1}{3}}} = {a^{\dfrac{2}{3}}}{b^{\dfrac{1}{3}}}$

  ${\text{And }}{G_2} = a{r^2} = a{\left( {\dfrac{b}{a}} \right)^{\dfrac{2}{3}}} = {a^{\dfrac{1}{3}}}{b^{\dfrac{2}{3}}}$ 

Hence,

\[\dfrac{{{G_1}^2}}{{{G_2}^2}} + \dfrac{{{G_2}^2}}{{{G_1}^2}} = \dfrac{{{G_1}^3 + {G_2}^3}}{{{G_1}{G_2}}} = \dfrac{{{a^2}b + a{b^2}}}{{ab}} = a + b = 2A\]

14. If \[{\theta _1},{\theta _2},{\theta _3},.......,{\theta _n}\] are in A.P., whose common difference is d, show that

$\sec {\theta _1}\sec {\theta _2} + \sec {\theta _2}\sec {\theta _3} + ...... + \sec {\theta _{n - 1}}\sec {\theta _n} = \dfrac{{\tan {\theta _n} - \tan {\theta _1}}}{{\sin d}}$.

Ans: Since \[{\theta _1},{\theta _2},{\theta _3},.......,{\theta _n}\] are in A.P.,

${\theta _2} - {\theta _1} = {\theta _3} - {\theta _2} = ... = {\theta _n} - {\theta _{n - 1}} = d{\text{  }}...\left( {\text{i}} \right)$

Now,

  $\sec {\theta _1}\sec {\theta _2} = \dfrac{1}{{\sin d}} \cdot \dfrac{{\sin d}}{{\cos {\theta _1}\cos {\theta _2}}}$

$= \dfrac{1}{{\sin d}} \cdot \dfrac{{\sin \left( {{\theta _2} - {\theta _1}} \right)}}{{\cos {\theta _1}\cos {\theta _2}}}{\text{   }}\left( {{\text{Using eq}}{\text{.}}\left( 1 \right)} \right)$

$= \dfrac{1}{{\sin d}} \cdot \dfrac{{\sin {\theta _2}\cos {\theta _1} - \cos {\theta _2}\sin {\theta _1}}}{{\cos {\theta _1}\cos {\theta _2}}}$

$= \dfrac{{\tan {\theta _2} - \tan {\theta _1}}}{{\sin d}}$ 

Similarly,

\[\sec {\theta _2}\sec {\theta _3} = \dfrac{{\tan {\theta _3} - \tan {\theta _2}}}{{\sin d}};\sec {\theta _3}\sec {\theta _4} = \dfrac{{\tan {\theta _4} - \tan {\theta _3}}}{{\sin d}}\] and so on.

Thus,

\[\sec {\theta _1}\sec {\theta _2} + \sec {\theta _2}\sec {\theta _3} + ... + \sec {\theta _{n - 1}}\sec {\theta _n} = \dfrac{{\tan {\theta _n} - \tan {\theta _1}}}{{\sin d}}\]

15. If the sum of $p$ terms of an A.P. is $q$ and the sum of $q$ terms is $p$, show that the

sum of $p+q$ terms is $-(p+q)$. Also, find the sum of first $p-q$ terms $(p>q)$.

Ans: Let first term and common difference of the A.P. be a and $d$, respectively.

Given, $S_{p}=q$

$\Rightarrow \dfrac{p}{2}\left[ {2a + \left( {p - 1} \right)d} \right] = q \Rightarrow 2a + \left( {p - 1} \right)d = \dfrac{{2q}}{p}{\text{   }}...\left( {\text{i}} \right)$

Also, $S_{q}=p$

\[ \Rightarrow \dfrac{q}{2}\left[ {2a + \left( {q - 1} \right)d} \right] = p \Rightarrow 2a + \left( {q - 1} \right)d = \dfrac{{2p}}{q}{\text{   }}...\left( {{\text{ii}}} \right)\]

On subtracting equation (ii) from equation (i),

  $\left[ {\left( {p - 1} \right) - \left( {q - 1} \right)} \right]d = \dfrac{{2q}}{p} - \dfrac{{2p}}{q} \Rightarrow \left( {p - q} \right)d = \dfrac{{2\left( {{q^2} - {p^2}} \right)}}{{pq}}$

 $\Rightarrow d = \dfrac{{ - 2\left( {p + q} \right)}}{{pq}}{\text{   }}...\left( {{\text{iii}}} \right)$ 

On substituting the value of d into equation (i),

  $2a + \left( {p - 1} \right)\left( {\dfrac{{ - 2\left( {p + q} \right)}}{{pq}}} \right) = \dfrac{{2q}}{p}$

 $\Rightarrow a = \dfrac{q}{p} + \dfrac{{\left( {p + q} \right)\left( {p - 1} \right)}}{{pq}}...\left( {{\text{iv}}} \right)$ 

Now,

  ${S_{p + q}} = \dfrac{{p + q}}{2}\left[ {2a + \left( {p + q - 1} \right)d} \right]$

$= \dfrac{{p + q}}{2}\left[ {\dfrac{{2q}}{p} + \dfrac{{2\left( {p + q} \right)\left( {p - 1} \right)}}{{pq}} - \dfrac{{2\left( {p + q - 1} \right)\left( {p + q} \right)}}{{pq}}} \right]$

$= \left( {p + q} \right)\left[ {\dfrac{q}{p} + \dfrac{{\left( {p + q} \right)\left( { - q} \right)}}{{pq}}} \right]$

$= \left( {p + q} \right)\left[ {\dfrac{q}{p} - \dfrac{{p + q}}{p}} \right]$

$= \left( {p + q} \right)\left[ {\dfrac{{q - p - q}}{p}} \right]$

$= \left( {p + q} \right)\left( {\dfrac{{ - p}}{p}} \right) =  - \left( {p + q} \right)$ 

Also,

  ${S_{p - q}} = \dfrac{{p - q}}{2}\left[ {2a + \left( {p - q - 1} \right)d} \right]$

$= \dfrac{{p - q}}{2}\left[ {\dfrac{{2q}}{p} + \dfrac{{2\left( {p + q} \right)\left( {p - 1} \right)}}{{pq}} - \dfrac{{\left( {p - q - 1} \right)2\left( {p + q} \right)}}{{pq}}} \right]$

$= \left( {p - q} \right)\left[ {\dfrac{q}{p} + \dfrac{{\left( {p + q} \right)\left( {p - 1 - p + q + 1} \right)}}{{pq}}} \right]$

$= \left( {p - q} \right)\left[ {\dfrac{q}{p} + \dfrac{{\left( {p + q} \right)q}}{{pq}}} \right]$

$= \left( {p - q} \right)\left[ {\dfrac{q}{p} + \dfrac{{p + q}}{p}} \right]$

$= \left( {p - q} \right)\left( {\dfrac{{p + 2q}}{p}} \right)$ 

16. If $p^{\text {th }}, q^{\text {th }}$, and $r^{\text {th }}$ terms of an A.P. and G.P. are both $a, b$ and c respectively, show that $a^{b-c} \cdot b^{c-a} \cdot c^{a-b}=1$.

Ans: Let A and d be the first term and common difference respectively of an A.P. and x and R

be the first term and common ratio respectively of the G.P.

$\therefore \mathrm{A}+(\mathrm{p}-1) \mathrm{d}=\mathrm{a} \ldots . .(\mathrm{i})$

$\mathrm{A}+(\mathrm{q}-1) \mathrm{d}=\mathrm{b} \ldots . . \text { (ii) }$

$\text { And } \mathrm{A}+(\mathrm{r}-1) \mathrm{d}=\mathrm{c} \ldots . .(\mathrm{iii})$

$\text { For G.P., we have }$

$\mathrm{xR}^{\mathrm{p}-1}=\mathrm{a} \ldots . . \text { (iv) }$

$\mathrm{xR}^{\mathrm{q}-1}=\mathrm{b} \ldots . .(\mathrm{v})$

$\text { and } \mathrm{xR}^{\mathrm{r}-1}=\mathrm{c} \ldots . . \text { (vi) }$

Subtracting eq. (ii) from eq. (i)

(p – q)d = a – b ….(vii)

Similarly, 

(q – r)d = b – c ….(viii)

And (r – p)d = c – a ….(ix)

Now we have to prove that

$a^{b-c} \cdot b^{c-a} \cdot c^{a-b}=1$

L.H.S. $=a^{b-c} \cdot b^{c-a} \cdot c^{a-b}$

$= {\left( {x{R^{p - 1}}} \right)^{\left( {q - r} \right)d}} \cdot {\left( {x{R^{q - 1}}} \right)^{\left( {r - p} \right)d}} \cdot {\left( {x{R^{r - 1}}} \right)^{\left( {p - q} \right)d}}{\text{   }}\left( {{\text{Using equations from }}\left( {\text{i}} \right){\text{to }}\left( {{\text{ix}}} \right)} \right)$

$= {x^{\left( {q - r} \right)d}}{x^{\left( {r - p} \right)d}}{x^{\left( {p - q} \right)d}}{R^{\left( {p - 1} \right)\left( {q - r} \right)d}}{R^{\left( {q - 1} \right)\left( {r - p} \right)d}}{R^{\left( {r - 1} \right)\left( {p - q} \right)d}}$

$= {x^{\left( {q - r + r - p + p - q} \right)d}}{R^{\left[ {\left( {p - 1} \right)\left( {q - r} \right) + \left( {q - 1} \right)\left( {r - p} \right) + \left( {r - 1} \right)\left( {p - q} \right)} \right]d}}$

$= {x^0}{R^{\left( {pq - pr - q + r + qr - pq - r + p + pr - qr - p + q} \right)d}}$

$= {x^0}{R^0}$

$= 1$ 

L.H.S. = R.H.S. Hence proved.

Objective Type Questions

Choose the correct answer out of the four given options in each of the Exercises 17 to

26 (M.C.Q.).

17. If the sum of n terms of an A.P. is given by $S_{n}=3 n+2 n^{2}$, then the common difference of the A.P. is

(A) 3 

(B) 2 

(C) 6 

(D) 4

Ans: Correct Option - D

To find: Common Difference of A.P that is d

Consider,

$\mathrm{S}_{\mathrm{n}}=3 \mathrm{n}+2 \mathrm{n}^{2}$

Putting $\mathrm{n}=1$, we get

$\mathrm{S}_{1}=3(1)+2(1)^{2}=3+2$

$\mathrm{~S}_{1}=5$

Putting $\mathrm{n}=2$, we get

$\mathrm{S}_{2}=3(2)+2(2)^{2}=6+2(4)$

$=6+8$

$\mathrm{~S}_{2}=14$

Now, we know that,

$\mathrm{S}_{1}=\mathrm{a}_{1}$

$\Rightarrow \mathrm{a}_{1}=5$

And $\mathrm{a}_{2}=\mathrm{S}_{2}-\mathrm{S}_{1}=14-5$

$=9$

$\therefore$ Common Difference, $\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}$

$=9-5$

$= 4$

18. The third term of G.P. is 4. The product of its first 5 terms is

(A) $4^{3}$

(B) $4^{4}$

(C) $4^{5}$

(D) None of these

Ans: Correct Option - C

Given the third term of G.P, $T_{3}=4$

To find the product of first five terms

We know that,

$T_{\mathrm{n}}=a r^{\mathrm{n}-1}$

It is given that, $T_{3}=4$

$\Rightarrow a r^{3-1}=4$

$\Rightarrow a r^{2}=4 \ldots$ (i)

Product of first 5 terms $=a \times a r \times a r^{2} \times a r^{3} \times a r^{4}$

$=a^{5} r^{1+2+3+4}$

$=a^{5} r^{10}$

$=\left(a r^{2}\right)^{5}$

$=(4)^{5}[$ from equation (i) $]$

19. If 9 times the $9^{\text {th }}$ term of an A.P. is equal to 13 times the $13^{\text {th }}$ term, then the 22 nd term of the A.P. is

(A) 0 

(B) 22 

(C) 220 

(D) 198

Ans: Correct Option - A

$T_{n} =\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$

$\therefore \mathrm{T}_{9}=\mathrm{a}+8 \mathrm{~d}$

and $\mathrm{T}_{13}=\mathrm{a}+12 \mathrm{~d}$

As per the given condition

$9[\mathrm{a}+8 \mathrm{~d}]=13[\mathrm{a}+12 \mathrm{~d}]$

As per the given condition

$9[a+8 d]=13[a+12 d]$

$\Rightarrow 9 a+72 d=13 a+156 d$

$\Rightarrow-4 a=84 d$

$\Rightarrow a=-21 d \ldots .(i)$

Now $T_{22}=a+21 d=-21 d+21 d=0[$ from eq. (i)]

20. If x, 2y, 3z are in A.P., where the distinct numbers x, y, z are in G.P. then the

common ratio of the G.P. is

(A) 3 

(B) $\dfrac{1}{3}$

(C) 2 

(D) $\dfrac{1}{2}$

Ans: Correct Option - B

Since $x, 2 y, 3 z$ are in A.P.

$\therefore 2 y-x=3 z-2 y$

$\Rightarrow 4 y=x+3 z \ldots$ (i)

Now $x, y, z$ are in G.P.

Common ratio, $r = \dfrac{y}{x} = \dfrac{z}{x}$

${y^2} = {\text{ }}xz{\text{  }}...\left( {{\text{ii}}} \right)$

Putting the value of x from eq. (i), we get

$y^{2}=(4 y-3 z) z \Rightarrow y^{2}=4 y z-3 z^{2}$

$\Rightarrow 3 z^{2}-4 y z+y^{2}=0 \Rightarrow 3 z^{2}-3 y z-y z+y^{2}=0$

$\Rightarrow 3 z(z-y)-y(z-y)=0 \Rightarrow(3 z-y)(z-y)=0$

$\Rightarrow 3 z-y=0$ and $z-y=0$

$\Rightarrow 3 z=y$ and $z \neq y[z$ and $y$ are distinct numbers $]$

 $\Rightarrow \dfrac{z}{y} = \dfrac{1}{3}$

 $\Rightarrow r = \dfrac{1}{3}\left( {{\text{From equation }}\left( {{\text{ii}}} \right)} \right)$ 

21. If in an A.P., $S_{n}=q n^{2}$ and $S_{m}=q m^{2}$, where $S_{r}$ denotes the sum of $r$ terms of the A.P., then $\mathrm{S}_{\mathrm{q}}$ equals

(A) $\dfrac{{{q^3}}}{2}$

(B) mnq 

(C) $q^{3}$

(D) $(m+n) q^{2}$

Ans: Correct Option - C

The given series is A.P. whose first term is a and common difference is d

  ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = q{n^2}$

 $\Rightarrow 2a + \left( {n - 1} \right)d = 2qn{\text{  }}...\left( {\text{i}} \right)$

  ${S_m} = \dfrac{m}{2}\left[ {2a + \left( {m - 1} \right)d} \right] = q{m^2}$

 $\Rightarrow 2a + \left( {m - 1} \right)d = 2qm{\text{  }}...\left( {{\text{ii}}} \right)$ 

Subtract equation (i) from equation (ii),

  $\left( {m - 1} \right)d - \left( {n - 1} \right)d = 2q\left( {m - n} \right)$

  $\left( {m - 1 - n + 1} \right)d = 2q\left( {m - n} \right)$

  $\left( {m - n} \right)d = 2q\left( {m - n} \right)$

  $d = 2q$  

Putting the value of d in eq. (ii) we get

$2 a+(m-1) \cdot 2 q=2 q m$

$\Rightarrow 2 a=2 q m-(m-1) 2 q$

$\Rightarrow 2 a=2 q(m-m+1)$

$\Rightarrow 2 a=2 q \Rightarrow a=q$

  ${S_q} = \dfrac{q}{2}\left[ {2a + \left( {q - 1} \right)d} \right] = \dfrac{q}{2}\left[ {2q + \left( {q - 1} \right)2q} \right]$

$= \dfrac{q}{2}\left[ {2q + 2{q^2} - 2q} \right] = \dfrac{q}{2} \times 2{q^2} = {q^3}$ 

22. Let $S_{n}$ denote the sum of the first $n$ terms of an A.P. If $S_{2 n}=3 S_{n}$ then $S_{3 n}: S_{n}$ is equal to

(A) 4 

(B) 6 

(C) 8 

(D) 10

Ans: Correct Option - B

As, ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

 $\Rightarrow {S_{2n}} = \dfrac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right]$

 $\Rightarrow {S_{3n}} = \dfrac{{3n}}{2}\left[ {2a + \left( {3n - 1} \right)d} \right]$ 

According to the question,

$\mathrm{S}_{2 \mathrm{n}}=3 \cdot \mathrm{S}_{\mathrm{n}}$

$\dfrac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right] = 3 \cdot \dfrac{n}{2}\left[ {2a + \left( {3n - 1} \right)d} \right]$

$\Rightarrow 2[2 a+(2 n-1) d]=3[2 a+(n-1) d]$

$\Rightarrow 4 a+(4 n-2) d=6 a+(3 n-3) d$

$\Rightarrow 6 a+(3 n-3) d-4 a-(4 n-2) d=0$

$\Rightarrow 2 a+(3 n-3-4 n+2) d=0$

$\Rightarrow 2 a+(-n-1) d=0$

$\Rightarrow 2 a-(n+1) d=0$

$\Rightarrow 2 a=(n+1) d \ldots$ (i)

Now,

  ${S_{3n}}:{S_n} = \dfrac{{3n}}{2}\left[ {2a + \left( {3n - 1} \right)d} \right]:\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$

$= \dfrac{{\dfrac{{3n}}{2}\left[ {2a + \left( {3n - 1} \right)d} \right]}}{{\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]}}$

$= \dfrac{{3\left[ {2a + \left( {3n - 1} \right)d} \right]}}{{2a + \left( {n - 1} \right)d}}$

$= \dfrac{{3d\left[ {\left( {n + 1} \right)d + \left( {3n - 1} \right)d} \right]}}{{\left( {n + 1} \right)d + \left( {n - 1} \right)d}}{\text{   }}\left( {{\text{Using equation }}\left( {\text{i}} \right)} \right)$

$= \dfrac{{3d\left( {n + 1 + 3n - 1} \right)}}{{d\left( {n + 1 + n - 1} \right)}} = \dfrac{{3\left( {4n} \right)}}{{2n}} = 6$ 

23. The minimum value of $4^{x}+4^{1-x}$, $x \in \mathbb{R}$, is

(A) 2 

(B) 4 

(C) 1 

(D) 0

Ans: Correct Option - B

Since AM ≥ GM.

  $\therefore \dfrac{{{4^x} + {4^{1 - x}}}}{2} \geqslant \sqrt {{4^x} \cdot {4^{1 - x}}}$

 $\Rightarrow {4^x} + {4^{1 - x}} \geqslant 2\sqrt {{4^{x + 1 - x}}}$

 $\Rightarrow {4^x} + {4^{1 - x}} \geqslant 2 \cdot 2$

 $\Rightarrow {4^x} + {4^{1 - x}} \geqslant 4$ 

24. Let $\mathrm{S}_{\mathrm{n}}$ denote the sum of the cubes of the first n natural numbers and sn denote the sum of the first n natural numbers. Then $\sum\limits_{r = 1}^n {\dfrac{{{S_r}}}{{{s_r}}}} $ equals

(A) $\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}$

(B) $\dfrac{{n\left( {n + 1} \right)}}{2}$

(C) $\dfrac{{{n^2} + 3n + 2}}{2}$

(D) None of these

Ans: Correct Option - A

Given $\sum\limits_{i = 1}^n {\dfrac{{{S_r}}}{{{s_r}}}}  = \dfrac{{{S_1}}}{{{s_1}}} + \dfrac{{{S_2}}}{{{s_2}}} + \dfrac{{{S_3}}}{{{s_3}}} + ... + \dfrac{{{S_n}}}{{{s_n}}}$

Let $\mathrm{T}_{\mathrm{n}}$ be the nth term of the above series

${T_n} = \dfrac{{{S_n}}}{{{s_n}}} = \dfrac{{\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]}}{{\dfrac{{n\left( {n + 1} \right)}}{2}}} = \dfrac{{n\left( {n + 1} \right)}}{2} = \dfrac{{{n^2} + n}}{2}$

Now sum of the given series

  $\sum\limits_{}^{} {{T_n}}  = \dfrac{1}{2}\sum\limits_{}^{} {\left( {{n^2} + n} \right)}$  $= \dfrac{1}{2}\left( {\sum\limits_{}^{} {{n^2}}  + \sum\limits_{}^{} n } \right)$

$= \dfrac{1}{2}\left[ {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \dfrac{{n\left( {n + 1} \right)}}{2}} \right]$

$= \dfrac{1}{2} \cdot \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{2n + 1}}{3} + 1} \right] = \dfrac{{n\left( {n + 1} \right)}}{4}\left[ {\dfrac{{2n + 1 + 3}}{3}} \right]$

$= \dfrac{{n\left( {n + 1} \right)}}{4} \cdot \dfrac{{\left( {2n + 4} \right)}}{3} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}$ 

25. If $t_{n}$ denotes the $n$th term of the series $2+3+6+11+18+\ldots$ then $t_{50}$ is

(A) $49^{2}-1$

(B) $49^{2}$

(C) $50^{2}+1$

(D) $49^{2}+2$

Ans: Correct Option - D

Let $S_{n}=2+3+6+11+18+\ldots+t 50$

Using method of difference,

$S_{n}=2+3+6+11+18+\ldots+t_{50} \ldots .$ (i)

And $S_{n}=0+2+3+6+11+\ldots+t_{49}+t_{50} \ldots$ (ii)

Subtracting eq. (ii) from eq. (i),

$0=2+1+3+5+7+\ldots-t_{50}$ terms

$\Rightarrow t_{50}=2+(1+3+5+7+\ldots$ up to 49 terms)

 $\Rightarrow {t_{50}} = 2 + \dfrac{{49}}{2}\left[ {2 \times 1 + \left( {49 - 1} \right)2} \right] = 2 + \dfrac{{49}}{2}\left[ {2 + 96} \right]$

$= 2 + \dfrac{{49}}{2} \times 98 = 2 + 49 \times 49 = {49^2} + 2$ 

26. The lengths of three unequal edges of a rectangular solid block are in G.P. The

volume of the block is 216 $\mathrm{cm}^{3}$ and the total surface area is 252$\mathrm{cm}^{2}$. The length

of the longest edge is

(A) 12 cm 

(B) 6 cm 

(C) 18 cm 

(D) 3 cm

Ans: Correct Option - A

Let the length, breadth and height of a rectangular block be $\dfrac{a}{r}$, a and ar.

$\left[\text{Since they are is G.P.}\right]$

∴ Volume = l × b × h

  $216 = \dfrac{a}{r} \times a \times ar$

  ${a^3} = 216 = {6^3}$

  $a = 6$ 

Now total surface area $= 2\left[lb + bh + lh\right]$

  $252 = 2\left[ {\dfrac{a}{r} \cdot a + a \cdot ar + \dfrac{a}{r} \cdot ar} \right]$

  $126 = \left[ {\dfrac{{{a^2}}}{r} + {a^2}r + {a^2}} \right]$

  $126 = {a^2}\left[ {\dfrac{{1 + {r^2} + r}}{r}} \right]$

  $126 = {\left( 6 \right)^2}\left[ {\dfrac{{1 + {r^2} + r}}{r}} \right]$

  $\dfrac{{21}}{6} = \dfrac{{1 + {r^2} + r}}{r}$

  $\dfrac{7}{2} = \dfrac{{1 + {r^2} + r}}{r}$ 

$\Rightarrow 2+2 r+2 r^{2}=7 r$

$\Rightarrow 2 r^{2}-5 r+2=0 \Rightarrow 2 r^{2}-4 r-r+2=0$

$\Rightarrow 2 r(r-2)-1(r-2)=0$

$\Rightarrow(r-2)(2 r-1)=0$

$\Rightarrow r-2=0$ and $2 r-1=0$

∴ $r = 2,\dfrac{1}{2}$

Therefore, the three edge are:

If r = 2 then edges are 3, 6, 12

If $r = \dfrac{1}{2}$ then edges are 12, 6, 3

So, the length of the longest edge = 12

Fill in the blanks in the Exercises 27 to 29.

27. For a, b, c to be in G.P. the value of $\dfrac{{a - b}}{{b - c}}$is equal to ...............

Ans: Since a, b and c are in G.P

$\therefore$ $\dfrac{b}{a} = \dfrac{c}{b} = r$(constant)

$\Rightarrow \mathrm{b}=\mathrm{ar}$ and $\mathrm{c}=\mathrm{br}$

$\Rightarrow \mathrm{c}=\mathrm{ar} \cdot \mathrm{r}=\mathrm{ar}^{2}$

So, $\dfrac{{a - b}}{{b - c}} = \dfrac{{a - ar}}{{ar - a{r^2}}} = \dfrac{{a\left( {1 - r} \right)}}{{ar\left( {1 - r} \right)}} = \dfrac{1}{r} = \dfrac{a}{b} = \dfrac{b}{c}$

Hence, the correct value of the filler is $\dfrac{a}{b}$ or $\dfrac{b}{c}$.

28. The sum of terms equidistant from the beginning and end in an A.P. is equal to

............ .

Ans: Let A.P be $a, a+d, a+2 d, a+3 d, \ldots, a+(n-1) d$

Taking first and last term

$a_{1}+a_{n}=a+a+(n-1) d=2 a+(n-1) d$

Taking second and second last term,

$a_{2}+a_{n-2}=(a+d)+[a+(n-2) d]=2 a+(n-1) d=a_{1}+a_{n}$

Taking third from the beginning and the third from the end

$a_{3}+a_{n-2}=(a+2 d)+[a+(n-3) d]=2 a+(n-1) d=a_{1}+a_{n}$

From the above pattern, we observe that the sum of terms equidistant from the beginning and the end in an A.P is equal to the [first term + last term]

Hence, the correct value of the filler is first team + last term.

29. The third term of a G.P. is 4, the product of the first five terms is .................

Ans: Given $\mathrm{T}_{3}=4$

$\therefore$ ar $^{2}=4 \ldots$..(i)

Product of first five terms $=\mathrm{a} \cdot \mathrm{ar} \cdot \mathrm{ar}^{2} \cdot \mathrm{ar}^{3} \cdot \mathrm{ar}^{4}$

$=\mathrm{a}^{5} \mathrm{r}^{10}=\left(\mathrm{ar}^{2}\right)^{5}=(4)^{5}$   $\left[\text{from equation (i)}\right]$

Hence, the correct value of the filter is $(4)^{5}$.

State whether statements in Exercises 30 to 34 are True or False.

30. Two sequences cannot be in both A.P. and G.P. together.

Ans: True

Let us consider G.P, $a$, ar and $a r^{2}$ 

If it is in A.P then $a r-a \neq a r^{2}-a r$

Hence, the given statement is True.

31. Every progression is a sequence but the converse, i.e., every sequence is also a

progression need not necessarily be true.

Ans: True

Let us consider a sequence of prime numbers 2, 3, 5, 7, 11, …

It is clear that this progression is a sequence but the sequence is not a progression because it does not follow a specific pattern. Here, the given statement is True.

32. Any term of an A.P. (except first) is equal to half the sum of terms which are

equidistant from it.

Ans: True

Let us consider an A.P $a, a+d, a+2 d$. 

$\therefore a_{2}+a_{4}=a+d+a+3 d=2 a+4 d=2 a_{3}$

$\Rightarrow {{\text{a}}_3} = \dfrac{{{{\text{a}}_2} + {{\text{a}}_4}}}{2}$

\[\dfrac{{{{\text{a}}_3} + {{\text{a}}_5}}}{2} = \dfrac{{{\text{a}} + 2d + {\text{a}} + 4d}}{2} = \dfrac{{2{\text{a}} + 6d}}{2}\]

$\Rightarrow=a+3 d=a_{4}$

Hence, the given statement is True.

33. The sum or difference of two G.P.s, is again a G.P.

Ans: False

Let two G.P.s are $a,a{r_1},a{r_1}^2,a{r_1}^3,...;{\text{ and }}b,b{r_2},b{r_2}^2,b{r_2}^3,...$

No sum of two G.P.s is $a + b,a{r_1} + b{r_2},a{r_1}^2 + b{r_2}^2,a{r_1}^3 + b{r_2}^3...$

Clearly $\dfrac{{a{r_1} + b{r_2}}}{{a + b}} \ne \dfrac{{a{r_1}^2 + b{r_2}^2}}{{a{r_1} + b{r_2}}}$

Similarly, for difference of two G.P.s, $\dfrac{{a{r_1} - b{r_2}}}{{a - b}} \ne \dfrac{{a{r_1}^2 - b{r_2}^2}}{{a{r_1} - b{r_2}}}$

So, the sum or difference of two G.P.s is not a G.P.

34. If the sum of n terms of a sequence is quadratic expression then it always represents an A.P.

Ans: False

Let $S_{n}=a n^{2}+b n+c$ (quadratic expression)

$S_{1}=a+b+c$

$\therefore a_{1}=a+b+c$

$S_{2}=4 a+2 b+c$

$a_{2}=S_{2}-S_{1}=(4 a+2 b+c)-(a+b+c)=3 a+b$

$S_{3}=9 a+3 b+c$

$\Rightarrow a_{3}=S_{3}-S_{2}=(9 a+3 b+c)-(4 a+2 b+c)=5 a+b$

Common difference, $d=a_{2}-a_{1}=(3 a+b)-(a+b+c)$

$=2 a-c$

and $d=a_{3}-a_{2}=(5 a+b)-(3 a+b)=2 a$

Here, we observe that $a_{2}-a_{1} \neq a_{3}-a_{2}$

So, it does not represent an A.P.

Hence, the given statement is False.

Match the questions given under Column I with their appropriate answers given

under the Column II.

35. 

Ans: (a) $4,1,\dfrac{1}{4},\dfrac{1}{{16}}$

Here, $\dfrac{{{a_2}}}{{{a_1}}} = \dfrac{1}{4},\dfrac{{{a_3}}}{{{a_2}}} = \dfrac{{\dfrac{1}{4}}}{1}{\text{ and }}\dfrac{{{a_4}}}{{{a_3}}} = \dfrac{{\dfrac{1}{{16}}}}{{\dfrac{1}{4}}} = \dfrac{1}{4}$ 

Hence it is G.P

$\therefore(\mathrm{a}) \leftrightarrow(\mathrm{iii})$

(b) $2,3,5,7$

Here $a_{2}-a_{1}=3-2=1$

$a_{3}-a_{2}=5-3=2$

$\therefore a_{2}-a_{1} \neq a_{3}-a_{2}$

Hence it is not A.P.

So, $\dfrac{3}{2} \ne \dfrac{5}{3}$

So, it is not G.P

Hence it is sequence.

∴ Hence, (b) $\leftrightarrow$ (ii)

(c) $13,8,3,-2,-7$

Here $a_{2}-a_{1}=8-13=-5$

$a_{3}-a_{2}=3-8=-5$

So, $a_{2}-a_{1}=a_{3}-a_{2}=-5$

So, it is an A.P.

Hence, (c) $\leftrightarrow$ (i)

36.

Ans: (a) Let $S_{n}=1^{2}+2^{2}+3^{2}+\ldots+n^{2}$

$K^{3}-(K-1)^{3}=3 K^{2}-3 k+1$

For $K=1,1^{3}-0^{3}=3(1)^{2}-3(1)+1$

For $K=2,2^{3}-1^{3}=3(2)^{2}-3(2)+1$

For $K=3,3^{3}-2^{3}=3(3)^{2}-3(3)+1$

For $K=n$,

Adding Column wise, we get

$n^{3}-0^{3}=3\left(1^{2}+2^{2}+3^{2}+\ldots+n^{2}\right)-3(1+2+3+\ldots+n)+n$

 $\Rightarrow {n^3} = 3 \cdot {S_n} - \dfrac{{3n\left( {n + 1} \right)}}{2} + n$

 $\Rightarrow {n^3} + \dfrac{{3n\left( {n + 1} \right)}}{2} - n = 3 \cdot {S_n}$

 $\Rightarrow \dfrac{{2{n^3} + 3{n^2} + 3n - 2n}}{2} = 3 \cdot {S_n}$

 $\Rightarrow 6{S_n} = 2{n^3} + 3{n^2} + n$

 $\Rightarrow 6{S_n} = n\left( {2{n^2} + 3n + 1} \right)$

 $\Rightarrow 6{S_n} = n\left( {2{n^2} + 2n + n + 1} \right)$

 $\Rightarrow 6{S_n} = n\left( {n + 1} \right)\left( {2n + 1} \right)$

 $\Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$ 

Here $(a) \leftrightarrow(i i i) .$

(b) Let $S_{n}=1^{3}+2^{3}+3^{3}+\ldots+n^{3}$

$K^{4}-(K-1)^{4}=4 K^{3}-6 k^{2}+4 k-1$

For $K=1,1^{4}-0^{4}=4(1)^{3}-6(1)^{2}+4(1)-1$

For $K=2,2^{4}-1^{4}=4(2)^{3}-6(2)^{2}+4(2)-1$

For $K=3,3^{4}-2^{4}=4(3)^{3}-6(3)^{2}+4(3)-1$

For $K=n, 3^{4}-2^{4}=4(n)^{3}-6(n)^{2}+4(n)-1$

Adding column wise,

$\left.\Rightarrow n^{4}-0^{4}=4\left(1^{3}+2^{3}+3^{3}+\ldots n^{3}\right)-6\left(1^{2}+2^{2}+3^{2}+\ldots+n^{2}\right)+4(1+2+3+\ldots n)-n\right)$

 $\Rightarrow {n^4} + 4{S_n} - \dfrac{{6n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \dfrac{{4n\left( {n + 1} \right)}}{2} - n$

 $\Rightarrow {n^4} + n\left( {n + 1} \right)\left( {2n + 1} \right) - 2n\left( {n + 1} \right) + n = 4{S_n}$

 $\Rightarrow n\left[ {{n^3} + \left( {n + 1} \right)\left( {2n + 1} \right) - 2\left( {n + 1} \right) + 1} \right] = 4{S_n}$

 $\Rightarrow n\left( {{n^3} + 2{n^2} + 3n + 1 - 2n - 1} \right) = 4{S_n}$

 $\Rightarrow n\left( {{n^3} + 2{n^2} + n} \right) = 4{S_n}$

 $\Rightarrow \dfrac{{{n^2}\left( {{n^2} + 2n + 1} \right)}}{4} = {S_n}$

 $\Rightarrow \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} = {S_n}$

 $ {\text{So,}}{S_n} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}$ 

Hence $(b) \leftrightarrow(i)$

(c) Let $S_{n}=2+4+6+\ldots+2 n$

$=2(1+2+3+\ldots+n)$

$ = 2\dfrac{{n\left( {n + 1} \right)}}{2}$

$=n(n+1)$

Hence $(c) \leftrightarrow($ ii $)$

(d) Let $S_{n}=1+2+3+\ldots+n=$ $\dfrac{{n\left( {n + 1} \right)}}{2}$

Hence $(d) \leftrightarrow(iv)$

Although There Are Many Types Of Series In Class 11, Two Main Types Of Sequences And Series Have Been Included

• Arithmetic Sequences and Series/Progression- This type of series is when the successive terms, except the first term, have the common term between two numbers found either through addition or subtraction. For example, 2, 4, 6, 8 …. is an arithmetic sequence and 2+4+6+8 ….. is an arithmetic progression/series.

• Geometric Sequences and Series/Progression- This type of series is when the successive terms have a common ratio by multiplying or dividing the common term between two terms. For example, 2, 4, 8, 16 …… is a geometric sequence and 2+ 4+ 8+ 16 …. is a geometric progression/series.

Each of these progressions can be solved through formulas derived for specific series only. There are formulas even to find the last term of the series (both finite and infinite).