NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

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Exercise 5.3

1. Find the sum of the following APs.

I.  $$\mathbf{2},\mathbf{7},\mathbf{12},....$$ to $$\mathbf{10}$$ terms.

Ans: Given, the first Term, $a=2$  ….. (1)

Given, the common Difference, $$d=7-2=5$$ …..(2)

Given, the number of Terms, $$n=10$$  …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, $S_n={\displaystyle \frac{10}{2}}[2\left(2\right)+(10-1)\left(5\right)]$

$\Rightarrow S_n=5[4+45]$

$\therefore S_n=245$ 

II. $$-\mathbf{37},-\mathbf{33},-\mathbf{29},...$$ to $$\mathbf{12}$$ terms

Ans: Given, the first Term, $a=-37$  ….. (1)

Given, the common Difference, $$d=-33-(-37)=4$$ …..(2)

Given, the number of Terms, $$n=12$$  …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, $S_n={\displaystyle \frac{12}{2}}[2(-37)+(12-1)\left(4\right)]$

$\Rightarrow S_n=6[-74+44]$

$\therefore S_n=-180$ 

iii. $$\mathbf{0}.\mathbf{6},\mathbf{1}.\mathbf{7},\mathbf{2}.\mathbf{8},......$$ to $$\mathbf{100}$$ terms

Ans: Given, the first Term, $a=0.6$  ….. (1)

Given, the common Difference, $$d=1.7-0.6=1.1$$ …..(2)

Given, the number of Terms, $$n=100$$  …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, $S_n={\displaystyle \frac{100}{2}}[2\left(0.6\right)+(100-1)\left(1.1\right)]$

$\Rightarrow S_n=50[1.2+108.9]$

$\therefore S_n=5505$ 

iv. ${\displaystyle \frac{1}{15}},{\displaystyle \frac{1}{12}},{\displaystyle \frac{1}{10}},.....$ to 11 terms

Ans: Given, the first Term, $a={\displaystyle \frac{1}{15}}$  ….. (1)

Given, the common Difference, $$d={\displaystyle \frac{1}{12}}-{\displaystyle \frac{1}{15}}={\displaystyle \frac{1}{60}}$$ …..(2)

Given, the number of Terms, $$n=11$$  …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, $S_n={\displaystyle \frac{11}{2}}[2\left({\displaystyle \frac{1}{15}}\right)+(11-1)\left({\displaystyle \frac{1}{60}}\right)]$

$\Rightarrow S_n={\displaystyle \frac{11}{2}}\left[{\displaystyle \frac{4+5}{30}}\right]$

$$\therefore S_n={\displaystyle \frac{33}{20}}$$ 

2. Find the sums given below

I. $7+10{\displaystyle \frac{1}{2}}+14+.....+84$ 

Ans: Given, the first Term, $a=7$  ….. (1)

Given, the common Difference, $$d=10{\displaystyle \frac{1}{2}}-7={\displaystyle \frac{7}{2}}$$ …..(2)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (3)

Substituting the values from (1) and (2) in (3) we get, 

$a_n=7+{\displaystyle \frac{7}{2}}(n-1)={\displaystyle \frac{7}{2}}(n+1)$  ….. (4)

Given, last term of the series, $$a_n=84$$  …..(5)

Substituting (5) in (4) we get, $84={\displaystyle \frac{7}{2}}(n+1)$

$\Rightarrow 24=(n+1)$ 

$$\therefore n=23$$  ……(6)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by $S_n={\displaystyle \frac{n}{2}}[a+l]$     …..(7)

Substituting the values from (1), (5) and (6) in (7) we get, $S_n={\displaystyle \frac{23}{2}}[7+84]$

$\Rightarrow S_n={\displaystyle \frac{23}{2}}\left(91\right)$

$$\therefore S_n=1046{\displaystyle \frac{1}{2}}$$ 

ii. $$\mathbf{34}+\mathbf{32}+\mathbf{30}+.....+\mathbf{10}$$ 

Ans: Given, the first Term, $a=34$  ….. (1)

Given, the common Difference, $$d=32-34=-2$$ …..(2)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (3)

Substituting the values from (1) and (2) in (3) we get, 

$a_n=34-2(n-1)=36-2n$  ….. (4)

Given, last term of the series, $$a_n=10$$  …..(5)

Substituting (5) in (4) we get, $10=36-2n$

$\Rightarrow 2n=26$ 

$$\therefore n=13$$  ……(6)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by $S_n={\displaystyle \frac{n}{2}}[a+l]$     …..(7)

Substituting the values from (1), (5) and (6) in (7) we get, $S_n={\displaystyle \frac{13}{2}}[34+10]$

$\Rightarrow S_n={\displaystyle \frac{13}{2}}\left(44\right)$

$$\therefore S_n=286$$ 

iii. $$-5+(-8)+(-11)+.....+(-230)$$ 

Ans: Given, the first Term, $a=-5$  ….. (1)

Given, the common Difference, $$d=-8-(-5)=-3$$ …..(2)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (3)

Substituting the values from (1) and (2) in (3) we get, 

$a_n=-5-3(n-1)=-2-3n$  ….. (4)

Given, last term of the series, $$a_n=-230$$  …..(5)

Substituting (5) in (4) we get, $-230=-2-3n$

$\Rightarrow -228=-3n$ 

$$\therefore n=76$$  ……(6)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by $S_n={\displaystyle \frac{n}{2}}[a+l]$     …..(7)

Substituting the values from (1), (5) and (6) in (7) we get, $S_n={\displaystyle \frac{76}{2}}[-5+(-230)]$

$\Rightarrow S_n={\displaystyle \frac{76}{2}}(-235)$

$$\therefore S_n=-8930$$ 

3. In an AP

i. Given $$a=5$$, $$d=3$$, $${\mathbf{a}}_{\mathbf{n}}=\mathbf{50}$$, find $$\mathbf{n}$$ and $${\mathbf{S}}_{\mathbf{n}}$$.

Ans: Given, the first Term, $a=5$  ….. (1)

Given, the common Difference, $$d=3$$ …..(2)

Given, $n^{th}$ term of the A.P., $${\mathbf{a}}_{\mathbf{n}}=\mathbf{50}$$  …..(3)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (4)

Substituting the values from (1), (2) and (3) in (4) we get, 

$50=5+3(n-1)=2+3n$ 

Simplifying it further we get, 

$n={\displaystyle \frac{50-2}{3}}$ 

$\therefore n=16$   …..(5)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(6)

Substituting the values from (1), (2) and (5) in (6) we get, $S_n={\displaystyle \frac{16}{2}}[2\left(5\right)+(16-1)\left(3\right)]$

$\Rightarrow S_n=8[10+45]$

$$\therefore S_n=440$$ 

ii. Given $$a=7$$, $$a_{13}=35$$, find $$d$$ and $${\mathbf{S}}_{13}$$.

Ans: Given, the first Term, $a=7$  ….. (1)

Given, ${13}^{th}$ term of the A.P., $$a_{13}=35$$  …..(2)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (3)

Substituting the values from (1), (2) in (3) we get, 

$35=7+(13-1)d=7+12d$ 

Simplifying it further we get, 

$d={\displaystyle \frac{28}{12}}$ 

$$\therefore d={\displaystyle \frac{7}{3}}$$   …..(4)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(5)

Substituting the values from (1) and (4) in (5) we get, $S_{13}={\displaystyle \frac{13}{2}}[2\left(7\right)+(13-1)\left({\displaystyle \frac{7}{3}}\right)]$

$\Rightarrow S_{13}={\displaystyle \frac{13}{2}}[14+28]$

$$\therefore S_{13}=273$$ 

iii. Given $$d=3$$, $${\mathbf{a}}_{12}=37$$, find $$a$$ and $${\mathbf{S}}_{12}$$.

Ans: Given, the common difference, $d=3$  ….. (1)

Given, ${12}^{th}$ term of the A.P., $${\mathbf{a}}_{12}=37$$  …..(2)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (3)

Substituting the values from (1), (2) in (3) we get, 

$37=a+3(12-1)=a+33$ 

Simplifying it further we get,  

$$\therefore a=4$$   …..(4)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(5)

Substituting the values from (1) and (4) in (5) we get, $S_{12}={\displaystyle \frac{12}{2}}[2\left(4\right)+(12-1)\left(3\right)]$

$\Rightarrow S_{12}=6[8+33]$

$$\therefore S_{12}=246$$

iv. Given $${\mathbf{a}}_3=1\mathbf{5}$$, $${\mathbf{S}}_{10}=125$$ find $$a_{10}$$ and $$d$$.

Ans: Given, $3^{rd}$ term of the A.P., $${\mathbf{a}}_3=1\mathbf{5}$$  …..(1)

Given, the sum of terms, $${\mathbf{S}}_{10}=125$$  ….. (2)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (3)

Substituting the values from (1) in (3) we get, 

$15=a+(3-1)d=a+2d$  …..(4)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(5)

Substituting the values from (1) in (5) we get, $125={\displaystyle \frac{10}{2}}[2a+(10-1)d]$

$\Rightarrow 125=5[2a+9d]$

$$\therefore 25=2a+9d$$  …..(5)

Let us solve equations (4) and (5) by subtracting twice of (4) from (5) we get,

$$25-30=(2a+9d)-(2a+4d)$$

$\Rightarrow -5=5d$ 

$\therefore d=-1$    …..(6)

From (4) and (6) we get, $a=17$   …..(7)

From (3), (6) and (7) for $n=10$ we get,

$a_{10}=17-(10-1)$

$\therefore a_{10}=8$  

v. Given $${\mathbf{S}}_9=75$$, $$d=5$$ find $$a$$ and $$a_9$$.

Ans: Given, common difference, $$d=5$$  …..(1)

Given, the sum of terms, $${\mathbf{S}}_9=75$$  ….. (2)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(3)

Substituting the values from (1), (2) in (3) we get, $75={\displaystyle \frac{9}{2}}[2a+5(9-1)]$

$\Rightarrow 25=3[a+20]$

$\Rightarrow 3a=-35$

$$\therefore a=-{\displaystyle \frac{35}{3}}$$  …..(4)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (5)

Substituting the values from (1), (4) in (5) we get, 

$a_9=-{\displaystyle \frac{35}{3}}+5(9-1)$ 

$\Rightarrow a_9=-{\displaystyle \frac{35}{3}}+40$ 

$\therefore a_9={\displaystyle \frac{85}{3}}$ 

vi. Given $$a=2$$, $$d=8$$, $$S_{\mathbf{n}}=9\mathbf{0}$$, find $$\mathbf{n}$$ and $$a_{\mathbf{n}}$$.

Ans: Given, common difference, $$d=8$$  …..(1)

Given, first term, $$a=2$$  …..(2)

Given, the sum of terms, $$S_{\mathbf{n}}=9\mathbf{0}$$  ….. (3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(3)

Substituting the values from (1), (2), (3) in (4) we get, $90={\displaystyle \frac{n}{2}}[2\left(2\right)+8(n-1)]$

$\Rightarrow 45=n[2n-1]$

$\Rightarrow 2n^2-n-45=0$

$\Rightarrow 2n^2-10n+9n-45=0$

$\Rightarrow 2n(n-5)+9(n-5)=0$

$\Rightarrow (n-5)(2n+9)=0$

$$\therefore n=5$$  …..(4)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (5)

Substituting the values from (1), (2), (4) in (5) we get, 

$a_5=2+8(5-1)$ 

$\Rightarrow a_5=2+32$ 

$\therefore a_5=34$ 

vii. Given $$a=8$$, $$S_{\mathbf{n}}=21\mathbf{0}$$, $${\mathbf{a}}_{\mathbf{n}}=62$$, find $$\mathbf{n}$$ and $$d$$.

Ans: Given, first term, $$a=8$$  …..(1)

Given, the sum of terms, $$S_{\mathbf{n}}=21\mathbf{0}$$  ….. (2)

Given, the $n^{th}$ term, $${\mathbf{a}}_{\mathbf{n}}=62$$ …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(4)

Substituting the values from (1), (2) in (4) we get, $210={\displaystyle \frac{n}{2}}[2\left(8\right)+d(n-1)]$

$\Rightarrow 420=n[16+(n-1)d]$  …..(4)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (5)

Substituting the values from (1), (3) in (5) we get, 

$62=8+(n-1)d$   …..(6) 

Let us solve equations (4) and (6) by subtracting $n$ times of (6) from (4) we get,

$420-62n=(16n+n(n-1)d)-(8n+n(n-1)d)$

$\Rightarrow 420-62n=8n$ 

$\Rightarrow 420=70n$

$\therefore n=6$  ……(7)

Substituting the values from (7) in (6) we get, 

 $62=8+(6-1)d$

$\Rightarrow 54=5d$ 

$\therefore d={\displaystyle \frac{54}{5}}$ 

viii. Given $$S_{\mathbf{n}}=-14$$, $$d=2$$, $${\mathbf{a}}_{\mathbf{n}}=4$$, find $$\mathbf{n}$$ and $$a$$.

Ans: Given, common difference, $$d=2$$  …..(1)

Given, the sum of terms, $$S_{\mathbf{n}}=-14$$  ….. (2)

Given, the $n^{th}$ term, $${\mathbf{a}}_{\mathbf{n}}=4$$ …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(4)

Substituting the values from (1), (2) in (4) we get, $-14={\displaystyle \frac{n}{2}}[2a+2(n-1)]$

$\Rightarrow -14=n[a+n-1]$  …..(5)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (6)

Substituting the values from (1), (3) in (6) we get, 

$4=a+2(n-1)$   …..(7) 

Let us solve equations (5) and (7) by substituting the value of $a$ from (7) in (5) we get,

$-14=n[(4-2(n-1))+n-1]$

$$\Rightarrow -14=n[5-n]$$

$\Rightarrow n^2-5n-14=0$ 

$\Rightarrow n^2-7n+2n-14=0$

$\Rightarrow (n-7)(n+2)=0$

$\therefore n=7$ (Since $n$ cannot be negative)  ……(8)

Substituting the values from (8) in (7) we get, 

 $4=a+2(7-1)$

$\Rightarrow 4=a+12$ 

$\therefore a=-8$ 

ix. Given $$a=3$$, $$n=8$$, $$S=192$$, find $$d$$.

Ans: Given, first term, $$a=3$$  …..(1)

Given, the sum of terms, $$S_{\mathbf{n}}=192$$  ….. (2)

Given, the number of terms, $$n=8$$ …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(4)

Substituting the values from (1), (2) in (4) we get, $192={\displaystyle \frac{8}{2}}[2\left(3\right)+d(8-1)]$

$\Rightarrow 192=4[6+7d]$

$\Rightarrow 48=6+7d$

$\Rightarrow 42=7d$

$\therefore d=6$ 

x. Given $$l=28$$, $$S=144$$ and there are total $9$ terms. Find $$a$$.

Ans: Given, last term, $$l=28$$  …..(1)

Given, the sum of terms, $$S_{\mathbf{n}}=144$$  ….. (2)

Given, the number of terms, $$n=9$$ …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by $S_n={\displaystyle \frac{n}{2}}[a+l]$     …..(4)

Substituting the values from (1), (2) in (4) we get, $144={\displaystyle \frac{9}{2}}[a+28]$

$\Rightarrow 32=a+28$

$\therefore a=4$ 

4. How many terms of the A.P. $$9,17,25...$$ must be taken to give a sum of $$\mathbf{636}$$?

Ans: Given, common difference, $$d=17-9=8$$  …..(1)

Given, first term, $$a=9$$  …..(2)

Given, the sum of terms, $$S_{\mathbf{n}}=636$$  ….. (3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(3)

Substituting the values from (1), (2), (3) in (4) we get, $636={\displaystyle \frac{n}{2}}[2\left(9\right)+8(n-1)]$

$\Rightarrow 636=n(5+4n)$

$\Rightarrow 4n^2+5n-636=0$

$\Rightarrow 4n^2+53n-48n-636=0$

$\Rightarrow n(4n+53)-12(4n+53)=0$

$\Rightarrow (n-12)(4n+53)=0$

$\Rightarrow n=12or-{\displaystyle \frac{53}{4}}$

Since $n$ can only be a natural number $$\therefore n=12$$

5. The first term of an AP is $5$, the last term is $45$ and the sum is $400$. Find the number of terms and the common difference.

Ans: Given, first term, $$a=5$$  …..(1)

Given, the sum of terms, $$S_{\mathbf{n}}=400$$  ….. (2)

Given, the $n^{th}$ term, $${\mathbf{a}}_{\mathbf{n}}=45$$ …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by $S_n={\displaystyle \frac{n}{2}}[a+l]$     …..(4)

Substituting the values from (1), (2), (3) in (4) we get, $400={\displaystyle \frac{n}{2}}[5+45]$

$\Rightarrow 400=25n$ 

$\therefore n=16$ 

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (5)

Substituting the values from (1), (3) in (5) we get, 

$45=5+(16-1)d$   

$\Rightarrow 40=15d$ 

$\therefore d={\displaystyle \frac{8}{3}}$

6. The first and the last term of an AP are $17$ and $350$ respectively. If the common difference is $9$, how many terms are there and what is their sum? 

Ans: Given, first term, $$a=17$$  …..(1)

Given, the common difference, $$d=9$$  ….. (2)

Given, the $n^{th}$ term, $${\mathbf{a}}_{\mathbf{n}}=350$$ …..(3)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (4)

Substituting the values from (1), (2), (3) in (4) we get, 

$350=17+9(n-1)$   

$\Rightarrow 333=9(n-1)$ 

$\Rightarrow 37=(n-1)$

$\therefore n=38$ ……(5)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by $S_n={\displaystyle \frac{n}{2}}[a+l]$     …..(6)

Substituting the values from (1), (5), (3) in (6) we get, $S_{38}={\displaystyle \frac{38}{2}}[17+350]$

$\Rightarrow S_{38}=19\left(367\right)$ 

$\therefore S_{38}=6973$ 

7. Find the sum of first $$22$$ terms of an AP in which $$d=7$$ and $${22}^{nd}$$ term is $$149$$.

Ans: Given, the common difference, $$d=7$$  ….. (1)

Given, the ${22}^{nd}$ term, $${\mathbf{a}}_{22}=149$$ …..(2)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (3)

Substituting the values from (1), (2) in (3) we get, 

$149=a+7(22-1)$   

$\Rightarrow 149=a+147$ 

$\therefore a=2$  ……(4)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by $S_n={\displaystyle \frac{n}{2}}[a+l]$     …..(5)

Substituting the values from (1), (2), (4) in (5) we get, $S_{22}={\displaystyle \frac{22}{2}}[2+149]$

$\Rightarrow S_{22}=11\left(151\right)$ 

$\therefore S_{22}=1661$ 

8. Find the sum of first $$51$$ terms of an AP whose second and third terms are $$14$$ and $$18$$ respectively.

Ans: Given, the $2^{nd}$ term, $${\mathbf{a}}_2=14$$  ….. (1)

Given, the $3^{rd}$ term, $${\mathbf{a}}_3=18$$ …..(2)

We know that the $n^{th}$ term of the A.P. with first term $a$ and common difference $d$ is given by $a_n=a+(n-1)d$   ….. (3)

Substituting the values from (1) in (3) we get, 

$14=a+d$   …..(4)

Substituting the values from (2) in (3) we get, 

$18=a+2d$   …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

$18-14=(a+2d)-(a+d)$ 

$\therefore d=4$  ……(6)

Substituting the value from (6) in (4) we get $a=10$.   …..(7) 

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     …..(8)

Substituting the values from (7), (6) in (8) we get for $n=51$ ,

$S_{51}={\displaystyle \frac{51}{2}}[2\left(10\right)+4(51-1)]$

$\Rightarrow S_{51}={\displaystyle \frac{51}{2}}[20+200]$ 

$\therefore S_{51}=5610$ 

9. If the sum of first $$7$$ terms of an AP is $$49$$ and that of $$17$$ terms is $$289$$, find the sum of first $$n$$ terms.

Ans: Given, the sum of first $7$ terms, $$S_7=49$$  ….. (1)

Given, the sum of first $17$ terms, $$S_{17}=289$$   …..(2)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     ….. (3)

Substituting the values from (1) in (3) we get, 

$$49={\displaystyle \frac{7}{2}}[2a+(7-1)d]$$

$$\Rightarrow 7=a+3d$$   …..(4)

Substituting the values from (2) in (3) we get, 

$$289={\displaystyle \frac{17}{2}}[2a+(17-1)d]$$

$$\Rightarrow 17=a+8d$$   …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

$17-7=(a+8d)-(a+3d)$ 

$\Rightarrow 10=5d$ 

$\therefore d=2$  ……(6)

Substituting the value from (6) in (4) we get $a=1$.   …..(7) 

Substituting the values from (7), (6) in (3) we get,

 $S_n={\displaystyle \frac{n}{2}}[2+2(n-1)]$

$\therefore S_n=n^2$ 

10. Show that $$a_1,a_2...,a_n,...$$ form an AP where $a_n$ is defined as below. Also find the sum of the first $15$ terms in each case. 

i. $$a_n=3+4n$$ 

Ans:  Consider two consecutive terms of the given sequence. Say $a_n,a_{n+1}$. Difference between these terms will be 

$a_{n+1}-a_n=[3+4(n+1)]-[3+4n]$ 

$\Rightarrow a_{n+1}-a_n=4(n+1)-4n$ 

$\Rightarrow a_{n+1}-a_n=4$

Which is a constant $\mathrm{\forall }n\in \mathbb{N}$. 

For $n=1$, $a_1=3+4=7$ 

Therefore, it is an A.P. with first term $7$ and common difference $4$. 

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     

Therefore, $S_{15}={\displaystyle \frac{15}{2}}[2\left(7\right)+4(15-1)]$

$\Rightarrow S_{15}={\displaystyle \frac{15}{2}}\left[14\left(5\right)\right]$ 

$\therefore S_{15}=525$ 

ii. $$a_n=9-5n$$ 

Ans: Consider two consecutive terms of the given sequence. Say $a_n,a_{n+1}$. Difference between these terms will be 

$a_{n+1}-a_n=[9-5(n+1)]-[9-5n]$ 

$\Rightarrow a_{n+1}-a_n=-5(n+1)+5n$ 

$\Rightarrow a_{n+1}-a_n=-5$

Which is a constant $\mathrm{\forall }n\in \mathbb{N}$. 

For $n=1$, $a_1=9-5=4$ 

Therefore, it is an A.P. with first term $4$ and common difference $-5$. 

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     

Therefore, $S_{15}={\displaystyle \frac{15}{2}}[2\left(4\right)-5(15-1)]$

$$\Rightarrow S_{15}=15[-31]$$ 

$\therefore S_{15}=-465$

11. If the sum of the first $n$ terms of an AP is $$4n-n^2$$, what is the first term (that is $S_1$)? What is the sum of first two terms? What is the second term? Similarly find the $$3^{rd}$$, the $${10}^{th}$$ and the $$n^{th}$$ terms. 

Ans:  Given, the sum of the first $n$ terms of an A.P. is $$4n-n^2$$.

First term $=S_1=4-1=3$. …..(1)

Sum of first two terms $=S_2=8-{\left(2\right)}^2=4$  …..(2) 

From (1) and (2), $2^{nd}$ term $=S_2-S_1=4-3=1$. 

Sum of first three terms $=S_3=12-{\left(3\right)}^2=3$  …..(3) 

From (3) and (2), $3^{rd}$ term $=S_3-S_2=3-4=-1$. 

Similarly, 

Sum of first $n$ terms $=S_n=4n-n^2$  …..(4) 

Sum of first $n-1$ terms $=S_{n-1}=4(n-1)-{(n-1)}^2=-n^2+6n-5$  …..(5) 

From (4) and (5), $n^{th}$ term $=S_n-S_{n-1}=(4n-n^2)-(-n^2+6n-5)=5-2n$  …..(6)

From (6), ${10}^{th}$ term is $5-2\left(10\right)=-15$. 

12. Find the sum of first $40$ positive integers divisible by $6$. 

Ans:  First positive integer that is divisible by $6$ is $6$ itself. 

Second positive integer that is divisible by $6$ is $$6+6=12$$.

Third positive integer that is divisible by $6$ is $$12+6=18$$.  

Hence, it is an A.P. with first term and common difference both as $6$.  

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     

Therefore, for $n=40$,

$S_{40}={\displaystyle \frac{40}{2}}[2\left(6\right)+6(40-1)]$

$$\Rightarrow S_{40}=120\left[41\right]$$ 

$\therefore S_{40}=4920$

13. Find the sum of first $15$ multiples of $8$. 

Ans:  First positive integer that is divisible by $8$ is $8$ itself. 

Second positive integer that is divisible by $8$ is $$8+8=16$$.

Third positive integer that is divisible by $8$ is $$16+8=24$$.  

Hence, it is an A.P. with first term and common difference both as $8$.  

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$     

Therefore, for $n=15$,

$S_{15}={\displaystyle \frac{15}{2}}[2\left(8\right)+8(15-1)]$

$$\Rightarrow S_{15}=60\left[16\right]$$ 

$\therefore S_{15}=960$

14. Find the sum of the odd numbers between $0$ and $50$.

Ans: The odd numbers between $0$ and $50$ are $1,3,5,...,49$.