Concise Mathematics Class 10 ICSE Solutions for Chapter 11 - Geometric Progression

Exercise 11A

1. Find, which of the following sequences form a G.P:

(i) 8, 24, 72, 216, .......

Ans: The series is 8,24,72,216,

The common ratios can be obtained as follows,

${r}=\dfrac{24}{8}=3$

${r}=\dfrac{72}{24}=3$

${r}=\dfrac{216}{72}=3$

The common ratio of the consecutive terms is equal. Therefore, it is a G.P.

(ii) $\dfrac{{1}}{{8}}{,}\dfrac{{1}}{{24}}{,}\dfrac{{1}}{{72}}{,}\dfrac{{1}}{{216}}{,}$.......

Ans: The common ratios can be obtained as follows,

${r}=\dfrac{\dfrac{1}{24}}{\dfrac{1}{8}}=\dfrac{8}{24}=\dfrac{1}{3}$

${r}=\dfrac{\dfrac{1}{72}}{\dfrac{1}{24}}=\dfrac{24}{72}=\dfrac{1}{3}$

${r}=\dfrac{\dfrac{1}{216}}{\dfrac{1}{72}}=\dfrac{72}{216}=\dfrac{1}{3}$

The common ratio of the consecutive terms is equal. Therefore, it is a G.P.

(iii) 9, 12, 16, 24, .......

Ans: The common ratios can be obtained as follows,

${r}=\dfrac{12}{9}=\dfrac{4}{3}$

${r}=\dfrac{16}{12}=\dfrac{4}{3}$

${r}=\dfrac{24}{16}=\dfrac{3}{2}$

The common ratio of the consecutive terms is not equal. Therefore, it is not a G.P.

2. Find the ${{{9}}^{{th}}}$term of the series 1, 4, 16, 64, .......

Ans: The series is 1,4,16,64,

First term, a=1.

Common ratio, r=4.

Ninth term ${{{a}}_{9}}$ a can be calculated as follows,

${{{a}}_{9}}={a}{{{r}}^{8}}$

$\Rightarrow {{{a}}_{9}}=1{{(4)}^{8}}$

$\Rightarrow {{{a}}_{9}}=65536$

3. Find the seventh term of the G.P ${1,}\sqrt{{3}}{,3,3}\sqrt{{3}}{,}.......$

Ans: First term, a=1

Common ratio, r=$\sqrt{3}$

Seventh term ${{{a}}_{7}}$can be calculated as follows,

${{{a}}_{7}}={a}{{{r}}^{6}}$

$\Rightarrow {{{a}}_{7}}=1{{(\sqrt{3})}^{6}}$

$\Rightarrow {{{a}}_{7}}=27$

4.Find the ${{{8}}^{{th}}}$term of the sequence $\dfrac{{3}}{{4}}{,1}\dfrac{{1}}{{2}}{,3,}.......$

Ans: First term, a=$\dfrac{3}{4}$

Common ratio, ${r}=\dfrac{\dfrac{3}{2}}{\dfrac{3}{4}}=\dfrac{3}{2}\times \dfrac{4}{3}=2$

Eight term ${{{a}}_{8}}$ a can be calculated as follows,

${{{a}}_{8}}={a}{{{r}}^{7}}$

$\Rightarrow {{{a}}_{8}}=\dfrac{3}{4}\times {{(2)}^{7}}$

$\Rightarrow {{{a}}_{8}}=96$

5. Find the ${1}{{{0}}^{{th}}}$ term of the G.P, ${12,4,1}\dfrac{{1}}{{3}}{,}.......$

Ans: First term, a=12

Common ratio, r=$\dfrac{4}{12}=\dfrac{1}{3}$

Tenth term ${{{a}}_{10}}$can be calculated as follows,

${{{a}}_{10}}={a}{{{r}}^{9}}$

$\Rightarrow {{{a}}_{10}}=12{{(\dfrac{1}{3})}^{9}}$

$\Rightarrow {{{a}}_{10}}=\dfrac{4}{6561}$

6. Find the ${{\mathbf{n}}^{\mathbf{th}}}$term of the series 1, 2, 4, 8, .......

Ans: First term, a=1

Common ratio, r=2

${{{N}}^{{th}}}$term ${{{a}}_{{n}}}$can be calculated as following

${{{a}}_{{n}}}={a}.{{{r}}^{\left( {n}-1 \right)}}$

$\Rightarrow {{{a}}_{{n}}}={{1.2}^{\left( {n}-1 \right)}}$

$\Rightarrow {{{a}}_{{n}}}={{2}^{{n}-1}}$

7. Find the next three terms of the sequence $\sqrt{{5}}{,5,5}\sqrt{{5}}{,}.......$

Ans: First term, ${a}=\sqrt{5}$

Common ratio, ${r}=\sqrt{5}$

Fourth term ${{{a}}_{4}}$can be calculated as follows

${{{a}}_{4}}={a}{{{r}}^{3}}$

$\Rightarrow {{{a}}_{4}}=\sqrt{5}{{(\sqrt{5})}^{3}}$

$\Rightarrow {{{a}}_{4}}=25$

Fifth term ${{{a}}_{5}}$can be calculated as follows

${{{a}}_{5}}={a}{{{r}}^{4}}$

$\Rightarrow {{{a}}_{5}}=\sqrt{5}{{(\sqrt{5})}^{4}}$

$\Rightarrow {{{a}}_{5}}=25\sqrt{5}$

Sixth term ${{{a}}_{6}}$can be calculated as follows

${{{a}}_{6}}={a}{{{r}}^{5}}$

$\Rightarrow {{{a}}_{6}}=\sqrt{5}{{(\sqrt{5})}^{5}}$

$\Rightarrow {{{a}}_{6}}=125$

Therefore the next three terms are $25,25\sqrt{5},125.$

8. Find the sixth term of the series, ${{{2}}^{{2}}}{,}{{{2}}^{{3}}}{,}{{{2}}^{{4}}}{,}.......$

Ans: First term, a=4

Common ratio, r=2 

Sixth term ${{{a}}_{6}}$can be calculated as follows,

${{{a}}_{6}}={a}{{{r}}^{5}}$

$\Rightarrow {{{a}}_{6}}=4{{(2)}^{5}}$

$\Rightarrow {{{a}}_{6}}=128$

9. Find the seventh term of the G.P $\sqrt{{3}}{+1,1,}\dfrac{\sqrt{{3}}{-1}}{{2}}{,}.......$

Ans: First term, ${a}=\sqrt{3}+1$

Common ratio, ${r}=\dfrac{1}{\sqrt{3}+1}$

Seventh term ${{{a}}_{7}}$can be calculated as follows

${{{a}}_{7}}={a}{{{r}}^{6}}$

$\Rightarrow {{{a}}_{7}}=\left( \sqrt{3}+1 \right){{(\dfrac{1}{\sqrt{3}+1})}^{6}}$

$\Rightarrow {{{a}}_{7}}={{(\dfrac{1}{\sqrt{3}+1})}^{5}}$

10. Find the G.P whose first term is 64 and the next term is 32.

Ans: First term, ${a}=64$

Second term, ${{{a}}_{2}}=32$

Common ratio, ${r}=\dfrac{32}{64}=\dfrac{1}{2}$

Third term ${{{a}}_{3}}$can be calculated as follows

${{{a}}_{3}}={a}{{{r}}^{2}}$

$\Rightarrow {{{a}}_{3}}=64{{(\dfrac{1}{2})}^{2}}$

$\Rightarrow {{{a}}_{3}}=16$

Fourth term ${{{a}}_{4}}$can be calculated as follows

${{{a}}_{4}}={a}{{{r}}^{3}}$

$\Rightarrow {{{a}}_{4}}=64{{(\dfrac{1}{2})}^{3}}$

$\Rightarrow {{{a}}_{4}}=8$

Fifth term ${{{a}}_{5}}$can be calculated as follows

${{{a}}_{5}}={a}{{{r}}^{4}}$

$\Rightarrow {{{a}}_{5}}=64{{(\dfrac{1}{2})}^{4}}$

$\Rightarrow {{{a}}_{5}}=4$

Therefore the G.P is 64, 32, 16, 8, 4, .......

11.Find the next three terms of the series $\dfrac{{2}}{{27}}{,}\dfrac{{2}}{{9}}{,}\dfrac{{2}}{{3}}{,}.......$

Ans: First term, ${a}=\dfrac{2}{27}$

Common ratio, ${r}=\dfrac{\dfrac{2}{9}}{\dfrac{2}{27}}=\dfrac{2}{9}\times \dfrac{27}{2}=3$

Fourth term ${{{a}}_{4}}$can be calculated as follows

${{{a}}_{4}}={a}{{{r}}^{3}}$

$\Rightarrow {{{a}}_{4}}=\dfrac{2}{27}{{(3)}^{3}}$

$\Rightarrow {{{a}}_{4}}=2$

Fifth term ${{{a}}_{5}}$can be calculated as follows

${{{a}}_{5}}={a}{{{r}}^{4}}$

$\Rightarrow {{{a}}_{5}}=\dfrac{2}{27}{{(3)}^{4}}$

$\Rightarrow {{{a}}_{5}}=6$

Sixth term ${{{a}}_{6}}$can be calculated as follows

${{{a}}_{6}}={a}{{{r}}^{5}}$

$\Rightarrow {{{a}}_{6}}=\dfrac{2}{27}{{(3)}^{5}}$

$\Rightarrow {{{a}}_{6}}=18$

Therefore the next three terms are 2, 6, 18.

12. Find the next two terms of the series ${2,-6,18,-54}$

Ans: First term, ${a}=2$

Common ratio, ${r}=\dfrac{-6}{2}=-3$

Fifth term ${{{a}}_{5}}$can be calculated as follows

${{{a}}_{5}}={a}{{{r}}^{4}}$

$\Rightarrow {{{a}}_{5}}=2{{(-3)}^{4}}$

$\Rightarrow {{{a}}_{5}}=162$

Sixth term ${{{a}}_{6}}$can be calculated as follows

${{{a}}_{6}}={a}{{{r}}^{5}}$

$\Rightarrow {{{a}}_{6}}=2{{(-3)}^{5}}$

$\Rightarrow {{{a}}_{6}}=-486$

Therefore the next two terms are 162 and -486.

Exercise 11B

1. Which term of the G.P ${-10,}\dfrac{{5}}{\sqrt{{3}}}{,}\dfrac{{-5}}{{6}}{,}.........$is $\dfrac{{-5}}{{72}}$?

Ans: First term, ${a}=-10$

Common ratio, ${r}=\dfrac{\dfrac{5}{\sqrt{3}}}{-10}=\dfrac{5}{\sqrt{3}}\times \dfrac{-1}{10}=\dfrac{-1}{2\sqrt{3}}$

Given ${{{a}}_{{n}}}=\dfrac{-5}{72}$ and we have to find n.

We know that ${{{a}}_{{n}}}={a}{{{r}}^{{n}-1}}$

$\Rightarrow \dfrac{-5}{72}=\left( -10 \right).{{(\dfrac{-1}{2\sqrt{3}})}^{{n}-1}}$

$\Rightarrow \dfrac{-5}{72}\times \dfrac{-1}{10}={{(\dfrac{-1}{2\sqrt{3}})}^{{n}-1}}$

$\Rightarrow \dfrac{1}{144}={{(\dfrac{-1}{2\sqrt{3}})}^{{n}-1}}$

$\Rightarrow \dfrac{1}{144}={{(\dfrac{-1}{2\sqrt{3}})}^{{n}}}\times (-2\sqrt{3)}$            \[ \left( \text{Since }{a^{n-1}}=\dfrac{{{{a}}^{{n}}}}{{a}} \right) \]

$\Rightarrow \dfrac{1}{144}\times \dfrac{-1}{2\sqrt{3}}={{(\dfrac{-1}{2\sqrt{3}})}^{{n}}}$

$\Rightarrow {{(\dfrac{1}{-2\sqrt{3}})}^{4}}\times \dfrac{-1}{2\sqrt{3}}={{(\dfrac{-1}{2\sqrt{3}})}^{{n}}}$

$\Rightarrow {{(\dfrac{-1}{2\sqrt{3}})}^{5}}={{(\dfrac{-1}{2\sqrt{3}})}^{{n}}}$

By comparing, we get

${n}=5$

Therefore the fifth term of the given series is $\dfrac{-5}{72}.$

2. The fifth term of a G.P is 81 and its second term is 24. Find the geometric progression.

Ans: Given fifth term, ${{{a}}_{5}}=81$and second term, ${{{a}}_{2}}=24$

We know that ${{{a}}_{5}}={a}{{{r}}^{4}}=81$and ${{{a}}_{2}}={a}{{{r}}^{1}}=24$

$\dfrac{{{{a}}_{5}}}{{{{a}}_{2}}}=\dfrac{{a}{{{r}}^{4}}}{{ar}}=\dfrac{81}{24}$

$\Rightarrow {{{r}}^{3}}=\dfrac{27}{8}$

$\Rightarrow {r}=\dfrac{3}{2}$

Substituting the value of r in ${{{a}}_{2}}$

${ar}=24$

$\Rightarrow {a}.\dfrac{3}{2}=24$

$\Rightarrow {a}=16$

First term, ${a}=16$

Second term, ${{{a}}_{2}}=24$

Third term, ${{{a}}_{3}}={a}{{{r}}^{2}}=16\times {{(\dfrac{3}{2})}^{2}}=36$

Fourth term, ${{{a}}_{4}}={a}{{{r}}^{3}}=16\times {{(\dfrac{3}{2})}^{3}}=54$

Therefore the G.P is $16,24,36,54,81,.......$

3. Fourth and seventh terms of a G.P are $\dfrac{{1}}{{18}}$and $\dfrac{{-1}}{{486}}$respectively. Find the G.P.

Ans: Given fourth term, ${{{a}}_{4}}=\dfrac{1}{18}$and seventh term, ${{{a}}_{7}}=\dfrac{-1}{486}$

We know that ${{{a}}_{4}}={a}{{{r}}^{3}}=\dfrac{1}{18}$and ${{{a}}_{7}}={a}{{{r}}^{6}}=\dfrac{-1}{486}$

$\dfrac{{{{a}}_{4}}}{{{{a}}_{7}}}=\dfrac{{a}{{{r}}^{3}}}{{a}{{{r}}^{6}}}=\dfrac{1}{18}\times -486$

$\Rightarrow {{{r}}^{3}}=\dfrac{-1}{27}$

$\Rightarrow {r}=\dfrac{-1}{3}$

Substituting the value of r in ${{{a}}_{4}}$

${a}{{{r}}^{3}}=\dfrac{1}{18}$

$\Rightarrow {a}.\dfrac{-1}{27}=\dfrac{1}{18}$

$\Rightarrow {a}=\dfrac{-3}{2}$

First term, ${a}=\dfrac{-3}{2}$

Second term, ${{{a}}_{2}}={ar}=\dfrac{1}{2}$

Third term, ${{{a}}_{3}}={a}{{{r}}^{2}}=\dfrac{-1}{6}$

Fourth term, ${{{a}}_{4}}=\dfrac{1}{18}$

Fifth term, ${{{a}}_{5}}=\dfrac{-1}{54}$

Sixth term, ${{{a}}_{6}}=\dfrac{1}{162}$

Seventh term, ${{{a}}_{7}}=\dfrac{-1}{486}$

Therefore the G.P is $\dfrac{-3}{2},\dfrac{1}{2},\dfrac{-1}{6},\dfrac{1}{18},\dfrac{-1}{54},\dfrac{1}{162},\dfrac{-1}{486},.......$

4. If the first and third term of a G.P are 2 and 8 respectively, find its second term.

Ans: Given first term, ${a}=2$ and third term, ${{{a}}_{3}}=8$

We know that ${{{a}}_{3}}={a}{{{r}}^{2}}$

$\Rightarrow 8=2{{({r})}^{2}}$

$\Rightarrow {{{r}}^{2}}=4$

$\Rightarrow {r}=2$

Second term, ${{{a}}_{2}}={ar}$

$\Rightarrow {{{a}}_{2}}=2\left( 2 \right)=4$

Therefore the second term is equal to 4.

5. The product of ${{{3}}^{{rd}}}$and ${{{8}}^{{th}}}$terms of a G.P is 243. If it's ${{{4}}^{{th}}}$term is 3, find its ${{{7}}^{{th}}}$term.

Ans: We know that ${{3}^{{rd}}}$term ${{{a}}_{3}}={a}{{{r}}^{2}}$, ${{8}^{{th}}}$term ${{{a}}_{8}}={a}{{{r}}^{7}}$and ${{4}^{{th}}}$term ${{{a}}_{4}}={a}{{{r}}^{3}}$

Given ${{{a}}_{3}}\times {{{a}}_{8}}=243$

$\Rightarrow {a}{{{r}}^{2}}\times {a}{{{r}}^{7}}=243$

$\Rightarrow {{{a}}^{2}}{{{r}}^{9}}=243$

We know that ${{{a}}_{4}}={a}{{{r}}^{3}}=3$

$\Rightarrow {a}=\dfrac{3}{{{{r}}^{3}}}$

Substituting the value of ${a}$in ${{{a}}^{2}}{{{r}}^{9}}$

$\Rightarrow {{(\dfrac{3}{{{{r}}^{3}}})}^{2}}{{{r}}^{9}}=243$

$\Rightarrow \dfrac{9}{{{{r}}^{6}}}.{{{r}}^{9}}=243$

$\Rightarrow {{{r}}^{3}}=27$

$\Rightarrow {r}=3$

Substituting the value of ${r}$in ${{{a}}_{4}}$

$\Rightarrow {a}{{{r}}^{3}}=3$

$\Rightarrow {a}=\dfrac{1}{9}$

${{7}^{{th}}}$term ${{{a}}_{7}}={a}{{{r}}^{6}}$

$\Rightarrow {{{a}}_{7}}=\dfrac{1}{9}{{(3)}^{6}}$

$\Rightarrow {{{a}}_{7}}=81$

Therefore the seventh term is equal to 81.

6. Find the geometric progression with fourth term = 54 and seventh term = 1458.

Ans: Given ${{4}^{{th}}}$term ${{{a}}_{4}}={a}{{{r}}^{3}}=54$and ${{7}^{{th}}}$term ${{{a}}_{7}}={a}{{{r}}^{6}}=1458$

$\dfrac{{{{a}}_{7}}}{{{{a}}_{4}}}=\dfrac{{a}{{{r}}^{6}}}{{a}{{{r}}^{3}}}$

$\Rightarrow {{{r}}^{3}}=\dfrac{1458}{54}$

$\Rightarrow {{{r}}^{3}}=27$

$\Rightarrow {r}=3$

Substituting the value of ${r}$in ${{{a}}_{4}}$

${{{a}}_{4}}=54$

$\Rightarrow {a}{{{r}}^{3}}=54$

$\Rightarrow {a}{{(3)}^{3}}=54$

$\Rightarrow {a}=2$

First term, ${a}=2$

Second term, ${{{a}}_{2}}={ar}=6$

Third term, ${{{a}}_{3}}={a}{{{r}}^{2}}=18$

Fourth term, ${{{a}}_{4}}={a}{{{r}}^{3}}=54$

Fifth term, ${{{a}}_{5}}={a}{{{r}}^{4}}=162$

Sixth term, ${{{a}}_{6}}={a}{{{r}}^{5}}=486$

Seventh term, ${{{a}}_{7}}={a}{{{r}}^{6}}=1458$

Therefore the G.P is $2,6,18,54,162,486,1458,.......$

7. Second term of a geometric progression is 6 and its fifth term is 9 times more than its third term. Find the geometric progression. Consider that all the terms of the G.P are positive.

Ans: Given ${{{a}}_{2}}=6$ and ${{{a}}_{5}}=9\left( {{{a}}_{3}} \right)$

${{{a}}_{2}}={ar}=6$

${{{a}}_{5}}=9\left( {{{a}}_{3}} \right)$

$\Rightarrow {a}{{{r}}^{4}}=9\left( {a}{{{r}}^{2}} \right)$

$\Rightarrow {{{r}}^{2}}=9$

$\Rightarrow {r}=3$

Substituting ${r}$in ${{{a}}_{2}}$

$\Rightarrow {a}\left( 3 \right)=6$

$\Rightarrow {a}=2$

First term, ${a}=2$

Second term, ${{{a}}_{2}}={ar}=6$

Third term, ${{{a}}_{3}}={a}{{{r}}^{2}}=18$

Fourth term, ${{{a}}_{4}}={a}{{{r}}^{3}}=54$

Fifth term, ${{{a}}_{5}}={a}{{{r}}^{4}}=162$

Therefore the G.P is $2,6,18,54,162,.......$

8. The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.

Ans: Given ${{{a}}_{4}}=10,{{{a}}_{7}}=80$and ${{{a}}_{{n}}}=2560$

$\dfrac{{{{a}}_{7}}}{{{{a}}_{4}}}=\dfrac{80}{10}$

$\Rightarrow \dfrac{{a}{{{r}}^{6}}}{{a}{{{r}}^{3}}}=8$

$\Rightarrow {{{r}}^{3}}=8$

$\Rightarrow {r}=2$

Substitute the value of ${r}$in ${{{a}}_{4}}$

${{{a}}_{4}}={a}{{{r}}^{3}}=10$

$\Rightarrow {a}{{(2)}^{3}}=10$

$\Rightarrow {a}=\dfrac{5}{4}$

Given ${{{a}}_{{n}}}=2560$

$\Rightarrow {a}{{{r}}^{{n}-1}}=2560$

$\Rightarrow \left( \dfrac{5}{4} \right){{(2)}^{{n}-1}}=2560$

$\Rightarrow {{2}^{{n}-1}}=2560\times \dfrac{4}{5}$

$\Rightarrow {{2}^{{n}-1}}=2048$

$\Rightarrow {{2}^{{n}-1}}={{2}^{11}}$

$\Rightarrow {n}-1=11$

$\Rightarrow {n}=12$

Therefore first term is $\dfrac{5}{4}$, common ratio is 2 and the total number of terms in the sequence is 12.

9. If the fourth and ninth terms of a G.P are 54 and 13122 respectively, find the G.P. Also find its general term.

Ans: Given ${{{a}}_{4}}=54$and ${{{a}}_{9}}=13122$

$\dfrac{{{{a}}_{9}}}{{{{a}}_{4}}}=\dfrac{13122}{54}$

$\Rightarrow \dfrac{{a}{{{r}}^{8}}}{{a}{{{r}}^{3}}}=243$

$\Rightarrow {{{r}}^{5}}={{3}^{5}}$

$\Rightarrow {r}=3$

Substituting ${r}$value in ${{{a}}_{4}}$

$\Rightarrow {a}{{{r}}^{3}}=54$

$\Rightarrow {a}{{(3)}^{3}}=54$

$\Rightarrow {a}=2$

First term, ${a}=2$

Second term, ${{{a}}_{2}}={ar}=6$

Third term, ${{{a}}_{3}}={a}{{{r}}^{2}}=18$

Fourth term, ${{{a}}_{4}}={a}{{{r}}^{3}}=54$

Fifth term, ${{{a}}_{5}}={a}{{{r}}^{4}}=162$

Sixth term, ${{{a}}_{6}}={a}{{{r}}^{5}}=486$

G.P is 2, 6, 18, 54, 162, 486, .......

General term, ${{{a}}_{{n}}}={a}{{{r}}^{{n}-1}}$

$\Rightarrow {{{a}}_{{n}}}=2{{(3)}^{{n}-1}}$

10. The fifth, eight and eleventh terms of a geometric progression are p, q and r respectively. Show that ${{\mathbf{q}}^{2}}=\mathbf{pr}.$

Ans: Fifth term, ${{{a}}_{5}}={a}{{{r}}^{4}}={p}$

Eight term, ${{{a}}_{8}}={a}{{{r}}^{7}}={q}$

Eleventh term, ${{{a}}_{11}}={a}{{{r}}^{10}}={r}$

${{{q}}^{2}}={{({a}{{{r}}^{7}})}^{2}}={{{a}}^{2}}{{{r}}^{14}}$

${pr}={a}{{{r}}^{4}}\times {a}{{{r}}^{10}}$

$\Rightarrow {pr}={{{a}}^{2}}{{{r}}^{14}}$

Hence proved.

Exercise 11C

1. Find the seventh term from the end of the series $\sqrt{{2}}{,2,2}\sqrt{{2}}{,}..........{,32}{.}$

Ans: We can write the given G.P as $32,..........,2\sqrt{2},2,\sqrt{2}$

Now the first term, ${a}=32$

Common ratio, ${r}=\dfrac{1}{\sqrt{2}}$

Seventh term from the end will be the seventh term from the beginning for the new series($32,..........,2\sqrt{2},2,\sqrt{2}$)

Now we know that ${{{a}}_{7}}={a}{{{r}}^{6}}$

$\Rightarrow {{{a}}_{7}}=32{{(\dfrac{1}{\sqrt{2}})}^{6}}$

$\Rightarrow {{{a}}_{7}}=32\times \dfrac{1}{8}$

$\Rightarrow {{{a}}_{7}}=4$

Therefore the seventh term from the end of the series $\sqrt{2},2,2\sqrt{2},..........,32$ is 4.

2. Find the third term from the end of the G.P $\dfrac{{2}}{{27}}{,}\dfrac{{2}}{{9}}{,}\dfrac{{2}}{{3}}{,}.......{,162}$

Ans: We can write the given G.P as $162,..........,\dfrac{2}{3},\dfrac{2}{9},\dfrac{2}{27}$

Now the first term, ${a}=162$

Common ratio, ${r}=\dfrac{1}{3}$

Third term from the end will be the third term from the beginning for the new series($162,..........,\dfrac{2}{3},\dfrac{2}{9},\dfrac{2}{27}$)

Now we know that ${{{a}}_{3}}={a}{{{r}}^{2}}$

$\Rightarrow {{{a}}_{3}}=162{{(\dfrac{1}{3})}^{2}}$

$\Rightarrow {{{a}}_{3}}=162\times \dfrac{1}{9}$

$\Rightarrow {{{a}}_{3}}=18$

Therefore the third term from the end of the series $\dfrac{2}{27},\dfrac{2}{9},\dfrac{2}{3},.......,162$ is 18.

3. For the G.P $\dfrac{{1}}{{27}}{,}\dfrac{{1}}{{9}}{,}\dfrac{{1}}{{3}}{,}.......{,81;}$find the product of the fourth term from the beginning and the fourth term from the end.

Ans: First term, ${a}=\dfrac{1}{27}$

Common ratio, ${r}=3$

Fourth term from the beginning, ${{{a}}_{4}}={a}{{{r}}^{3}}$

$\Rightarrow {{{a}}_{4}}=\dfrac{1}{27}{{(3)}^{3}}=1$

Given the final term ${{{a}}_{{n}}}=81$

$\Rightarrow {a}{{{r}}^{{n}-1}}=81$

$\Rightarrow \dfrac{1}{27}{{(3)}^{{n}-1}}=81$

$\Rightarrow {{3}^{{n}-1}}={{3}^{4}}\times {{3}^{3}}$

$\Rightarrow {{3}^{{n}}}={{3}^{4}}\times {{3}^{4}}$

$\Rightarrow {n}=8$

So there are a total of 8 terms in the series

Fourth term from the end will be the fifth term

${{{a}}_{5}}={a}{{{r}}^{4}}$

$\Rightarrow {{{a}}_{5}}=\dfrac{1}{27}{{(3)}^{4}}$

$\Rightarrow {{{a}}_{5}}=3$

Therefore the product of the fourth term from the beginning and from the end is ${{{a}}_{4}}\times {{{a}}_{5}}=1\times 3=3$

4. If for a G.P ${{\mathbf{p}}^{\mathbf{th}}},{{\mathbf{q}}^{\mathbf{th}}}$and ${{\mathbf{r}}^{\mathbf{th}}}$terms are a,b and c respectively; prove that:

$\left( \mathbf{q}-\mathbf{r} \right)\mathbf{log}~\mathbf{a}+\left( \mathbf{r}-\mathbf{p} \right)\mathbf{log}~\mathbf{b}+\left( \mathbf{p}-\mathbf{q} \right)\mathbf{log}~\mathbf{c}=0$

Ans: ${{{p}}^{{th}}}$term is a

$\Rightarrow {{{A}}_{{p}}}={A}{{{R}}^{{p}-1}}={a}$

${{{q}}^{{th}}}$term is b

$\Rightarrow {{{A}}_{{q}}}={A}{{{R}}^{{q}-1}}={b}$.

${{{r}}^{{th}}}$term is c

$\Rightarrow {{{A}}_{{r}}}={A}{{{R}}^{{r}-1}}={c}$

${{{a}}^{{q}-{r}}}\times {{{b}}^{{r}-{p}}}\times {{{c}}^{{p}-{q}}}={{({A}{{{R}}^{{p}-1}})}^{{q}-{r}}}\times {{({A}{{{R}}^{{q}-1}})}^{{r}-{p}}}\times {{({A}{{{R}}^{{r}-1}})}^{{p}-{q}}}$

$\Rightarrow {{{a}}^{{q}-{r}}}\times {{{b}}^{{r}-{p}}}\times {{{c}}^{{p}-{q}}}={{{A}}^{{q}-{r}+{r}-{p}+{p}-{q}}}{{{R}}^{\left( {p}-1 \right)\left( {q}-{r} \right)+\left( {q}-1 \right)\left( {r}-{p} \right)+\left( {r}-1 \right)\left( {p}-{q} \right)}}$

$\Rightarrow {{{a}}^{{q}-{r}}}\times {{{b}}^{{r}-{p}}}\times {{{c}}^{{p}-{q}}}={{{A}}^{0}}{{{R}}^{{pq}-{pr}-{q}+{r}+{qr}-{pq}-{r}+{p}+{rp}-{rq}-{p}+{q}}}$

$\Rightarrow {{{a}}^{{q}-{r}}}\times {{{b}}^{{r}-{p}}}\times {{{c}}^{{p}-{q}}}={{{A}}^{0}}{{{R}}^{0}}$

$\Rightarrow {{{a}}^{{q}-{r}}}\times {{{b}}^{{r}-{p}}}\times {{{c}}^{{p}-{q}}}=1$

Applying log on both sides

$\Rightarrow {log}\left( {{{a}}^{{q}-{r}}}\times {{{b}}^{{r}-{p}}}\times {{{c}}^{{p}-{q}}} \right)={log}\left( 1 \right)$

$\Rightarrow {log}\left( {{{a}}^{{q}-{r}}} \right)+{log}\left( {{{b}}^{{r}-{p}}} \right)+{log}\left( {{{c}}^{{p}-{q}}} \right)=0$

\[ \Rightarrow (q-r)log a + (r-p)log b +(p-q)log c = 0 \]

5. If a,b and c are in G.P, prove that: $\mathbf{log}~\mathbf{a},~\mathbf{log}~\mathbf{b}$ and $\mathbf{log}~\mathbf{c}$ are in A.P.

Ans: Let the first term be ${a}={A}$

Second term be ${b}={AR}$

Third term be ${c}={A}{{{R}}^{2}}$

Now \[ log a , log b ,log c \] will become

\[\text{log A, log (AR), log (AR^{2})}\]

Let \[ log A \]be x, ${log}\left( {AR} \right)$be y and ${log}\left( {A}{{{R}}^{2}} \right)$be z.

We know that in an A.P

${d}={y}-{x}$

$\Rightarrow {d}={log}\left( {AR} \right)-{log}\left( {A} \right)$

$\Rightarrow {d}={log}\left( {A} \right)+{log}\left( {R} \right)-{log}\left( {A} \right)$

$\Rightarrow {d}={log}\left( {R} \right)$

${d}={z}-{y}$

$\Rightarrow {d}={log}\left( {A}{{{R}}^{2}} \right)-{log}\left( {AR} \right)$

$\Rightarrow {d}={log}\left( {A} \right)+2{log}\left( {R} \right)-{log}\left( {A} \right)-{log}\left( {R} \right)$

$\Rightarrow {d}={log}\left( {R} \right)$

Therefore ${y}-{x}={z}-{y}$

Therefore \[ \text{log a , log b and log c} \]

6. If each term of the G.P is raised to the power $\mathbf{x},$show that the resulting sequence is also a G.P.

Ans: Let the G.P be ${a},{ar},{a}{{{r}}^{2}},........$

If the terms are raised to a power x, new series will be

${{{a}}^{{x}}},{{({ar})}^{{x}}},{{({a}{{{r}}^{2}})}^{{x}}},.........$

Common ratio, ${r}=\dfrac{{{({ar})}^{{x}}}}{{{{a}}^{{x}}}}={{{r}}^{{x}}}$ for the first and second term

Common ratio, ${r}=\dfrac{{{({a}{{{r}}^{2}})}^{{x}}}}{{{\left( {ar} \right)}^{{x}}}}=\dfrac{{{{a}}^{{x}}}{{{r}}^{2{x}}}}{{{{a}}^{{x}}}{{{r}}^{{x}}}}={{{r}}^{{x}}}$ for the second and third term

Common ratio is the same for all the terms

So the series will be in G.P even if all the terms are raised to the power x.

7. If a,b,c are in A.P, a,x,b are in G.P, whereas b,y,c are also in G.P. Show that: ${{\mathbf{x}}^{2}},{{\mathbf{b}}^{2}},{{\mathbf{y}}^{2}}$ are in A.P.

Ans: If a,b,c are in A.P then $2{b}={a}+{c}$

If a,x,b are in G.P then ${{{x}}^{2}}={ab}$

If b,y,c are in G.P then ${{{y}}^{2}}={bc}$

For ${{{x}}^{2}},{{{b}}^{2}},{{{y}}^{2}}$to be in A.P

$2{{{b}}^{2}}={{{x}}^{2}}+{{{y}}^{2}}$

Consider ${{{x}}^{2}}+{{{y}}^{2}}$

$\Rightarrow {{{x}}^{2}}+{{{y}}^{2}}={ab}+{bc}$

$\Rightarrow {{{x}}^{2}}+{{{y}}^{2}}={b}\left( {a}+{c} \right)$

$\Rightarrow {{{x}}^{2}}+{{{y}}^{2}}={b}\left( 2{b} \right)$

$\Rightarrow {{{x}}^{2}}+{{{y}}^{2}}=2{{{b}}^{2}}$

Hence proved.

8. If a,b,c are in G.P and a,x,b,y,c are in A.P. Prove that

(i) $\dfrac{1}{\mathbf{x}}+\dfrac{1}{\mathbf{y}}=\dfrac{2}{\mathbf{b}}$       (ii) $\dfrac{\mathbf{a}}{\mathbf{x}}+\dfrac{\mathbf{c}}{\mathbf{y}}=2$

Ans: If x,b,y, are in G.P, then ${{{b}}^{2}}={ac}\Rightarrow {c}=\dfrac{{{{b}}^{2}}}{{a}}$

If a,x,b,y,c are in A.P, then $2{x}={a}+{b}$

$\Rightarrow {x}=\dfrac{{a}+{b}}{2}$ and

$2{y}={b}+{c}\Rightarrow {y}=\dfrac{{b}+{c}}{2}$

(i) $\dfrac{{1}}{{x}}{+}\dfrac{{1}}{{y}}{=}\dfrac{{2}}{{a+b}}{+}\dfrac{{2}}{{b+c}}$

$\Rightarrow \dfrac{1}{{x}}+\dfrac{1}{{y}}=\dfrac{2}{{a}+{b}}+\dfrac{2}{{b}+\dfrac{{{{b}}^{2}}}{{a}}}$

$\Rightarrow \dfrac{1}{{x}}+\dfrac{1}{{y}}=\dfrac{2}{{a}+{b}}+\dfrac{2{a}}{{ab}+{{{b}}^{2}}}$

$\Rightarrow \dfrac{1}{{x}}+\dfrac{1}{{y}}=\dfrac{2}{{a}+{b}}\left( 1+\dfrac{{a}}{{b}} \right)$

$\Rightarrow \dfrac{1}{{x}}+\dfrac{1}{{y}}=\dfrac{2}{{a}+{b}}\left( \dfrac{{b}+{a}}{{b}} \right)$

$\Rightarrow \dfrac{1}{{x}}+\dfrac{1}{{y}}=\dfrac{2}{{b}}$

(ii) $\dfrac{{a}}{{x}}+\dfrac{{c}}{{y}}=\dfrac{{a}}{\dfrac{{a}+{b}}{2}}+\dfrac{{c}}{\dfrac{{b}+{c}}{2}}$

$\Rightarrow \dfrac{{a}}{{x}}+\dfrac{{c}}{{y}}=\dfrac{2{a}}{{a}+{b}}+\dfrac{2{c}}{{b}+{c}}$

$\Rightarrow \dfrac{{a}}{{x}}+\dfrac{{c}}{{y}}=\dfrac{2\dfrac{{{{b}}^{2}}}{{c}}}{\dfrac{{{{b}}^{2}}}{{c}}+{b}}+\dfrac{2{c}}{{b}+{c}}$

$\Rightarrow \dfrac{{a}}{{x}}+\dfrac{{c}}{{y}}=\dfrac{2{{{b}}^{2}}}{{{{b}}^{2}}+{bc}}+\dfrac{2{c}}{{b}+{c}}$

$\Rightarrow \dfrac{{a}}{{x}}+\dfrac{{c}}{{y}}=\dfrac{2{{{b}}^{{}}}}{{{{b}}^{{}}}+{c}}+\dfrac{2{c}}{{b}+{c}}$

$\Rightarrow \dfrac{{a}}{{x}}+\dfrac{{c}}{{y}}=\dfrac{2\left( {b}+{c} \right)}{{b}+{c}}$

$\Rightarrow \dfrac{{a}}{{x}}+\dfrac{{c}}{{y}}=2$

Hence proved.

9. If a,b,c are in A.P and also in G.P, show that: $\mathbf{a}=\mathbf{b}=\mathbf{c}.$

Ans: If ${a},{b},{c}$ are in A.P then $2{b}={a}+{c}\Rightarrow {b}=\dfrac{{a}+{c}}{2}$

If ${a},{b},{c}$ are in G.P then ${{{b}}^{2}}={ac}$

Substitute b value in  ${{{b}}^{2}}={ac}$

$\Rightarrow {{(\dfrac{{a}+{c}}{2})}^{2}}={ac}$

$\Rightarrow {{{a}}^{2}}+{{{c}}^{2}}+2{ac}=4{ac}$

$\Rightarrow {{{a}}^{2}}+{{{c}}^{2}}-2{ac}=0$

$\Rightarrow {{({a}-{c})}^{2}}=0$

$\Rightarrow {a}={c}$

Substitute ${a}={c}$ in $2{b}={a}+{c}$

$\Rightarrow 2{b}={c}+{c}$

$\Rightarrow {b}={c}$

Therefore ${a}={b}={c}$

Exercise 11D

1. Find the sum of G.P

(i) ${1+3+9+27+}........\text{12 }{ terms}$

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

Here a=1, r=3 and n=12

${{{S}}_{{n}}}=\dfrac{1\left( {{3}^{12}}-1 \right)}{3-1}$

$\Rightarrow {{{S}}_{{n}}}=\dfrac{{{3}^{12}}-1}{2}$

(ii) \[{0}{.3+0}{.03+0}{.003+}.......\text{ 8 } terms\]

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

Here a=0.3, r=0.1 and n=8

${{{S}}_{{n}}}=\dfrac{0.3\left( {{0.1}^{8}}-1 \right)}{0.1-1}$

$\Rightarrow {{{S}}_{{n}}}=\dfrac{0.3\left( 1-{{0.1}^{8}} \right)}{0.9}$

(iii) \[{1-}\dfrac{{1}}{{2}}{+}\dfrac{{1}}{{4}}{-}\dfrac{{1}}{{8}}{+}........\text{9 } terms \]

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

Here a=1, r=$\dfrac{-1}{2}$ and n=9

${{{S}}_{{n}}}=\dfrac{1\left( {{\dfrac{-1}{2}}^{9}}-1 \right)}{\dfrac{-1}{2}-1}$

$\Rightarrow {{{S}}_{{n}}}=\dfrac{1+\dfrac{1}{{{2}^{9}}}}{\dfrac{3}{2}}$

$\Rightarrow {{{S}}_{{n}}}=\dfrac{{{2}^{9}}+1}{3\times {{2}^{8}}}$

(iv)\[{1-}\dfrac{{1}}{{3}}{+}\dfrac{{1}}{{{{3}}^{{2}}}}{-}\dfrac{{1}}{{{{3}}^{{3}}}}{+}.......\text{ n } terms \]

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

Here a=1, r=$\dfrac{-1}{3}$

${{{S}}_{{n}}}=\dfrac{1\left( {{\dfrac{-1}{3}}^{{n}}}-1 \right)}{\dfrac{-1}{3}-1}$

$\Rightarrow {{{S}}_{{n}}}=\dfrac{(1-\left( -\dfrac{1}{3}{{)}^{{n}}} \right)}{\dfrac{4}{3}}$

$\Rightarrow {{{S}}_{{n}}}=\dfrac{3}{4}(1-\left( -\dfrac{1}{3}{{)}^{{n}}} \right)$

(v) $\dfrac{{x+y}}{{x-y}}{+1+}\dfrac{{x-y}}{{x+y}}{+}........\text{n }{ terms}$

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

Here a=$\dfrac{{x}+{y}}{{x}-{y}}$, r=$\dfrac{{x}-{y}}{{x}+{y}}$

${{{S}}_{{n}}}=\dfrac{\dfrac{{x}+{y}}{{x}-{y}}(\left( \dfrac{{x}-{y}}{{x}+{y}}{{)}^{{n}}}-1 \right)}{\dfrac{{x}-{y}}{{x}+{y}}-1}$

(vi) $\sqrt{{3}}{+}\dfrac{{1}}{\sqrt{{3}}}{+}\dfrac{{1}}{{3}\sqrt{{3}}}{+}.......\text{n }{ terms}$

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

Here a=$\sqrt{3}$, r=$\dfrac{1}{3}$

${{{S}}_{{n}}}=\dfrac{\sqrt{3}\left( {{\dfrac{1}{3}}^{{n}}}-1 \right)}{\dfrac{1}{3}-1}$

$\Rightarrow {{{S}}_{{n}}}=\dfrac{\sqrt{3}(\left( \dfrac{1}{3}{{)}^{{n}}}-1 \right)}{\dfrac{-2}{3}}$

$\Rightarrow {{{S}}_{{n}}}=\dfrac{3\sqrt{3}}{2}(1-\left( \dfrac{1}{3}{{)}^{{n}}} \right)$

2. How many terms of the G.P ${1+4+16+64+}........$must be added to get a sum equal to 5461?

Ans: Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

a=1, r=4 and ${{{S}}_{{n}}}=5461$

$\Rightarrow 5461=\dfrac{1\left( {{4}^{{n}}}-1 \right)}{4-1}$

$\Rightarrow 5461\times 3={{4}^{{n}}}-1$

$\Rightarrow 16383+1={{4}^{{n}}}$

$\Rightarrow {{4}^{{n}}}=16384$

$\Rightarrow {{4}^{{n}}}={{4}^{7}}$

$\Rightarrow {n}=7$

Therefore 7 terms of the G.P must be added to get the sum 5461.

3. The first term of a G.P is 27 and its ${{{8}}^{{th}}}$term is $\dfrac{{1}}{{81}}{.}$ Find the sum of its first 10 terms.

Ans: Given first term, ${a}=27$and ${{{a}}_{8}}=\dfrac{1}{81}$

$\Rightarrow {a}{{{r}}^{7}}=\dfrac{1}{81}$

$\Rightarrow {{{r}}^{7}}=\dfrac{1}{{{3}^{4}}\times {{3}^{3}}}$

$\Rightarrow {r}=\dfrac{1}{3}$

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

${{{S}}_{10}}=\dfrac{27(\left( \dfrac{1}{3}{{)}^{10}}-1 \right)}{\dfrac{1}{3}-1}$

${{{S}}_{10}}=\dfrac{27\left( \dfrac{1-{{3}^{10}}}{{{3}^{10}}} \right)}{\dfrac{-2}{3}}$

$\Rightarrow {{{S}}_{10}}=27\times \left( \dfrac{1-{{3}^{10}}}{{{3}^{10}}} \right)\times \left( \dfrac{-3}{2} \right)$

$\Rightarrow {{{S}}_{10}}={{3}^{3}}\times \left( \dfrac{{{3}^{10}}-1}{{{3}^{9}}} \right)\times \left( \dfrac{1}{2} \right)$

$\Rightarrow {{{S}}_{10}}=\dfrac{{{3}^{10}}-1}{2\times {{3}^{6}}}$

4. A boy spends 10rs on the first day, 20rs on the second day, 40rs on the third day and so on. Find how much in all, will he spend in 12 days?

Ans: According to the given information, G.P will be

10,20,40,.......,12 terms

First term, ${a}=10$

Common ratio, ${r}=2$

${n}=12$

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

$\Rightarrow {{{S}}_{12}}=\dfrac{10\left( {{2}^{12}}-1 \right)}{2-1}$

$\Rightarrow {{{S}}_{12}}=10\left( 1024-1 \right)$

$\Rightarrow {{{S}}_{12}}=10230$

So, the boy will spend a total of 10230rs in 12 days.

5. The ${{{4}}^{{th}}}$ and ${{{7}}^{{th}}}$ terms of a G.P are $\dfrac{{1}}{{27}}$and $\dfrac{{1}}{{729}}$respectively. Find the sum of n terms of this G.P.

Ans: ${{4}^{{th}}}$term ${{{a}}_{4}}={a}{{{r}}^{3}}=\dfrac{1}{27}$ and

${{7}^{{th}}}$term ${{{a}}_{7}}={a}{{{r}}^{6}}=\dfrac{1}{729}$

$\dfrac{{a}{{{r}}^{6}}}{{a}{{{r}}^{3}}}=\dfrac{1}{729}\times 27$

$\Rightarrow {{{r}}^{3}}=\dfrac{1}{27}$

$\Rightarrow {r}=\dfrac{1}{3}$

Substitute r value in ${a}{{{r}}^{3}}=\dfrac{1}{27}$

$\Rightarrow {a}\times \dfrac{1}{{{3}^{3}}}=\dfrac{1}{27}$

$\Rightarrow {a}=1$

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

$\Rightarrow {{{S}}_{{n}}}=\dfrac{1(\left( \dfrac{1}{3}{{)}^{{n}}}-1 \right)}{\dfrac{1}{3}-1}$

$\Rightarrow {{{S}}_{{n}}}=\dfrac{1-{{3}^{{n}}}}{{{3}^{{n}}}}\times \dfrac{3}{-2}$

$\Rightarrow {{{S}}_{{n}}}=\dfrac{1-{{3}^{{n}}}}{{{3}^{{n}-1}}}\times \dfrac{1}{-2}$

$\Rightarrow {{{S}}_{{n}}}=\dfrac{{{3}^{{n}}}-1}{2\times {{3}^{{n}-1}}}$

6. A geometric progression has a common ratio=3 and last term=486. If the sum of its terms is 728; find its first term.

Ans: Given ${r}=3,{{{a}}_{{n}}}=486,{{{S}}_{{n}}}=728$

We know that ${{{a}}_{{n}}}={a}{{{r}}^{{n}-1}}=486$and ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

$\Rightarrow \dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}=728$

$\Rightarrow \dfrac{{a}{{{r}}^{{n}}}-{a}}{{r}-1}=728$

$\Rightarrow \dfrac{{a}{{{r}}^{{n}-1}}.{r}-{a}}{{r}-1}=728$

$\Rightarrow \dfrac{486\times 3-{a}}{3-1}=728$

$\Rightarrow 1458-{a}=1456$

$\Rightarrow {a}=2$

Therefore, the first term is equal to 2.

7. Find the sum of G.P: ${3,6,12,}........{1536}{.}$

Ans: First term, ${a}=3$

Common ratio, ${r}=2$

Given last term, ${{{a}}_{{n}}}=1536$

$\Rightarrow {a}{{{r}}^{{n}-1}}=1536$

$\Rightarrow 3\times {{2}^{{n}-1}}=1536$

$\Rightarrow {{2}^{{n}-1}}=512$

$\Rightarrow {{2}^{{n}-1}}={{2}^{9}}$

$\Rightarrow {n}-1=9$

$\Rightarrow {n}=10$

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

$\Rightarrow {{{S}}_{10}}=\dfrac{3\left( {{2}^{10}}-1 \right)}{2-1}$

$\Rightarrow {{{S}}_{10}}=3\left( 1024-1 \right)$

$\Rightarrow {{{S}}_{10}}=3069$

Therefore the sum of the G.P is 3069.

8. How many terms of the series ${2+6+18+}.......$must be taken to make the sum equal to 728?

Ans: From the given data

First term, ${a}=2$

Common ratio, ${r}=3$

Sum of n terms, ${{{S}}_{{n}}}=728$

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

$\Rightarrow 728=\dfrac{2\left( {{3}^{{n}}}-1 \right)}{3-1}$

$\Rightarrow 728={{3}^{{n}}}-1$

$\Rightarrow {{3}^{{n}}}=729$

$\Rightarrow {{3}^{{n}}}={{3}^{6}}$

$\Rightarrow {n}=6$

Therefore 6 terms from the series must be taken to make the sum equal to 728.

9. In a G.P, the ratio between the sum of the first three terms and that of the first six terms is 125:152. Find its common ratio.

Ans: Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

Sum of first 3 terms, ${{{S}}_{3}}=\dfrac{{a}\left( {{{r}}^{3}}-1 \right)}{{r}-1}$

Sum of first 6 terms, ${{{S}}_{6}}=\dfrac{{a}\left( {{{r}}^{6}}-1 \right)}{{r}-1}$

$\Rightarrow \dfrac{\dfrac{{a}\left( {{{r}}^{6}}-1 \right)}{{r}-1}}{\dfrac{{a}\left( {{{r}}^{3}}-1 \right)}{{r}-1}}=\dfrac{152}{125}$

$\Rightarrow \dfrac{{{{r}}^{6}}-1}{{{{r}}^{3}}-1}=\dfrac{152}{125}$

$\Rightarrow \dfrac{{{({{{r}}^{3}})}^{2}}-1}{{{{r}}^{3}}-1}=\dfrac{152}{125}$

$\Rightarrow \dfrac{({{{r}}^{3}}-1)\left( {{{r}}^{3}}+1 \right)}{{{{r}}^{3}}-1}=\dfrac{152}{125}$

$\Rightarrow {{{r}}^{3}}+1=\dfrac{152}{125}$

$\Rightarrow {{{r}}^{3}}=\dfrac{152}{125}-1$

$\Rightarrow {{{r}}^{3}}=\dfrac{152-125}{125}$

$\Rightarrow {{{r}}^{3}}=\dfrac{27}{125}$

$\Rightarrow {r}=\dfrac{3}{5}$

Therefore the common ratio is $\dfrac{3}{5}.$

10. Find how many terms of G.P, $\dfrac{{2}}{{9}}{,}\dfrac{{-1}}{{3}}{,}\dfrac{{1}}{{2}}{,}.......$must be added to get the sum equal to $\dfrac{{55}}{{72}}$?

Ans: From the given data

First term, ${a}=\dfrac{2}{9}$

Common ratio, ${r}=\dfrac{-3}{2}$

Sum of n terms, ${{{S}}_{{n}}}=\dfrac{55}{72}$

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

$\Rightarrow \dfrac{55}{72}=\dfrac{\dfrac{2}{9}\left( {{\dfrac{-3}{2}}^{{n}}}-1 \right)}{\dfrac{-3}{2}-1}$

$\Rightarrow \dfrac{55}{72}=\dfrac{2}{9}\times \dfrac{-2}{5}\times (\left( \dfrac{-3}{2}{{)}^{{n}}}-1 \right)$

$\Rightarrow \dfrac{-55\times 9\times 5}{72\times 2\times 2}=(\left( \dfrac{-3}{2}{{)}^{{n}}}-1 \right)$

$\Rightarrow (\left( \dfrac{-3}{2}{{)}^{{n}}}-1 \right)=\dfrac{-275}{32}$

$\Rightarrow {{(\dfrac{-3}{2})}^{{n}}}=\dfrac{-275}{32}+1$

$\Rightarrow {{(\dfrac{-3}{2})}^{{n}}}=\dfrac{-275+32}{32}$

$\Rightarrow {{(\dfrac{-3}{2})}^{{n}}}=\dfrac{-243}{32}$

$\Rightarrow {{(\dfrac{-3}{2})}^{{n}}}={{(\dfrac{-3}{2})}^{5}}$

$\Rightarrow {n}=5$

Therefore 5 terms from the series must be taken to make the sum equal to $\dfrac{55}{72}$.

11. If the sum of ${1+2+}{{{2}}^{{2}}}{+}.......{+}{{{2}}^{{n-1}}}$is 255, find the value of n.

Ans: From the given information

First term, ${a}=1$

Common ratio, ${r}=2$ 

Sum of all terms, ${{{S}}_{{n}}}=255$

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

$\Rightarrow 255=\dfrac{1\left( {{2}^{{n}}}-1 \right)}{2-1}$

$\Rightarrow 255={{2}^{{n}}}-1$

$\Rightarrow {{2}^{{n}}}=256$

$\Rightarrow {{2}^{{n}}}={{2}^{8}}$

$\Rightarrow {n}=8$

12. Find the geometric mean between:

(i) $\dfrac{{4}}{{9}}$and $\dfrac{{9}}{{4}}$

Geometric mean of two numbers a and b is $\sqrt{{ab}}$

G.M=$\sqrt{\dfrac{4}{9}\times \dfrac{9}{4}}=1$

(ii) ${14}$and $\dfrac{{7}}{{32}}$

Geometric mean of two numbers a and b is $\sqrt{{ab}}$

G.M=$\sqrt{14\times \dfrac{7}{32}}=\sqrt{\dfrac{7\times 7}{16}}=\dfrac{7}{4}$

(iii) $2\mathbf{a}$and $8{{\mathbf{a}}^{3}}$

Geometric mean of two numbers a and b is $\sqrt{{ab}}$

G.M=$\sqrt{2{a}\times 8{{{a}}^{3}}}=\sqrt{16{{{a}}^{4}}}=4{{{a}}^{2}}$

13. The sum of 3 numbers in G.P is $\dfrac{{39}}{{10}}$ and their product is 1 Find the numbers.

Ans: Let the three numbers of G.P be $\dfrac{{a}}{{r}},{a},{ar}$

Product=$\dfrac{{a}}{{r}}\times {a}\times {ar}=1$

$\Rightarrow {{{a}}^{3}}=1$

$\Rightarrow {a}=1$

Sum=$\dfrac{{a}}{{r}}+{a}+{ar}=\dfrac{39}{10}$

$\Rightarrow {a}\left( \dfrac{1}{{r}}+1+{r} \right)=\dfrac{39}{10}$

$\Rightarrow 1\left( \dfrac{1}{{r}}+1+{r} \right)=\dfrac{39}{10}$

$\Rightarrow \dfrac{1}{{r}}+{r}=\dfrac{39}{10}-1$

$\Rightarrow 1+{{{r}}^{2}}=\dfrac{29{r}}{10}$

$\Rightarrow 10{{{r}}^{2}}-29{r}+10=0$

$\Rightarrow 10{{{r}}^{2}}-25{r}-4{r}+10=0$

$\Rightarrow 5{r}\left( 2{r}-5 \right)-2\left( 2{r}-5 \right)=0$

$\Rightarrow \left( 5{r}-2 \right)\left( 2{r}-5 \right)=0$

$\Rightarrow {r}=\dfrac{2}{5}$or $\dfrac{5}{2}$

Three numbers are $\dfrac{5}{2},1,\dfrac{2}{5}$ when common ratio, ${r}=\dfrac{2}{5}$

Three numbers are $\dfrac{2}{5},1,\dfrac{5}{2}$ when common ratio, ${r}=\dfrac{5}{2}$

14. The first term of a G.P is -3 and the square of the second term is equal to its fourth term. Find its seventh term.

Ans: From the given information

First term, ${a}=-3$

${{({{{a}}_{2}})}^{2}}={{{a}}_{4}}$

$\Rightarrow {{({ar})}^{2}}={a}{{{r}}^{3}}$

$\Rightarrow {{{a}}^{2}}{{{r}}^{2}}={a}{{{r}}^{3}}$

$\Rightarrow {a}={r}=-3$

Seventh term, ${{{a}}_{7}}={a}{{{r}}^{6}}$

$\Rightarrow {{{a}}_{7}}=\left( -3 \right){{(-3)}^{6}}$

$\Rightarrow {{{a}}_{7}}={{(-3)}^{7}}$

$\Rightarrow {{{a}}_{7}}=-2187$

Therefore the seventh term is equal to -2187.

15. Find the fifth term of the G.P, $\dfrac{{5}}{{2}}{,1,}.......$

Ans: From the given information

First term, ${a}=\dfrac{5}{2}$

Common ratio, ${r}=\dfrac{2}{5}$

Fifth term, ${{{a}}_{5}}={a}{{{r}}^{4}}$

$\Rightarrow {{{a}}_{5}}=\left( \dfrac{5}{2} \right){{(\dfrac{2}{5})}^{4}}$

$\Rightarrow {{{a}}_{5}}=\dfrac{5}{2}\times \dfrac{2}{5}\times {{(\dfrac{2}{5})}^{3}}$

$\Rightarrow {{{a}}_{5}}={{(\dfrac{2}{5})}^{3}}$

$\Rightarrow {{{a}}_{5}}=\dfrac{8}{125}$

Therefore the fifth term is equal to $\dfrac{8}{125}.$

16. The first two terms of a G.P are 125 and 25 respectively. Find the ${{5}^{\mathbf{th}}}$and ${{6}^{\mathbf{th}}}$terms of the G.P.

Ans: Given first term, ${a}=125$

Second term, ${{{a}}_{2}}=25$

${{{a}}_{2}}={ar}$

$\Rightarrow {ar}=25$

$\Rightarrow 125{r}=25$

$\Rightarrow {r}=\dfrac{1}{5}$

Fifth term, ${{{a}}_{5}}={a}{{{r}}^{4}}$

$\Rightarrow {{{a}}_{5}}=125{{(\dfrac{1}{5})}^{4}}$

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

Sixth term, ${{{a}}_{6}}={a}{{{r}}^{5}}$

$\Rightarrow {{{a}}_{6}}=125{{(\dfrac{1}{5})}^{5}}$

$\Rightarrow {{{a}}_{6}}=\dfrac{1}{25}$

Therefore the ${{5}^{{th}}}$and ${{6}^{{th}}}$terms are $\dfrac{1}{5}$and $\dfrac{1}{25}$respectively.

17. Find the sum of the sequence $\dfrac{{-1}}{{3}}{,1,-3,9,}.......{,8 }\!\!~\!\!{ terms}{.}$

Ans: Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

First term, ${a}=\dfrac{-1}{3}$

Common ratio, ${r}=-3$

${{{S}}_{8}}=\dfrac{\dfrac{-1}{3}(\left( -3{{)}^{8}}-1 \right)}{-3-1}$

$\Rightarrow {{{S}}_{8}}=\dfrac{-(\left( -3{{)}^{8}}-1 \right)}{-12}$

$\Rightarrow {{{S}}_{8}}=\dfrac{6561-1}{12}$

$\Rightarrow {{{S}}_{8}}=\dfrac{6560}{12}$

$\Rightarrow {{{S}}_{8}}=\dfrac{1640}{3}$

Therefore the sum of the sequence is equal to $\dfrac{1640}{3}.$

18. The first term of a G.P is 27. If the ${{{8}}^{{th}}}$ term is $\dfrac{{1}}{{81}}{.}$What will be the sum of 10 terms?

Ans: Given first term, ${a}=27$

${{8}^{{th}}}$term ${{{a}}_{8}}={a}{{{r}}^{7}}=\dfrac{1}{81}$

$\Rightarrow \left( 27 \right){{{r}}^{7}}=\dfrac{1}{81}$

$\Rightarrow {{{r}}^{7}}=\dfrac{1}{{{3}^{4}}\times {{3}^{3}}}$

$\Rightarrow {r}=\dfrac{1}{3}$

Sum of n terms in G.P, ${{{S}}_{{n}}}=\dfrac{{a}\left( {{{r}}^{{n}}}-1 \right)}{{r}-1}$

${{{S}}_{10}}=\dfrac{21(\left( \dfrac{1}{3}{{)}^{10}}-1 \right)}{\dfrac{1}{3}-1}$

$\Rightarrow {{{S}}_{10}}=21\times 3\times \dfrac{-1}{2}\times (\left( \dfrac{1}{3}{{)}^{10}}-1 \right)$

$\Rightarrow {{{S}}_{10}}=\dfrac{63(1-\left( \dfrac{1}{3}{{)}^{10}} \right)}{2}$

19. Find a G.P for which the sum of first two terms is -4 and the fifth term is 4 times the third term.

Ans: Given ${a}+{{{a}}_{2}}=-4$ and ${{{a}}_{5}}=4{{{a}}_{3}}$

${a}+{{{a}}_{2}}=-4$

$\Rightarrow {a}+{ar}=-4$

$\Rightarrow {a}\left( 1+{r} \right)=-4$

${{{a}}_{5}}=4{{{a}}_{3}}$

$\Rightarrow {a}{{{r}}^{4}}=4{a}{{{r}}^{2}}$

$\Rightarrow {r}=2$ 

Substituting ${r}$ value in the above equation

$\Rightarrow {a}\left( 1+2 \right)=-4$

$\Rightarrow {a}=\dfrac{-4}{3}$

First term, ${a}=\dfrac{-4}{3}$

Second term, ${{{a}}_{2}}={ar}=\dfrac{-4}{3}\times 2=\dfrac{-8}{3}$

Third term,  ${{{a}}_{3}}={a}{{{r}}^{2}}=\dfrac{-4}{3}\times 4=\dfrac{-16}{3}$

Fourth term, ${{{a}}_{4}}={a}{{{r}}^{3}}=\dfrac{-4}{3}\times 8=\dfrac{-32}{3}$

Therefore the series is $\dfrac{-4}{3},\dfrac{-8}{3},\dfrac{-16}{3},\dfrac{-32}{3},.......$.

Geometric Progression

In this chapter, you are going to read about one of the most important chapters of sequence and series, which is a geometric progression. Geometric Progression is the series in which every next term can be found by multiplying the previous one. A geometric Progression is also called a GP.

Here the ratio of every next term and the previous term is called the common ratio.

For example, let a,b,c,d are in GP then the,

b= ar ;

c = ar² ;

d = ar³ 

and so on.

Also, we can write a common GP as a, ar,ar², ar³,….arⁿ.    

The general term of the GP can be denoted by an. You can find out any of the terms using the formula of an 

an = ar^n-1

For example, a = 3, r= 2 

Then next three terms of the GP will be 

a₂ = ar¹ = 3*2 = 6 

a₃= ar² = 3*2² =12

a₄ = ar³ = 3* 2³ = 24

The common ratio 'or is defined as the ratio of the previous term and the next term.

r =  arⁿ/ ar^n-1

GP Can be Divided into Two Categories Such as :

1. Finite GP 

Finite GP is the GP that contains only a finite number of terms. For example; 3, 6,9,12.

2. Infinite GP

Infinite GP is the GP that contains an infinite number of terms. It is generally denoted by dots after the last term. For example the GP; 3,6, 9,12…

Sum of n Terms of a GP 

The sum of a GP which consists n terms is :

• Case -1: When the common ratio of the GP, r is less than one, i.e. r< 1 ; 

then the sum of the n terms of GP will be 

S_n = a (1-rⁿ)/ (1-r)

• Case -2: When the common ratio of the GP, r is more than one, i.e. r> 1 ;

then the sum of the n terms of GP will be 

S_n = an ( rⁿ -1)/(r-1)