2 Marks Questions

Question 12 Marks

Find the sum of the following series to infinity:

$1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}+\ ....\infty$

Answer

$\text{S}_{\infty}=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\ ...$
$\Rightarrow\text{a}=1, \text{r}=-\frac{1}{3}$
$\text{S}_{\infty}=\frac{\text{a}}{1-\text{r}}$
$=\frac{1}{1+\frac13}$
$\text{S}_{\infty}=\frac34$

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Question 22 Marks

Show the of the following progression is G.P. Also, find the common ratio in case:

$\frac{-2}{3},-6,-54,\dots$

Answer

$\frac{-2}{3},-6,-54,\dots$
$\frac{\text{t}_\text{n}}{\text{t}_\text{n}-1}=\text{r}=\text{comman ratio}\cdots(\text{i})$
$\frac{\text{t}_2}{\text{t}_1}=\frac{-6}{\frac{-2}{3}}=\frac{18}{2}=9$
$\frac{\text{t}_3}{\text{t}_2}=\frac{-54}{-6}=9$
$\therefore\text{r}=9$

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Question 32 Marks

Find the sum of the following geometric progrssions:

1, 3, 9, 27, ... to 8 terms

Answer

1, 3, 9, 27, ... to 8 terms

$\text{a}=1,\text{r}=\frac31=3,\text{n}=8$

$\text{S}_\text{n}=\text{a}\frac{(\text{r}^{\text{n}-1})}{\text{r}-1}$

$\text{S}_8=1\frac{(3^8-1)}{3-1}=3280$

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Question 42 Marks

Show the of the following progression is G.P. Also, find the common ratio in case:

$4,-2,1,\frac{-1}{2}.\dots$

Answer

$4,-2,1,\frac{-1}{2}.\dots$
$\frac{\text{t}_\text{n}}{\text{t}_\text{n}-1}=\text{r}=\text{comman ratio}\cdots(\text{i})$
$\frac{\text{t}_2}{\text{t}_1}=\frac{-2}{4}=\frac{-1}{2}$
$\frac{\text{t}_3}{\text{t}_2}=\frac{1}{-2}=\frac{-1}{2}$

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Question 52 Marks

Find the sum of the following geometric progrssions:

2, 6, 18, ... to 7 terms

Answer

2, 6, 18, ... to 7 term
$\text{a}=2,\text{r}=\frac62=3,\text{n}=7$
$\text{S}_\text{n}=\text{a}\frac{(\text{r}^{\text{n}-1})}{\text{r}-1}$
$\text{S}_7=2\frac{(3^7-1)}{3-1}=\frac{2}{2}(3^7-1)$
$=2187-1=2186$

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Question 62 Marks

Find the 4^th term from the end of the G.P. $\frac{2}{27},\frac{2}{9},\frac{2}{3},\ \dots ,162.$

Answer

$\frac{2}{27},\frac{2}{9},\frac{2}{3},\ \dots ,162.$
n^th term from the end
$\text{a}_\text{n}=\text{l}\Big(\frac{1}{\text{r}}\Big)^{\text{n}-1}$
l = 162, r = common ratio $=\frac{\text{t}_2}{\text{t}_1}$
$=\frac{\frac{2}{9}}{\frac{2}{27}}=3$
$\text{n}=4$
$\text{t}_4=(162)\Big(\frac{1}{3}\Big)^3$
$=\frac{162}{27}$
$=6$

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Question 72 Marks

Find the rational number whose decimal expansion is $0.\overline{423}.$

Answer

$0.\overline{4.23}=0.4+0.0232323\ \dots$
$=0.4+0.023+0.00023+\ \dots$
$=0.4+\frac{23}{10^3}+\frac{23}{10^5}+\dots$
$=0.4+\frac{23}{10^3}\Big(1+\frac{1}{10^2}+\frac{1}{10^4}+\dots\Big)$
$=0.4+\frac{23}{1000}\Bigg(\frac{1}{1-\frac{1}{100}}\Bigg)$
$=0.4+\frac{23}{1000}\Big(\frac{100}{99}\Big)$
$=\frac{4}{10}+\frac{23}{990}$
$=\frac{360+23}{990}$
$0.\overline{423}=\frac{419}{990}$

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Question 82 Marks

Find the sum of the following geometric progrssions:

$1,\frac{1}{2},\frac14,\frac{-1}{8},\ ...$

Answer

$1,\frac{1}{2},\frac14,\frac{-1}{8},\ ...9 \text{ terms}$
$\text{a}=1,\text{r}=\frac{\frac{-1}{2}}{1}=\frac{-1}{2},\text{n}=9$
$\text{S}_\text{n}=\text{a}\frac{(\text{r}^{\text{n}-1})}{\text{r}-1}$
$\text{S}_9=1\frac{\big(\frac{-1}{2}\big)-1}{\frac{-1}{2}-1}$
$=\frac{\frac{-1}{512}-1}{\frac{-1}{2}-1}$
$=\frac{\frac{-1-512}{512}}{\frac{-1-2}{2}}$
$=\frac{-513}{512}\times\frac{2}{-3}$
$=\frac{171}{256}$

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Question 92 Marks

Show the of the following progression is G.P. Also, find the common ratio in case:

$\text{a}\frac{3\text{a}^2}{4},\frac{9\text{a}^3}{16},\cdots$

Answer

$\text{a}\frac{3\text{a}^2}{4},\frac{9\text{a}^3}{16},\cdots$
$\frac{\text{t}_\text{n}}{\text{t}_\text{n}-1}=\text{r}=\text{comman ratio}\cdots(\text{i})$
$\frac{\text{t}_3}{\text{t}_2}=\frac{\frac{9\text{a}^3}{16}}{\frac{3\text{a}^2}{4}}=\frac{9\text{a}^3}{16}\times\frac{4}{3\text{a}^2}=\frac{3\text{a}^2}{4}$
$\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{3\text{a}^2}{4}}{\text{a}}=\frac{3\text{a}^2}{\text{a}}$
$\therefore\text{r}=\frac{3}{4}\text{a}$

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Question 102 Marks

Find the sum of the following series to infinit:

$10-9+8.1-7.29+\ ...\infty$

Answer

This infinity G.P. has first term a = 10 and common ratio $\text{r}=-\frac{9}{10}=-0.9$
Thus the sum of the infinity G.P. will be:
$10-9+8.9-7.29+\ ...\infty=\frac{\text{a}}{1-\text{r}}$ $[\text{Since }|\text{r}|<1]$
$=\frac{10}{1-(-0.9)}$
$=\frac{10}{1.9}$
$=\frac{100}{19}$
$=5.263$

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Question 112 Marks

Which term of the progression 0.004, 0.02, 0.1, ... is 12.5₹

Answer

0.004, 0.02, 0.1, ... is 12.5
Hare,
a = 0.004, t_n = 12.5
$\text{r}=\frac{\text{t}_2}{\text{t}_1}=\frac{0.02}{0.004}=5$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$12.5=(0.004)(5)^{\text{n}-1}$
^{$\frac{12.5}{0.004}=(5)^{\text{n}-1}$}
$\frac{125\times100}{4}=5^{\text{n}-1}$
$5^5=5^{\text{n}-1}$
$=\text{n}-1$
$\text{n}=6$

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Question 122 Marks

Show the of the following progression is G.P. Also, find the common ratio in case:

$\frac12,\frac13,\frac29,\frac{4}{27}\dots$

Answer

$\frac12,\frac13,\frac29,\frac{4}{27}\dots$
$\frac{\text{t}_\text{n}}{\text{t}_\text{n}-1}=\text{r}=\text{comman ratio}\cdots(\text{i})$
$\frac{\text{t}_3}{\text{t}_2}=\frac{\frac{2}{9}}{\frac{1}{3}}=\frac{2}{3}$
$\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{3}}{\frac12}=\frac{2}{3}$
$\therefore\text{r}=\frac{2}{3}$

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Question 132 Marks

Find the sum of the following series to infinity:

$8+4\sqrt{2}+4+\ ...\infty$

Answer

$\text{S}_{\infty}=8+4\sqrt{2}+4+\ ...$
$\Rightarrow\text{a}=8, \text{r}=\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\text{S}_{\infty}=\frac{\text{a}}{1-\text{r}}$
$=\frac{8}{1-\frac{1}{\sqrt{2}}}$
$=\frac{8\sqrt{2}}{\sqrt{2}-1}\times\frac{\big(\sqrt{2}+1\big)}{\big(\sqrt{2}+1\big)}$
$=\frac{8\big(2+\sqrt{2}\big)}{2+1}$
$\text{S}_{\infty}=8\big(2+\sqrt{2}\big)$

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Question 142 Marks

The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.

Answer

Let three numbers in A.P. be a - d, a + d
Here, a - d + a + a + d =21
3a = 21
a = 7
And,
(7 - d), (7 - 1), (7 + d) + 1 are in G.P.
(7 - d), 6, (8 + d) are in G.P.
(6)² = (7 - d) (8 + d)
36 = 56 + 7d - 8d - d²
d² + d - 20 = 0
(d + 5) (d - 4) = 0
d = 4, -5
So, Numbers are 3, 7, 11 or 12, 7, 2.

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Question 152 Marks

How many terms of the series 2 + 6 + 18 + ... must be make the sum equal to 728₹

Answer

2 + 6 + 18 + ...
$\text{S}_\text{n}=728$
Now,
$\text{S}_\text{n}=\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{r}}$
$\text{a}=2,\text{r}=\frac62=3$
$728=\frac{2(3^\text{n}-1)}{3-1}$
$728=\frac{2(3^\text{n}-1)}{2}=(3^{\text{n}}-1)$
$728+1=3^\text{n}$
$729=3^\text{n}$
$(3)^6=3\text{n}$
$\Rightarrow\text{n}=6$

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Question 162 Marks

Find:
The 10^th term of the G.P. $\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\dots$

Answer

10^th term of the G.P. $\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\dots$
$\text{a}=\sqrt{2}$
$\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac12$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{t}_{10}=\text{ar}^9$
$=\sqrt{2}\Big(\frac12\Big)^9$
$=\frac{1}{\sqrt{2}}\Big(\frac12\Big)^8$

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Question 172 Marks

Find:
The 10^th term of the G.P. $-\frac{3}{4},\frac12,-\frac{1}{3},\frac29,\ ...$

Answer

10^th term of G.P. $-\frac{3}{4},\frac12,-\frac{1}{3},\frac29,\ ...$
$\text{a}=\frac{-3}{4}$
Because it is G.P.
$\therefore\text{r}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{2}}{\frac{-3}{4}}=\frac{-2}{3}$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{t}_{10}=\text{ar}^9=\Big(\frac{-3}{4}\Big)\Big(\frac{-2}{3}\Big)^9=\frac12\Big(\frac23\Big)^8$

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Question 182 Marks

Find:
The 12^th term of the G.P. $\frac{1}{\text{a}^3\text{x}^3},\text{ax},\text{a}^5\text{x}^5\dots$

Answer

12^th term of the G.P. $\frac{1}{\text{a}^3\text{x}^3},\text{ax},\text{a}^5\text{x}^5\dots$
$\text{a}=\frac{1}{\text{a}^3\text{x}^3}$
$\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\text{a}\text{x}}{\frac{1}{\text{a}^3\text{x}^3}}=\text{a}^4\text{x}^4$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{t}_{12}=\text{ar}^{11}$
$=\Big(\frac{1}{\text{a}^3\text{x}^3}\Big)\big(\text{a}^4\text{x}^4\big)^{11}$
$=(\text{a}\text{x})^{41}$

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Question 192 Marks

Find the geometric means of the following pairs of numbers:
- 8 and - 2

Answer

- 8 and - 2
Geometric means between a and $\text{b}=\sqrt{\text{ab}}$
a = - 8, b = ab³
$\therefore\text{Geometric means}=\sqrt{-8\times-2}$
$=\sqrt{16}=4,-4$

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Question 202 Marks

Express the recurring decimal 0.125125125 ... as a rational number.

Answer

$0.125125125 \ \dots=0.\overline{125}$
$=0.125+0.000125+0.000000125+\ \dots$
$=\frac{125}{10}^3\Big(1+\frac{1}{10^3}+\frac{1}{10^6}+\ \dots\Big)$
$=\frac{125}{10^3}\Bigg(\frac{1}{1-\frac{1}{1000}}\Bigg)$
$=\frac{125}{1000}\Big(\frac{1000}{999}\Big)$
$0.125125125\ \dots=\frac{125}{999}$

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Question 212 Marks

If a, b, c are in G.P., Prove that $\log \text{a}, \log \text{b},\log\text{c}$ are in A.P.

Answer

Here, a, b, c are in G.P.
$\text{b}^2=\text{ac}\cdots(\text{i})$
Now, $2\log\text{b}=\log\text{b}^2$
$=\log\text{ ac}$
$2\log\text{b}=\log\text{a}+\log\text{c}$
$\log\text{b}-\log\text{a}=\log\text{c}-\log\text{b}$
$\Rightarrow\log\text{a},\log\text{b}\log\text{c},\text{ are in A.P.}$

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Question 222 Marks

Show that the sequence defined by $\text{a}_\text{n}=\frac{2}{3^\text{n}},\text{n}\in\text{N}$ is a G.P.

Answer

$\text{a}_\text{n}=\frac{2}{3^\text{n}},\text{n}\in\text{N}$
Put n = 1, 2, 3 ... because n is natural number
$\frac{2}{3},\frac{2}{3^2},\frac{2}{3^3},\dots$
$\frac{\text{t}_3}{\text{t}_2}=\frac{\frac{2}{3^3}}{\frac{2}{3^2}}=\frac13$
$\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{2}{3^2}}{\frac{2}{3}}=\frac13$
Ratio of consecutive terms is solve
$\therefore\frac{1}{3}$ is common ratio, Hence it is G.P. $\forall\text{n}\in\text{N}.$

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Question 232 Marks

Find:
The ninth term of the G.P. 1, 4, 16, 64, ...

Answer

9^th term of G.P. 1, 4, 16, 64, ...
t₁ = 1 = a
t₂ = 4
Because it is G.P.
$\frac{\text{t}_2}{\text{t}_1}=\text{common ratio}=\text{r}$
$\text{r}=\frac{4}{1}=4$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{t}_9=\text{ar}^8=1(4)^8=4^8$

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Question 242 Marks

Find the sum of the following geometric series:
$\text{x}^3,\text{x}^5,\text{x}^7,\ ...\ \text{to n terms}$

Answer

Here the first term of the G.P. is $\text{a}=\text{x}^3$ and the common ratio is $\text{r}=\frac{\text{x}^5}{\text{x}^3}=\text{x}^2$
Thus the sum of the G.P. is:
$\text{x}^3+\text{x}^5+\text{x}^7+\ ...\ \text{to n terms}=\frac{\text{x}^3\big((\text{x}^2)^\text{n}-1\big)}{\text{x}^2-1}=\frac{\text{x}^3(\text{x}^{2\text{n}}-1)}{\text{x}^2-1}$

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Question 252 Marks

Find:
n^th term of the G.P. $\sqrt{3},\frac{1}{\sqrt{3}},\frac{1}{3\sqrt{3}},\dots$

Answer

n^th term of the G.P. $\sqrt{3},\frac{1}{\sqrt{3}},\frac{1}{3\sqrt{3}},\dots$
$\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac13$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{t}_{\text{n}}=\sqrt{3}\Big(\frac{1}{3}\Big)^{\text{n}-1}$

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Question 262 Marks

If a is the G.M. of 2 and $\frac14,$ find a.

Answer

a is the G.M. between 2 and $\frac14,$
Then,
$\text{a}=\sqrt{2\times\frac14}$
$\text{a}=\sqrt{\frac12}=\frac{1}{\sqrt{2}}$

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Question 272 Marks

Find:
The 8^th term of the G.P. 0.3, 0.06, 0.012, ...

Answer

Here,
First term, a = 0.3
Common ratio, $\text{r}=\frac{\text{a}_2}{\text{a}_1}=\frac{0.06}{0.3}=0.2$
$\therefore$ 8^th term = a8 = ar^(8-1) = 0.3 (0.2)⁷
Thus, the 8th term of the given GP is 0.3 (0.2)⁷.

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Question 282 Marks

The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.

Answer

Let numbers are $\text{a},\text{ar},\text{ar}^2$
$\text{a}+\text{ar}+\text{ar}^2=56\cdots(1)$
$(\text{a}-1),(\text{ar}-7),(\text{ar}^2-21)\text{ are in A.P.}$
$\Rightarrow2(\text{ar}-7)=\text{a}-1+\text{ar}^2-21$
$=(\text{ar}^2+\text{a})-22$
$2\text{ar}-14=(56-\text{ar})-22$ [Using equation (1)]
$2\text{ar}-14=34-\text{ar}$
$3\text{ar}=48$
$\text{ar}=16\cdots(2)$
$\text{a}=\frac{16}{\text{r}}$
Put a in equation (1),
$\frac{16+16\text{r}+16\text{r}^2}{\text{r}}=56$
$16+16\text{r}+16\text{r}^2=56\text{r}$
$16\text{r}^2-40\text{r}+16 =0$
$2\text{r}^2-4\text{r}-\text{r}+2=0$
$2\text{r}(\text{r}-2)-1(\text{r}-2)=0$
$(\text{r}-2)(2\text{r}-1)=0$
$\text{r}=2, \frac12$
Put r in equation (2),
$\text{ar}=16$
For $\text{r}=\frac{2}{\text{a}}=8$
For $\text{r}=\frac12, \text{a}=32$
Thus, three numbers are 8, 16, 32
In both cases.

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Question 292 Marks

If a, b, c are in G.P., Prove that $\frac{1}{\log _\text{a}\text{m}},\frac{1}{ \log _\text{b}\text{m}},\frac{1}{\log_\text{c}\text{m}}$ are in A.P.

Answer

Here, a, b, c are in G.P., so
$\text{b}^2=\text{ac}$
Now, $\frac{2}{\log_\text{b}\text{m}}=2\log_\text{m}\text{b}$
$=\log_\text{m}\text{b}^2$
$=\log_\text{m}\text{ac}$
$=\log_\text{m}\text{a}+\log_\text{m}\text{c}$
$\frac{2}{\log_\text{b}\text{m}}=\frac{1}{\log_\text{a}\text{m}}+\frac{1}{\log_\text{c}\text{m}}$
$\Rightarrow\frac{1}{\log_\text{b}\text{m}}-\frac{1}{\log_\text{a}\text{m}}=\frac{1}{\log_\text{c}\text{m}}-\frac{1}{\log_\text{b}\text{m}}$
$\Rightarrow\frac{1}{\log_\text{a}\text{m}},\frac{1}{\log_\text{b}\text{m}},\frac{1}{\log_\text{c}\text{m}}\text{ are in A.P.}$

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Question 302 Marks

Three numbers are in A.P. and their sum is 15. If 1, 3, 9 be added to them respectively, they from a G.P. Find the numbers.

Answer

Let a - d, a, a + d be numbers in A.P.
Here, a - d + a + a + d = 15
3 a = 15
a = 5
Find,
[(5 - d) + 1], (5 + 3), [(5 + d) + 9] are in G.P.
$\Rightarrow(6-\text{d}),8(14 + \text{d})$ are in G.P.
$(8)^2=(6-\text{d})(14+\text{d})$
$64=84+6\text{d}-14\text{d}-\text{d}^2$
$\text{d}^2+8\text{d}-20=0$
$(\text{d}+10)(\text{d}-2)=0$
$\text{d}=2,-10$
So, Number are 3, 5, 7 or 15, 5, -5

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Question 312 Marks

Find the sum of the following series to infinity:

$\frac{2}{5}+\frac{3}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\ ... \infty$

Answer

$\text{S}_{\infty}=\frac{2}{5}+\frac{3}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\ ...$
$=\Big(\frac{2}{5}+\frac{3}{5^3}+\ ...\Big) +\Big(\frac{3}{5^2}+\frac{3}{5^4}+\ ...\Big)$
$\text{S}_\infty=\text{S}'_\infty+\text{S}"_\infty$
For
$\text{S}'_\infty=\frac{\text{a}}{1-\text{r}}$
$=\frac{\frac{2}{5}}{1-\frac{1}{25}}$
$=\frac12\times\frac{25}{24}$
$\text{S}'_\infty=\frac{5}{12}$
$\text{S}"_\infty=\frac{\frac{3}{25}}{1-\frac{1}{25}}$
$=\frac{3}{25}\times\frac{25}{24}$
$=\frac{3}{24}$
$\text{S}_\infty=\text{S}'_\infty+\text{S}"_\infty$
$=\frac{5}{12}+\frac{3}{24}$
$=\frac{13}{24}$
$\text{S}_\infty=\frac{13}{24}$

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Question 322 Marks

Insert 5 geometric means between $\frac{32}{9}$ and $\frac{81}{2}.$

Answer

5 geometric means between $\frac{32}{9}$ and $\frac{81}{2}.$

Let G₁, G₂, G₃, G₄, G₅be 5 geometric means between $\text{a}=\frac{32}{9}$ and $\text{b}=\frac{81}{2}$

Then, $\frac{32}{9}$, G₁, G₂, G₃, G₄, G₅ $\frac{81}{2}$ is a G.P. with $\text{a}=\frac{32}{9}$, $\text{b}=\frac{81}{2}$

$\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{\text{n}+1}}$

$=\Bigg(\frac{\frac{81}{2}}{\frac{32}{9}}\Bigg)^{\frac{1}{5+1}}=\Big(\frac{81}{2}\times\frac{9}{32}\Big)^{\frac{1}{6}}=\Big(\frac{{3}^6}{{2}^6}\Big)=\frac32$

$\therefore\text{G}_1=\text{ar}=\frac{32}{9}\times\frac32=\frac{16}{3}$

$\text{G}_2=\text{ar}^2=\frac{32}{9}\times\frac{9}{4}=8$

$\text{G}_3=\text{ar}^3=\frac{32}{9}\times\frac{27}{8}=12$

$\text{G}_4=\text{ar}^4=\frac{32}{9}\times\frac{{3}^4}{{2}^4}=2\times9=18$

$\text{G}_5=\text{ar}^5=\frac{32}{9}\times\frac{{5}^5}{{2}^5}=27$

Hence, $\frac{16}{3},8,12,18,27$ are 5 geometric means between $\frac{32}{9}$ and $\frac{81}{2}.$

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