Question 12 Marks
Find the sum of the following geometric progrssions:1, 3, 9, 27, ... to 8 terms
Answer1, 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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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.
AnswerLet three numbers in A.P. be $a-d, a+d$ Here, $a-d+a+a+d=213 a=21 a=7$ And, $(7-d),(7-1),(7+d)+1$ are in G.P. $(7-d), 6,(8+d)$ are in G.P. $(6)^2=(7-d)(8+d) 36=56+7 d-8 d-d^2 d^2+d-20=0(d+5)(d-4)=0 d=4,-5$
So, Numbers are $3,7,11$ or $12,7,2$.
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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.
AnswerLet 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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How many terms of the series 2 + 6 + 18 + ... must be make the sum equal to 728₹
Answer2 + 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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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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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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Insert 5 geometric means between $\frac{32}{9}$ and $\frac{81}{2}.$
Answer5 geometric means between $\frac{32}{9}$ and $\frac{81}{2}.$Let $G_1, G_2, G_3, G_4, G_5$ be 5 geometric means between $\text{a}=\frac{32}{9}$ and $\text{b}=\frac{81}{2}$
Then, $\frac{32}{9}$, $G_1, G_2, G_3, G_4, G_5$ $\frac{81}{2}$ 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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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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Find the geometric means of the following pairs of numbers: - 8 and - 2
Answer- 8 and - 2 Geometric means between a and $\text{b}=\sqrt{\text{ab}} a = - 8, b = ab^3$ $\therefore\text{Geometric means}=\sqrt{-8\times-2}$ $=\sqrt{16}=4,-4$
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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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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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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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Find: The ninth term of the G.P. 1, 4, 16, 64, ...
Answer$9^{th}$ term of G.P. $1, 4, 16, 64, ... t_1 = 1 = a$ $t_2 = 4$ 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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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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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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Find: The $8^{th}$ term of the G.P. 0.3, 0.06, 0.012, ...
AnswerHere, 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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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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If a, b, c are in G.P., Prove that $\log \text{a}, \log \text{b},\log\text{c}$ are in A.P.
AnswerHere, 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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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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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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Find the sum of the following series to infinit:$10-9+8.1-7.29+\ ...\infty$
AnswerThis 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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Find the sum of the following geometric progrssions:2, 6, 18, ... to 7 terms
Answer2, 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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Which term of the progression 0.004, 0.02, 0.1, ... is 12.5₹
Answer0.004, 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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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.
AnswerLet 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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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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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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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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Find the sum of the following geometric series: $\text{x}^3,\text{x}^5,\text{x}^7,\ ...\ \text{to n terms}$
AnswerHere 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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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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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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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.
AnswerHere, 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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If a is the G.M. of 2 and $\frac14,$ find a.
Answera 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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