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$\text{a}=1,\text{r}=\frac31=3,\text{n}=8$
$\text{S}_\text{n}=\text{a}\frac{(\text{r}^{\text{n}-1})}{\text{r}-1}$
$\text{S}_8=1\frac{(3^8-1)}{3-1}=3280$
View full question & answer→Question 22 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.
AnswerLet 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.
$(6)^2 = (7 - d) (8 + d)$
$36 = 56 + 7d - 8d - d^2$
$d^2 + d - 20 = 0$
$(d + 5) (d - 4) = 0$
$d = 4, -5$
So, Numbers are $3, 7, 11$ or $12, 7, 2.$
View full question & answer→Question 32 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.
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.
View full question & answer→Question 42 Marks
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,
$\text{S}_\text{n}=\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{r}}$
$\text{a}=2,\text{r}=\frac62=3$
$728=\frac{2(3^\text{n}-1)}{3-1}$
$728=\frac{2(3^\text{n}-1)}{2}=(3^{\text{n}}-1)$
$728+1=3^\text{n}$
$729=3^\text{n}$
$(3)^6=3\text{n}$
$\Rightarrow\text{n}=6$
View full question & answer→Question 52 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. $\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\dots$
$\text{a}=\sqrt{2}$
$\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac12$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{t}_{10}=\text{ar}^9$
$=\sqrt{2}\Big(\frac12\Big)^9$
$=\frac{1}{\sqrt{2}}\Big(\frac12\Big)^8$
View full question & answer→Question 62 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.
$\therefore\text{r}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{2}}{\frac{-3}{4}}=\frac{-2}{3}$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{t}_{10}=\text{ar}^9=\Big(\frac{-3}{4}\Big)\Big(\frac{-2}{3}\Big)^9=\frac12\Big(\frac23\Big)^8$
View full question & answer→Question 72 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 $G_1, G_2, G_3, G_4, G_5$ 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 $\frac{81}{2}.$
View full question & answer→Question 82 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$\text{S}_{\infty}=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\ ...$
$\Rightarrow\text{a}=1, \text{r}=-\frac{1}{3}$
$\text{S}_{\infty}=\frac{\text{a}}{1-\text{r}}$
$=\frac{1}{1+\frac13}$
$\text{S}_{\infty}=\frac34$
View full question & answer→Question 92 Marks
Find the geometric means of the following pairs of numbers:
$- 8$ and $- 2$
Answer$- 8$ and $- 2$
Geometric means between a and $\text{b}=\sqrt{\text{ab}}$
$a = - 8, b = ab^3$
$\therefore\text{Geometric means}=\sqrt{-8\times-2}$
$=\sqrt{16}=4,-4$
View full question & answer→Question 102 Marks
Show the of the following progression is G.P. Also, find the common ratio in case:$\frac{-2}{3},-6,-54,\dots$
Answer$\frac{-2}{3},-6,-54,\dots$
$\frac{\text{t}_\text{n}}{\text{t}_\text{n}-1}=\text{r}=\text{comman ratio}\cdots(\text{i})$
$\frac{\text{t}_2}{\text{t}_1}=\frac{-6}{\frac{-2}{3}}=\frac{18}{2}=9$
$\frac{\text{t}_3}{\text{t}_2}=\frac{-54}{-6}=9$
$\therefore\text{r}=9$
View full question & answer→Question 112 Marks
Find the rational number whose decimal expansion is $0.\overline{423}.$
Answer$0.\overline{4.23}=0.4+0.0232323\ \dots$
$=0.4+0.023+0.00023+\ \dots$
$=0.4+\frac{23}{10^3}+\frac{23}{10^5}+\dots$
$=0.4+\frac{23}{10^3}\Big(1+\frac{1}{10^2}+\frac{1}{10^4}+\dots\Big)$
$=0.4+\frac{23}{1000}\Bigg(\frac{1}{1-\frac{1}{100}}\Bigg)$
$=0.4+\frac{23}{1000}\Big(\frac{100}{99}\Big)$
$=\frac{4}{10}+\frac{23}{990}$
$=\frac{360+23}{990}$
$0.\overline{423}=\frac{419}{990}$
View full question & answer→Question 122 Marks
Find the sum of the following geometric progrssions:$1,\frac{1}{2},\frac14,\frac{-1}{8},\ ...$
Answer$1,\frac{1}{2},\frac14,\frac{-1}{8},\ ...9 \text{ terms}$
$\text{a}=1,\text{r}=\frac{\frac{-1}{2}}{1}=\frac{-1}{2},\text{n}=9$
$\text{S}_\text{n}=\text{a}\frac{(\text{r}^{\text{n}-1})}{\text{r}-1}$
$\text{S}_9=1\frac{\big(\frac{-1}{2}\big)-1}{\frac{-1}{2}-1}$
$=\frac{\frac{-1}{512}-1}{\frac{-1}{2}-1}$
$=\frac{\frac{-1-512}{512}}{\frac{-1-2}{2}}$
$=\frac{-513}{512}\times\frac{2}{-3}$
$=\frac{171}{256}$
View full question & answer→Question 132 Marks
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.
$\frac{\text{t}_2}{\text{t}_1}=\text{common ratio}=\text{r}$
$\text{r}=\frac{4}{1}=4$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{t}_9=\text{ar}^8=1(4)^8=4^8$
View full question & answer→Question 142 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. $\frac{1}{\text{a}^3\text{x}^3},\text{ax},\text{a}^5\text{x}^5\dots$
$\text{a}=\frac{1}{\text{a}^3\text{x}^3}$
$\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\text{a}\text{x}}{\frac{1}{\text{a}^3\text{x}^3}}=\text{a}^4\text{x}^4$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{t}_{12}=\text{ar}^{11}$
$=\Big(\frac{1}{\text{a}^3\text{x}^3}\Big)\big(\text{a}^4\text{x}^4\big)^{11}$
$=(\text{a}\text{x})^{41}$
View full question & answer→Question 152 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
$\text{S}'_\infty=\frac{\text{a}}{1-\text{r}}$
$=\frac{\frac{2}{5}}{1-\frac{1}{25}}$
$=\frac12\times\frac{25}{24}$
$\text{S}'_\infty=\frac{5}{12}$
$\text{S}"_\infty=\frac{\frac{3}{25}}{1-\frac{1}{25}}$
$=\frac{3}{25}\times\frac{25}{24}$
$=\frac{3}{24}$
$\text{S}_\infty=\text{S}'_\infty+\text{S}"_\infty$
$=\frac{5}{12}+\frac{3}{24}$
$=\frac{13}{24}$
$\text{S}_\infty=\frac{13}{24}$
View full question & answer→Question 162 Marks
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 $0.3 (0.2)^7$.
View full question & answer→Question 172 Marks
Express the recurring decimal 0.125125125 ... as a rational number.
Answer$0.125125125 \ \dots=0.\overline{125}$
$=0.125+0.000125+0.000000125+\ \dots$
$=\frac{125}{10}^3\Big(1+\frac{1}{10^3}+\frac{1}{10^6}+\ \dots\Big)$
$=\frac{125}{10^3}\Bigg(\frac{1}{1-\frac{1}{1000}}\Bigg)$
$=\frac{125}{1000}\Big(\frac{1000}{999}\Big)$
$0.125125125\ \dots=\frac{125}{999}$
View full question & answer→Question 182 Marks
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, $2\log\text{b}=\log\text{b}^2$
$=\log\text{ ac}$
$2\log\text{b}=\log\text{a}+\log\text{c}$
$\log\text{b}-\log\text{a}=\log\text{c}-\log\text{b}$
$\Rightarrow\log\text{a},\log\text{b}\log\text{c},\text{ are in A.P.}$
View full question & answer→Question 192 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. $\forall\text{n}\in\text{N}.$
View full question & answer→Question 202 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$\text{a}\frac{3\text{a}^2}{4},\frac{9\text{a}^3}{16},\cdots$
$\frac{\text{t}_\text{n}}{\text{t}_\text{n}-1}=\text{r}=\text{comman ratio}\cdots(\text{i})$
$\frac{\text{t}_3}{\text{t}_2}=\frac{\frac{9\text{a}^3}{16}}{\frac{3\text{a}^2}{4}}=\frac{9\text{a}^3}{16}\times\frac{4}{3\text{a}^2}=\frac{3\text{a}^2}{4}$
$\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{3\text{a}^2}{4}}{\text{a}}=\frac{3\text{a}^2}{\text{a}}$
$\therefore\text{r}=\frac{3}{4}\text{a}$
View full question & answer→Question 212 Marks
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:
$10-9+8.9-7.29+\ ...\infty=\frac{\text{a}}{1-\text{r}}$ $[\text{Since }|\text{r}|<1]$
$=\frac{10}{1-(-0.9)}$
$=\frac{10}{1.9}$
$=\frac{100}{19}$
$=5.263$
View full question & answer→Question 222 Marks
Find the sum of the following geometric progrssions:2, 6, 18, ... to 7 terms
Answer2, 6, 18, ... to 7 term
$\text{a}=2,\text{r}=\frac62=3,\text{n}=7$
$\text{S}_\text{n}=\text{a}\frac{(\text{r}^{\text{n}-1})}{\text{r}-1}$
$\text{S}_7=2\frac{(3^7-1)}{3-1}=\frac{2}{2}(3^7-1)$
$=2187-1=2186$
View full question & answer→Question 232 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,
$a = 0.004, t_n = 12.5$
$\text{r}=\frac{\text{t}_2}{\text{t}_1}=\frac{0.02}{0.004}=5$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$12.5=(0.004)(5)^{\text{n}-1}$
$\frac{12.5}{0.004}=(5)^{\text{n}-1}$
$\frac{125\times100}{4}=5^{\text{n}-1}$
$5^5=5^{\text{n}-1}$
$=\text{n}-1$
$\text{n}=6$
View full question & answer→Question 242 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.
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
View full question & answer→Question 252 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$4,-2,1,\frac{-1}{2}.\dots$
$\frac{\text{t}_\text{n}}{\text{t}_\text{n}-1}=\text{r}=\text{comman ratio}\cdots(\text{i})$
$\frac{\text{t}_2}{\text{t}_1}=\frac{-2}{4}=\frac{-1}{2}$
$\frac{\text{t}_3}{\text{t}_2}=\frac{1}{-2}=\frac{-1}{2}$
View full question & answer→Question 262 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 $\frac12,\frac13,\frac29,\frac{4}{27}\dots$
$\frac{\text{t}_\text{n}}{\text{t}_\text{n}-1}=\text{r}=\text{comman ratio}\cdots(\text{i})$
$\frac{\text{t}_3}{\text{t}_2}=\frac{\frac{2}{9}}{\frac{1}{3}}=\frac{2}{3}$
$\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{3}}{\frac12}=\frac{2}{3}$
$\therefore\text{r}=\frac{2}{3}$
View full question & answer→Question 272 Marks
Find the sum of the following series to infinity:$8+4\sqrt{2}+4+\ ...\infty$
Answer$\text{S}_{\infty}=8+4\sqrt{2}+4+\ ...$
$\Rightarrow\text{a}=8, \text{r}=\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\text{S}_{\infty}=\frac{\text{a}}{1-\text{r}}$
$=\frac{8}{1-\frac{1}{\sqrt{2}}}$
$=\frac{8\sqrt{2}}{\sqrt{2}-1}\times\frac{\big(\sqrt{2}+1\big)}{\big(\sqrt{2}+1\big)}$
$=\frac{8\big(2+\sqrt{2}\big)}{2+1}$
$\text{S}_{\infty}=8\big(2+\sqrt{2}\big)$
View full question & answer→Question 282 Marks
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:
$\text{x}^3+\text{x}^5+\text{x}^7+\ ...\ \text{to n terms}=\frac{\text{x}^3\big((\text{x}^2)^\text{n}-1\big)}{\text{x}^2-1}=\frac{\text{x}^3(\text{x}^{2\text{n}}-1)}{\text{x}^2-1}$
View full question & answer→Question 292 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. $\sqrt{3},\frac{1}{\sqrt{3}},\frac{1}{3\sqrt{3}},\dots$
$\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac13$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{t}_{\text{n}}=\sqrt{3}\Big(\frac{1}{3}\Big)^{\text{n}-1}$
View full question & answer→Question 302 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 $=\frac{\text{t}_2}{\text{t}_1}$
$=\frac{\frac{2}{9}}{\frac{2}{27}}=3$
$\text{n}=4$
$\text{t}_4=(162)\Big(\frac{1}{3}\Big)^3$
$=\frac{162}{27}$
$=6$
View full question & answer→Question 312 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.
AnswerHere, a, b, c are in G.P., so
$\text{b}^2=\text{ac}$
Now, $\frac{2}{\log_\text{b}\text{m}}=2\log_\text{m}\text{b}$
$=\log_\text{m}\text{b}^2$
$=\log_\text{m}\text{ac}$
$=\log_\text{m}\text{a}+\log_\text{m}\text{c}$
$\frac{2}{\log_\text{b}\text{m}}=\frac{1}{\log_\text{a}\text{m}}+\frac{1}{\log_\text{c}\text{m}}$
$\Rightarrow\frac{1}{\log_\text{b}\text{m}}-\frac{1}{\log_\text{a}\text{m}}=\frac{1}{\log_\text{c}\text{m}}-\frac{1}{\log_\text{b}\text{m}}$
$\Rightarrow\frac{1}{\log_\text{a}\text{m}},\frac{1}{\log_\text{b}\text{m}},\frac{1}{\log_\text{c}\text{m}}\text{ are in A.P.}$
View full question & answer→Question 322 Marks
If a is the G.M. of 2 and $\frac14,$ find a.
Answera is the G.M. between 2 and $\frac14,$
Then,
$\text{a}=\sqrt{2\times\frac14}$
$\text{a}=\sqrt{\frac12}=\frac{1}{\sqrt{2}}$
View full question & answer→