Question 12 Marks
Find the sum of G.P. $: 3, 6, 12, …… 1536.$
AnswerGiven G.P. $: 3, 6, 12, .........., 1536$
Here,
First trem, $a = 3$
$\text { common ratio, } r =\frac{6}{3}=2( r >1) $
$\text { Last term, } I =1536 $
$ \therefore \text { Required sum }=\frac{\mid r-a}{r-1}$
$=\frac{1536 \times 2-3}{2-1} $
$=3072-3 $
$ =3069$
View full question & answer→Question 22 Marks
The first term of a G.P. is -3 and the square of the second term is equal to its $4^{\text {th }}$ term. Find its $7^{\text {th }}$ term.
AnswerFor a $G.P.$
First term, $a = -3$
It is given that,
$(2^{nd} term)^2 = 4^{th}$ term
$\Rightarrow (ar)^2 = ar^3$
$\Rightarrow a^2r^2 = ar^3$
$\Rightarrow a = r$
$\Rightarrow r = -3$
Now, $7^{th}$ term $= ar^6 = -3 x (-3)^6 = -3 x 729 =-2187$
View full question & answer→Question 32 Marks
If a, b and c are in $G.P$., prove that:
log a, log b and log c are in $A.P.$
AnswerHere, $a, b, c$ are in $G.P.$
$\Rightarrow b^2 = ac$
Taking log on both sides, we get
$\log(b^2) = \log(ac)$
$\Rightarrow 2log b = \log a + \log c$
$\Rightarrow \log b + \log b = \log a + \log c$
$\Rightarrow \log b - \log a = \log c - \log b$
$\Rightarrow \log a, \log b and \log c are in A.P.$
View full question & answer→Question 42 Marks
Find the third term from the end of the G.P.
$\frac{2}{27}, \frac{2}{9}, \frac{2}{3}, \ldots \ldots \ldots 162$
AnswerGiven G.P. : $\frac{2}{27}, \frac{2}{9}, \frac{2}{3}, \ldots, 162$
Here,
Common ratio,$r=\frac{\frac{2}{9}}{\frac{2}{27}}=3$
Last term, l = 162
$\therefore 3^{rd}$ term from an end $=\frac{1}{ r ^2}=\frac{162}{(3)^2}=\frac{162}{9}=18$
View full question & answer→Question 52 Marks
Find the seventh term from the end of the series: $\sqrt{2}, 2,2 \sqrt{2}, \ldots \ldots, 32$
AnswerGiven series : $\sqrt{2}, 2,2 \sqrt{2}, \ldots \ldots ., 32$.
Now, $\frac{2}{\sqrt{=}} \sqrt{2}, \frac{2 \sqrt{2}}{2}=\sqrt{2}$
So, the given series is a G.P. with common ratio, $r=\sqrt{2}$
Here, last term, $I =32$
$\therefore 7^{\text {th }}$ term from an end $=\frac{1}{r^6}=\frac{32}{(\sqrt{2})^6}=\frac{32}{8}=4$
View full question & answer→Question 62 Marks
If the first and the third terms of a $G.P.$ are $2$ and $8$ respectively. find its second term.
AnswerLet the first term of the $G.P$. be a and its common ratio be r .
$\therefore 1^{\text {st }} \text { term }=a=2$
and, $3^{\text {rd }}$ term $=8$
$\Rightarrow a r^2=8$
Now, $\frac{a r^2}{a}=\frac{8}{2}$
$\Rightarrow r^2=4$
$\Rightarrow r=2$
when $a =2$ and $r =2$
$2^{\text {nd }} \text { term }=a r=2 \times 2=4$
View full question & answer→Question 72 Marks
Find the sixth term of the series :
$2^2, 2^3, 2^4,...................$
AnswerGiven sequence: $2^2, 2^3, 2^4,...................$
Now,
$\frac{2^3}{2^2}=2, \frac{2^4}{2^3}=2$
Since $\frac{2^3}{2^2}=\frac{2^4}{2^3}=\ldots \ldots . .=2$, the given sequence is a G.P. with first term, $a=2^2=4$ and common ratio, $r=2$.
Now, $t_n = ar^{n-1}$
$\therefore t_6 = 4 x (2)^5 = 4 x 32 = 128$
View full question & answer→Question 82 Marks
Find the $n^{th}$ term of the series:
$1, 2, 4, 8, .......................$
AnswerGiven series: $1, 2, 4, 8,...............$
Now,
$\frac{2}{1}=2, \frac{4}{2}=2, \frac{8}{4}=2$
Since $\frac{2}{1}=\frac{4}{2}=\frac{8}{4}=\ldots \ldots . .=2$,
The given sequence is $A G.P$. with first term, $a=1$ and common
ratio, $r = 2.$
Now, $t_n = ar^{n-1}$
$\Rightarrow t_n = 1 x 2^{n-1} = 2^{n-1}$
View full question & answer→Question 92 Marks
Find the $10^{th}$ term of the G.P. :
$12,4,1 \frac{1}{3}$,........
AnswerGiven G.P. : $12,4,1 \frac{1}{3}$,........
Here,
First term, a = 12
Common ratio, $r =\frac{4}{12}=\frac{1}{3}$
Now, $t_n = ar^{n-1}$
$\Rightarrow t _{10}=12 \times\left(\frac{1}{3}\right)^9=12 \times \frac{1}{19683}=\frac{4}{6561}$
View full question & answer→Question 102 Marks
Find the seventh term of the G.P. :
$1, \sqrt{3}, 3,3 \sqrt{3} \ldots$
AnswerGiven G.P. $1, \sqrt{3}, 3,3 \sqrt{3} \ldots \ldots \ldots$
Here,
First term, a = 1
Common ratio, $r =\frac{\sqrt{3}}{1}=\sqrt{3}$
Now, $t_n = ar^{n-1}$
$\Rightarrow t_7=1 \times(\sqrt{3})^6=27$
View full question & answer→Question 112 Marks
Find the $9^{th}$ term of the series :
$1, 4, 16, 64, ..........................$
AnswerGiven sequence: $1, 4, 16, 64...............$
Now,
$\frac{4}{1}=4, \frac{16}{4}=4, \frac{64}{16}=4$
Since $\frac{4}{1}=\frac{16}{4}=\frac{64}{16}=\ldots \ldots=4$ the given sequence is a G.P. with the first term, a = 1 and common ratio, $r = 4.$
Now, $t_n = ar^{n-1}$
$\Rightarrow t_9= 1 x 4^8 = 65536$
View full question & answer→Question 122 Marks
Find, which of the following sequence from a G.P. :
9, 12, 16, 24,................
AnswerGiven sequence: 9, 12, 16, 24,................
Now,
$\frac{12}{9}=\frac{4}{3}, \frac{16}{12}=\frac{4}{3}, \frac{24}{16}=\frac{3}{2}$
Since $\frac{24}{8}=\frac{72}{24} \neq \frac{216}{72}$, the given sequence is not a G.P.
View full question & answer→Question 132 Marks
Find the next two terms of the series :
$2 - 6 + 18 - 54............$
AnswerGiven series: $2 - 6 + 18 - 54............$
Now,
$\frac{-6}{2}=-3, \frac{18}{-6}=-3, \frac{-54}{18}=-3$
Since $\frac{-6}{2}=\frac{18}{-6}=\frac{-54}{18}=\ldots .=-3$, the given sequence is a G.P. with first term, $a=2$ and commom ratio, $r = -3.$
Now, $t_n = ar^{n-1}$
$\therefore $ Next two terms :
$5^{th}$ term $= 2 x (-3)^4 = 2 x 82 = 162$
$6^{th}$ term $= 2 x (-3)^5 = 2 x (-243) = -486$
View full question & answer→Question 142 Marks
Find, which of the following sequence from a G.P. :
$\frac{1}{8}, \frac{1}{24}, \frac{1}{72}, \frac{1}{216}, \ldots \ldots \ldots \ldots \ldots$
AnswerGiven sequence: $\frac{1}{8}, \frac{1}{24}, \frac{1}{72}, \frac{1}{216} \ldots \ldots . . .$.
Now,
$\frac{\frac{1}{24}}{\frac{1}{8}}=\frac{1}{3}, \frac{\frac{1}{72}}{\frac{1}{24}}=\frac{1}{3}, \frac{\frac{1}{216}}{\frac{1}{72}}=\frac{1}{3}$
Since $\frac{\frac{1}{24}}{\frac{1}{8}}=\frac{\frac{1}{72}}{\frac{1}{24}}=\frac{\frac{1}{216}}{\frac{1}{72}}=\ldots \ldots \ldots . .=\frac{1}{3}$, the given sequence is a G.P. with common ratio $\frac{1}{3}$.
View full question & answer→Question 152 Marks
Find, which of the following sequence from a G.P. :
8, 24, 72, 216,................
AnswerGiven sequence: 8, 24, 72, 216,................
Now,
$\frac{24}{8}=3, \frac{72}{24}=3, \frac{216}{72}=3$
Since $\frac{24}{8}=\frac{72}{24}=\frac{216}{72}=\ldots \ldots . .=3$, the given sequence is a G.P. with common ratio 3.
View full question & answer→Question 162 Marks
Find the G.P. whose first term is $64$ and next term is $32.$
AnswerFirst term, $a = 64$
Second term, $t_2 = 32$
$\Rightarrow ar = 32$
$\Rightarrow 64 x r = 32$
$\Rightarrow r=\frac{32}{64}=\frac{1}{2}$
$\therefore$ Required G.P. = $a, ar, ar^{n-1}, ar^{n-2},..........$
$=64,32,64 \times\left(\frac{1}{2}\right)^2, 64 \times\left(\frac{1}{2}\right)^3, \ldots \ldots \ldots . . .$
$=64,32,16,8, \ldots \ldots . .$
View full question & answer→