Question 12 Marks
If the second term of a $G.P$. is $4$ then find the product of the first three terms of the $G.P.$
AnswerLet $a$ and $r$ be the first term and common ratio.Here, the second term is $4.$ i.e. $T_{2}=4$.Now putting $n=1, n=2$ and $n=3$ in the general term $\mathrm{T}_{n}=a r^{n-1}$, we get $\mathrm{T}_{1}=a, \mathrm{~T}_{2}=a r$ and $\mathrm{T}_{3}=a r^{2}$Hence, $\mathrm{T}_{1} \times \mathrm{T}_{2} \times \mathrm{T}_{3}=a \times a r \times a r^{2}$$=a^{3} \times \mathrm{r}^{3}$$=(a r)^{3}$$=(4)^{3}$$=64$Hence, the product of the first three terms of the $G.P.$ is $64 .$
View full question & answer→Question 22 Marks
Find the 8th term of the G. P. $\frac{1}{8}, \frac{1}{4}, \frac{1}{2}, \ldots . .$
AnswerHere $a=\frac{1}{8}$ and $r=\frac{\frac{1}{4}}{\frac{1}{8}}=2$. Now, 8th term is to be found, so $n=8$
Putting the values of $a, r$ and $n$ in the general term $\mathrm{T}_{n}=a r^{n-1}$, we get$\begin{aligned}T_{8} &=\frac{1}{8} \times(2)^{g-1} \\&=\frac{1}{8} \times(2)^{7} \\&=\frac{1}{8} \times 128 \\&=16\end{aligned}$Hence, the 8 th term of the G. P. is 16 .
View full question & answer→Question 32 Marks
Find the 5th term of a GP. 9, —6, 4, ....
AnswerHere, $a=9$ and $r=\frac{-6}{9}=\frac{-2}{3} .$ The 5 th term is to be found, so $n=5$
Putting values of $a, r$ and $n$ in the general term $\mathrm{T}_{n}=a r^{n-1}$, we get
$
\begin{aligned}
\mathrm{T}_{5} &=9 \times\left(\frac{-2}{3}\right)^{5-1} \\
&=9 \times\left(\frac{-2}{3}\right)^{4} \\
&=9 \times \frac{16}{81} \\
&=\frac{16}{9}
\end{aligned}
$
Hence, the 5 th term of the G. P. is $\frac{16}{9}$.
View full question & answer→Question 42 Marks
The first term and the common ratio of a GP. are 4 and -2 respectively. If its nth term is -128, find the value of n.
AnswerFor the given G.P. first term $a=4$, common ratio $r=-2$ and the $n$-th term is $-128$ i.e. $\mathrm{T}_{n}=-128$. Here, $\mathrm{T}_{n}=-128$ $\therefore \quad a r^{n-1}=-128 \quad\left(\because \mathrm{T}_{n}=a r^{n-1}\right)$ $\therefore \quad 4 \times(-2)^{n-1}=-128 \quad(\because a=4$ and $r=-2)$ $\therefore \quad(-2)^{n-1}=\frac{-128}{4}=-32$ $\therefore \quad(-2)^{n-1}=(-2)^{5}$ Equating the powers on both the sides, we get $\therefore n-1=5$ $\therefore n=6$
View full question & answer→Question 52 Marks
If in a G. P., $S_{n}=\frac{4}{3}\left(3^{n}-1\right)$, find $T_{3}$.
AnswerWe know $\quad \mathrm{T}_{n+1}=\mathrm{S}_{n+1}-\mathrm{S}_{n}$$\begin{aligned} \therefore \quad \mathrm{T}_{3} &=\mathrm{S}_{3}-\mathrm{S}_{2} \\ &=\frac{4}{3}\left(3^{3}-1\right)-\frac{4}{3}\left(3^{2}-1\right) \\ &=\frac{4}{3}[(27-1)-(9-1)] \\ &=\frac{4}{3}[26-8] \\ &=\frac{4}{3}(18) \\ \therefore \quad \mathrm{T}_{3} &=24 \end{aligned}$
View full question & answer→Question 62 Marks
If in a G. P., $S_{n}=\frac{2}{3}\left(4^{n}-1\right)$, obtain $T_{n+1}$.
AnswerWe know $\mathrm{T}_{n+1}=\mathrm{S}_{n+1}-\mathrm{S}_{n}$$\begin{aligned} &=\frac{2}{3}\left[4^{n+1}-1\right]-\frac{2}{3}\left[4^{n}-1\right] \\ &=\frac{2}{3}\left[\left(4^{n+1}-1\right)-\left(4^{n}-1\right)\right] \\ &=\frac{2}{3}\left[4^{n+1}-1-4^{n}+1\right] \\ &=\frac{2}{3}\left[4^{n+1}-4^{n}\right] \\ &=\frac{2}{3} \times 4^{n}[4-1] \\ \therefore \quad \mathrm{T}_{n+1} &=2\left(4^{n}\right) \end{aligned}$
View full question & answer→Question 72 Marks
If for a G.P., $\mathbf{T}_{n}=2^{n+1}$, obtain $\mathrm{S}_{4}$.
AnswerHere, $\mathrm{T}_{n}=2^{n+1}$$\therefore \quad \mathrm{T}_{1}=2^{1+1}=4, \mathrm{~T}_{2}=2^{2+1}=8, \mathrm{~T}_{3}=2^{3+1}=16, \ldots$Here, the first term $a=4$ and the common ratio $r=\frac{8}{4}=2 .$ Sum of the first four terms is required i.e. $n=4$.Putting the values of $a, r$ and $n$ in $\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{(r-1)}$, we get$\begin{aligned}\mathrm{S}_{4} &=\frac{4\left(2^{4}-1\right)}{(2-1)} \\&=\frac{4(16-1)}{1} \\&=4 \times 15 \\&=60\end{aligned}$Thus, the sum of the first four terms of the G. P. is 60 .
View full question & answer→Question 82 Marks
If three positive numbers k + 1, 3k — 1, 5k + 1 are in GR, find the value of k.
AnswerHere, $k+1,3 k-1,5 k+1$ are in G. P., so $\frac{T_{2}}{T_{1}}=\frac{T_{3}}{T_{2}}=$ common ratio $r$$\begin{array}{ll}\therefore & \frac{3 k-1}{k+1}=\frac{5 k+1}{3 k-1} \\ \therefore & (3 k \quad 1)^{2}-(5 k+1)(k+1) \\ \therefore & 9 \mathrm{k}^{2}-6 k+1=5 k^{2}+6 k+1 \\ \therefore & 4 k^{2}-12 k=0 \\ \therefore & 4 k(k-3)=0 \\ \therefore & 4 k=0 \text { or } k-3=0 \\ \therefore & k=0 \text { or } k=3\end{array}$But $k=0$ is not possible because for $k=0$, the value of $(3 k-1)$ will be $-1$ which a negative number. Hence, $k=3$
View full question & answer→Question 92 Marks
A person deposits Rs. 20,000 in a bank at the compound interest rate of 8% per
annum. Find the amount the person receives after 5 years.
AnswerThe person deposits ₹ 20,000 i.e. $a=20,000=\mathrm{T}_{1}$Amount received after first year $20,000 \times 1.08=21,600=\mathrm{T}_{2}$The rate of interest is $8 \%$ so the common ratio $r=1.08$We want to find the amount received after 5 years,so $n=6$Putting the values of $a, r$ and $n$ in the general term $\mathrm{T}_{n}=a r^{n-1}$, we get,$\begin{aligned}\mathrm{T}_{6} &=20,000 \times(1.08)^{6-1} \\&=20,000\times(1.08)^{5} \\&=29386.5615 \\&=29386.56\end{aligned}$Thus, amount received after 5 years will be ₹ $29,386.56$.
View full question & answer→Question 102 Marks
If for a G.P., $S_{4}=10 S_{4}$ find ' $r$ '.
AnswerHere, $\mathrm{S}_{\mathrm{s}}=10 \mathrm{~S}_4$
$
\begin{array}{ll}
\therefore & \frac{a\left(r^2-1\right)}{(r-1)}=10\left[\frac{a\left(r^4-1\right)}{(r-1)}\right] \\
\therefore & r^3-1=10\left(r^4-1\right) \\
\therefore & \left(r^4-1\right)\left(r^4+1\right)=10\left(r^4-1\right) \\
\therefore & r^4+1-10 \\
\therefore & r^4=9 \\
\therefore & r^2-3 \\
\therefore & r= \pm \sqrt{3}
\end{array}
$
View full question & answer→Question 112 Marks
Find the sum of the first four terms of the GP. whose first term is 3 and
common ratio is 2.
AnswerHere, the first term $a=3$ and the common ratio $r=2$. Sum of the first four terms is required i.e. $n=4$.Putting the values of $a, r$ and $n$ in $\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{(r-1)}$, we get
$\begin{aligned}\mathrm{S}_{4} &=\frac{3\left[2^{4}-1\right]}{(2-1)} \\&=\frac{3(16-1)}{1} \\&=3 \times 15 \\&=45\end{aligned}$Thus, the sum of the first four terms of the G. P. is 45 .
View full question & answer→Question 122 Marks
A person deposits ₹ 10,000 in a bank in the year 2009, ₹ 20,000 in the year 2010,
₹ 40,000 in the year 2011. The amount of deposits in any given year is twice the amount
of deposit of the previous year. What would be the amount of deposit in the year 2014 ?
AnswerA person deposits $₹ 10,000$ in the first year i.e. 2009 , so the first term $a=10000$. Every year the amount of deposits is twice the previous year's amount, so the common ratio $(r)=2$. We want to find the amount deposited in the year 2014 i.e. the sixth year, so $n=6$.Putting the values of $a, r$ and $n$ in the general term $\mathrm{T}_{n}=a r^{n-1}$, we get$\begin{aligned}\mathrm{T}_{6} &=10000(2)^{6-1} \\&=10,000(2)^{5} \\&=10,000 \times 32 \\&=3,20,000\end{aligned}$Thus the amount deposited in the year 2014 is $₹ 3,20,000$.
View full question & answer→Question 132 Marks
For the numbers $a, b, c, d$, if $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$, show that the numbers $a, b, c, d$ are in G.P.
AnswerSuppose $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=r$, where $r$ is a non-zero constant.$\therefore \quad b=a r, c=b r, d=c r$$\therefore \quad c=(a r) r=a r^{2}$ and $d=\left(a r^{2}\right) r=a r^{3}$If we take $\mathrm{T}_{1}=a, \mathrm{~T}_{2}=b=a r, \mathrm{~T}_{3}=c=a r^{2}, \mathrm{~T}_{4}=d=a r^{3}$ then we see that $\mathrm{T}_{1}, \mathrm{~T}_{2}, \mathrm{~T}_{3}$ and $\mathrm{T}_{4}$ are in G.P. Hence $a, b, c, d$ are in G.P.
View full question & answer→Question 142 Marks
The numbers 2, G, 50 are in GP. Find the value of G.
AnswerHere $\mathrm{T}_{1}=2, \mathrm{~T}_{2}=\mathrm{G}$ and $\mathrm{T}_{3}=50$The common ratio of the G. P. is $=\frac{T_{2}}{T_{2}}=\frac{T_{3}}{T_{2}}$$\begin{array}{ll}\therefore \quad & \frac{G}{2}=\frac{50}{G} \\\therefore &\mathrm{G}^{2}=100 \\\therefore \quad & \mathrm{G}=\pm 10 \\\therefore \quad & \mathrm{G}=10 \text { or }-10\end{array}$
View full question & answer→Question 152 Marks
Find the required term of the following sequence using sequence formula: (i) $2,10,50, \ldots$ (6th term) (ii) $100,50,25, \ldots$ (7th term) (iii) $\frac{1}{3}, \frac{2}{9}, \frac{4}{27} \ldots \ldots$ (8th term) (iv) $2,2 \sqrt{2}, 4, \ldots$ (5th term)
Answer(i) $2,10,50, \ldots$ (6th term)
Answer:
$2,10,50, \ldots \mathrm{T}_{6}=$ ?
Here. $a=2, r=\frac{10}{2}=5, n=6$
$T_{6}=a \cdot r^{n-1} $
$\therefore T_{6}=2 \cdot(5)^{6-1} $
$=2(5)^{5} $
$=2 \times 3125 $
$=6250$
(ii) $100,50,25, \ldots$ (7th term)
Answer:
Here, $a=100 ; r=\frac{50}{100}=\frac{1}{2}=n=7$
$T_{n}=a \cdot r^{n-1}$
$\therefore T_{7}=100\left(\frac{1}{2}\right)^{7-1}$
$\left.=100\left(\frac{1}{2}\right)^{6}\right] $
$=\frac{100}{64}=\frac{25}{16}$
(iii) $\frac{1}{3}, \frac{2}{9}, \frac{4}{27} \ldots \ldots$ (8th term)
Answer:
Here, $a=\frac{1}{3} ; r=\frac{\frac{2}{9}}{\frac{1}{3}}=\frac{2}{9} \times \frac{3}{1}=\frac{2}{3} ; n=8$
$\begin{aligned}\mathrm{T}_{n} &=a \cdot r^{n-1} \\\therefore \mathrm{T}_{8} &=\frac{1}{3}\left(\frac{2}{3}\right)^{8-1} \\&=\frac{1}{3}\left(\frac{2}{3}\right)^{7} \\&=\frac{1}{3} \times \frac{128}{2187}=\frac{128}{6561}\end{aligned}$
(iv) $2,2 \sqrt{2}, 4, \ldots$ (5th term)
Answer:
Here. $\mathrm{a}=2 ; \mathrm{r}=\frac{2 \sqrt{2}}{2}=\sqrt{2} ; \mathrm{n}=5$
$T_{n}=a . r^{n-1}$
$\therefore T_{5}=2(\sqrt{2})^{5-1}=2(\sqrt{2})^{4}=\left(2^{\frac{1}{2}}\right)^{4}=2(2)^{2}=2 \times 4=8$
View full question & answer→Question 162 Marks
If the common ration of a geometric progression is $2$ , find the ration of its $7^{th}$ and $3$ rd terms.
Answer$r=2, \frac{T_7}{T_s}=?$
$T_n=a \cdot r^{n-1}$
$\therefore \frac{T_7}{T_s}=\frac{a \cdot(2)^{7-1}}{a(2)^{s-1}}=\frac{2^6}{2^2}=2^4=16$
View full question & answer→Question 172 Marks
For a $G.P.a =\frac{4}{9}$ and $r =-\frac{3}{2}$. Find $T_3$.
Answer$a=\frac{4}{9^{\prime}} r=-\frac{3}{2^2} T_3=?$
$T_n=a \cdot r^{n-1}$
$\therefore T_3=\left(\frac{4}{9}\right)\left(-\frac{3}{2}\right)^{3-1}=\left(\frac{4}{9}\right)\left(-\frac{3}{2}\right)^2=\frac{4}{9} \times \frac{9}{4}=1$
Hence, the third term of a $G.P.$ is $1.$
View full question & answer→Question 182 Marks
Which term of the $G.P. 4,12, 36, ...$ is $324 ?$
AnswerThe $G.P.$ is $4,12,36, \ldots$.
$\therefore a =4, r =\frac{12}{4}=3, T_{ n }=324, n =?$
$T_{ n }= a \cdot r ^{ n -1}$
$\therefore 324=4 \cdot(3)^{ n -1}$
$\therefore \frac{324}{4}=(3)^{ n -1}$
$\therefore 81=(3)^{ n -1}$
$\therefore(3)^4=(3)^{ n -1}$
$\therefore 4= n -1(\because$ Base are equal, then Index are equal)
$\therefore n =5$
Hence, the fifth term of the $G.P.$ is $324.$
View full question & answer→Question 192 Marks
For a $G.P., a =2$ and $r =3$; then find the sum of first four terms.
Answer$a=2, r=3, n=4$
$S_n=\frac{a\left[r^n-1\right]}{r-1}$
Putting $n=4 ; a=2 ; r=3$ in the formula, $S_4=\frac{2\left[(3)^4-1\right]}{3-1}=\frac{2[81-1\rfloor}{2}=80$
Hence, the sum of first four terms of a $G.P.$ is $80\ .$
View full question & answer→Question 202 Marks
For a $G.P., T_1=2$ and the product of the first three terms is $1000$ . Find the common ratio.
AnswerFor a $G.P., T_1=2 \therefore a =2$
Now, $T _1 \times T _2 \times T _3= ax$ ar $x ar ^2$
$\therefore 1000=2 \times 2 r \times 2 r^3$
$\therefore 1000=8 r^3$
$\therefore r^3=\frac{1000}{8}=125=(5)^3$
$\therefore r=5(\because$ Index are equal then base are equal.)
Hence, Common ratio $r=5$
View full question & answer→Question 212 Marks
If in a $G.P$., common ratio is $1$ and $S_8=24$, find the first term of the $G.P.$
Answerif in a $G.P.$ common ratio $r=1$, then
$S_n=\text { n.a }$
$\therefore S_8=8 a$
$\therefore 24=8 a$
$\therefore a=3$
Hence, the first term of the $G.P.$ is $3.$
View full question & answer→Question 222 Marks
AnswerIf the consecutive terms of a geometric progression are expressed in terms of $a$ and $r$ as $a, a r, a r^2, \ldots$, , then $a+a r+a r^2$ $+\ldots .+a \cdot r^{n-1}$ is called geometric series. It is denoted by $S_n$, i.e., $S_n=a+a r+a r^2+\ldots .+a \cdot r^{n-1}$.
View full question & answer→Question 232 Marks
Define Geometric progression.
Answerif a and $r$ non-zero real numbers, the sequence whose $n$th term is $T_n=$ a.r $r^{n-1}$ for an integer $n \geq 1$, is called the geometric progression. Where. $a =$ first term of sequence and $r =$ common ratio of sequence.
View full question & answer→Question 242 Marks
In a geometric progression if the ratio of the sum of 3rd term and 2nd term to the difference of 3rd term and 2nd term is 3 : 2, Find the common ratio.
Question 252 Marks
The common ratio of a G.P. is -3 and the sum of its first five terms is 183. Write the G.P.
Answera = 3; Sequence : 3, -9, 27, -81, 243, ……
View full question & answer→Question 262 Marks
The common ratio of a G.P. is 3 and the sum of its first four terms is 400, write the G.P.
Question 272 Marks
The sum of first five terms of a G.P. is 3875. If these five terms are respectively 16a, 8a, 4a, 2a, a, find the value of ‘a’.
Question 282 Marks
The first five terms of a G.P. are such that the product of second and fifth term is 243. Find the product of its third and fourth term.
Question 292 Marks
The first four terms of G.P. are such that the product of its second and third term is 1326. Find the product of its first and fourth term.
Question 302 Marks
The second term of a G.P. is 1.2. Find the product of its first three terms.
Question 312 Marks
The third term of a G.P. is 2, find the product of its first five terms.
Question 322 Marks
If the numbers k, 2.5, 6.25 are in G.P., find k.
Question 332 Marks
The sum of first 4 terms of a G.P. is 40 and its first terms is 1. Find the common ratio of the progression.
Question 342 Marks
If the sum of first 6 terms of a G.P. is 28 times the sum of Its first 3 terms, find the common ratio.
Question 352 Marks
The sum of how many terms of G.P. 1, 4, 16,... will be 341?
Question 362 Marks
The sum of first n terms of a G.P. is 255. If its last term Is 128 and common ratio is 2, find n.
Question 372 Marks
The first term of a G.P. is 900 and its common ratio is 0.03. Find the sum of first four terms of the sequence.
Question 382 Marks
The first term of a G.P. is $\frac{1}{2}$ and Its common ratio is $\frac{2}{3}$. Find the sum of first 6 terms of the progression.
View full question & answer→Question 392 Marks
The first term of a G.P. is 1000 and its common ratio is 0.1. Find the sum of first five terms of the series.
Question 402 Marks
Obtain the sum of the terms as required in the following G.P: $\frac{1}{8}+\frac{1}{4}+\frac{1}{2}+1+\ldots(10$ terms $)$
View full question & answer→Question 412 Marks
Obtain the sum of the terms as required in the following G.P:: $625+125+25+\ldots(7$ terms $)$
View full question & answer→Question 422 Marks
Obtain the sum of the terms as required in the following G.P:: $\sqrt{2}+\sqrt{6}+\sqrt{18}+\ldots \ldots(10$ terms $)$
Answer$\frac{\sqrt{2}(242}{\sqrt{3}-1}$
View full question & answer→Question 432 Marks
Obtain the sum of the terms as required in the following G.P.: $1 \frac{1}{2}+2 \frac{1}{4}+3 \frac{3}{8}+\ldots \ldots .(6$ terms $)$
View full question & answer→Question 442 Marks
Obtain the sum of the terms as required in the following G.P.: $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots .(8$ terms $)$
View full question & answer→Question 452 Marks
Obtain the sum of the terms as required in the following G.P.: $1024+512+256+\ldots \ldots(12$ terms)
View full question & answer→Question 462 Marks
Obtain the sum of the terms as required in the following G.P. $1+3+9+27+\ldots . .(8$ terms $)$
View full question & answer→Question 472 Marks
Find the nth term of G.P. $1.1,11,110,1100, \ldots \ldots \ldots$. and hence obtain its 8th term.
Answer$T_{n}=1.1(10)^{n-1}, T_{8}=11000000$
View full question & answer→Question 482 Marks
Find the nth term of G.P. $32,-16,8,-4$, and hence obtain its 15 th term.
Answer$\bar{T}_{n}=32\left(-\frac{1}{2}\right)^{n-1}, T_{15}=\frac{1}{512} .$
View full question & answer→Question 492 Marks
Find the nth term of G.P. $0.243,0.081,0.027, \ldots \ldots .$ and hence obtain the 8th term.
Answer$\overline{T_{n}}=0.243\left(\frac{1}{3}\right)^{n-1}, T_{8}=\frac{1}{9000} .$
View full question & answer→Question 502 Marks
Find the nth term of G.P. $\frac{5}{8}, \frac{5}{4}, \frac{5}{2}, 5, \ldots \ldots$ and hence obtain the 9th term.
Answer$\overline{T_{n}}=\frac{5}{8}(2)^{n-1} \cdot T_{9}=160$
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