MCQ 11 Mark
Which of the following is/are correct?
Statement A: If the first term and the common ratio of a G.P. are 5 and 2 respectively, then the G.P. is $5,10,20,40 \ldots$
Statement B: Three numbers $a, b, c$ are in G.P. iff $\frac{b}{a}=\frac{c}{b}$
Statement C: In G.P.
$a=1, a_3+a_5=90$ and common ratio $r= \pm 3$
- A
- B
- C
- ✓
All A, B and C are correct
AnswerCorrect option: D. All A, B and C are correct
(d) All A, B and C are correct
Explanation:
Statement A: First term $a=5, r=2$
Then G. P. is $5,10,20,40, \ldots$
Hence, statement A is true.
Statement B: $a, b, c$ are in G.P.
Then, $\frac{\text { Second term }}{\text { First term }}=\frac{\text { Third term }}{\text { Second term }}$
$\frac{b}{a}=\frac{c}{b}$
Hence, statement B is true.
Statement C: Given, $a=1$ and $a_3+a_5=90$
$\begin{array}{l}\Rightarrow a r^2+a r^4=90 \\ \Rightarrow r^2+r^4=90 \\ \Rightarrow r^4+r^2-90=0 \\ \Rightarrow\left(r^2-9\right)\left(r^2+10\right)=0 \\ \Rightarrow r^2=9,-10\end{array}$
(but $r$ is real, so $r^2 \neq-10$ )
$\Rightarrow r^2=9 \Rightarrow r= \pm 3$
Hence, statement C is true.
View full question & answer→MCQ 21 Mark
Which of the following is/are correct?
Statement (A): The next term of an G.P. $\frac{1}{6}, \frac{1}{3}, \frac{2}{3}, \ldots$ is $\frac{4}{3}$
Statement (B): The nth term of this G.P. is $\frac{1}{3}(2)^{-2}$.
Statement (C): The 6th term of this G.P. is $\frac{16}{3}$
- A
- B
- C
- ✓
All A, B and C are correct
AnswerCorrect option: D. All A, B and C are correct
(d) All A, B and C are correct
Explanation:
We have, $\frac{1}{6}, \frac{1}{3}, \frac{2}{3}, \ldots$
$\begin{array}{l}a_1=\frac{1}{6}, a_2=\frac{1}{3}, a_3=\frac{2}{3}, a_4=? \\ \frac{a_2}{a_1}=\frac{\frac{1}{3}}{\frac{1}{6}}=2, \frac{a_3}{a_2}=\frac{\frac{2}{3}}{\frac{1}{3}}=2\end{array}$
Then, $\frac{a_4}{a_3}=2 a_4=2 \times a_3$
$\begin{array}{l}a_4=2 \times \frac{2}{3} \Rightarrow a_4=\frac{4}{3} \\ \ldots a=\frac{1}{6}, r=2\end{array}$
Then, $a n=a r^{n-1}=\frac{1}{6}(2)^{n-1}=\frac{1}{3}(2)^{n-2} \quad . .( i )$
For 20th term,
Putting $n=6$ in equation (i)
$\begin{array}{l}a_{20} \Rightarrow a_6=\frac{1}{3}(2)^{6-2}=\frac{1}{3}(2)^4 \\ \Rightarrow a_6=\frac{16}{3}\end{array}$
View full question & answer→MCQ 31 Mark
Which of the following is/are an geometric progression (G.P.)?
1. $1,-\frac{1}{2},+\frac{1}{4},-\frac{1}{8},+\ldots+\ldots$
2. $\sqrt{3}, 3 \sqrt{3}, 3,3 \sqrt{3}, \ldots$
3. $18,-12,-6, \ldots \ldots$.
Answer(a) Only 1
Explanation:
1. We have, $1,-\frac{1}{2},+\frac{1}{4},-\frac{1}{8},+\ldots+\ldots$
$\begin{array}{l}\frac{a_2}{a_1}=\frac{\frac{-1}{2}}{1}=\frac{-1}{2} \\ \frac{a_3}{a_2}=\frac{\frac{1}{4}}{\frac{-1}{2}}=\frac{-1}{2}, \text { and } \frac{a_4}{a_3}=\frac{\frac{-1}{8}}{\frac{1}{4}}=\frac{-1}{2} \\ \ldots \frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}\end{array}$
So, the given list of numbers is an G.P. with common ratio is $\frac{-1}{2}$.
2. We have, $\sqrt{3}, 3 \sqrt{3}, 3,3 \sqrt{3} \ldots$
$\begin{array}{l}\frac{a_2}{a_1}=\frac{3 \sqrt{3}}{\sqrt{3}}=3, \frac{a_3}{a_2}=\frac{3}{3 \sqrt{3}}=\frac{1}{\sqrt{3}} \\ \ldots \frac{a_2}{a_1} \neq \frac{a_3}{a_2}\end{array}$
So, the given list of numbers does not an G.P.
3. We have, $18,-12,-6, \ldots$
$\frac{a_2}{a_1}=\frac{-12}{18}=\frac{-2}{3}, \frac{a_3}{a_2}=\frac{-6}{-12}=\frac{1}{2}$
$\ldots \frac{a_2}{a_1} \neq \frac{s_3}{a_2}$
So, the given list of numbers doe snot an G.P.
View full question & answer→MCQ 41 Mark
Statement (A): The first and last terms of a G.P. are 3 and 96 respectively. If the common ratio is 2 ; then the number of terms of the G.P. is 6 .
Statement (B): The sum of the n terms is 186.
Which of the statement is valid?
Answer(a) Only A
Explanation:
We have, $a=3, l=96, r=2$
We know that last term $=l=a r^{n-1}$
$\begin{array}{l}\Rightarrow 96=3 \times 2^{n-1} \Rightarrow 2^{n-1}=32 \\ \Rightarrow 2^{n-1}=2^5 \Rightarrow n-1=5=1 n=6\end{array}$
Hence, the number of terms $=6$
Statement A is true
Sum of n terms i.e., 6 terms $=a\left(\frac{r^{n-1}}{r-1}\right)=3\left(\frac{2^6-1}{2-1}\right)$
$=3\left(\frac{64-1}{1}\right)=3 \times 63=189$
Statement B is false
View full question & answer→MCQ 51 Mark
Statement (A): The nth term of the of the G.P. $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots$ is $\frac{5}{2^{\prime \prime}}$
Statement (B): The 20th term of the G.P. $\frac{5}{2}, \frac{5}{9}, \frac{5}{8}, \ldots$ is $\frac{5}{2^{20}}$.
Which of the statement is valid?
Answer(c) Both A and B
Explanation:
We have, $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}$,
$a=\frac{5}{2}, r=\frac{\frac{5}{5}}{\frac{5}{2}}=\frac{1}{2}$
$a_n=a r^{n-1}=\frac{5}{2}\left(\frac{1}{2}\right)^{n-1}=\frac{5}{2^n} \ldots$ (i)
For 20th term,
Putting $n=20$ in equation (i)
$a_{20}=\frac{5}{(2)^{20}}$
View full question & answer→MCQ 61 Mark
The sum of 10 terms of G.P. $1,-\frac{1}{2}, \frac{1}{4},-8, \ldots$ is:
- ✓
$\frac{1025}{1536}$
- B
$\frac{1125}{1236}$
- C
$\frac{1135}{1536}$
- D
$\frac{1236}{1025}$
AnswerCorrect option: A. $\frac{1025}{1536}$
(a) $\frac{1025}{1536}$
Explanation:
Here, $a=1, r=\frac{-1}{2}+1=\frac{-1}{2}, n=10$
Sum of n term of G.P. $=\frac{n\left(1-r^*\right)}{1-r}$
$\begin{array}{l}=1\left[1-\left(\frac{-1}{2}\right)^{10}\right] \\ 1-\left(-\frac{1}{2}\right) \\ =\frac{1+\frac{1}{1024}}{\frac{3}{2}}=\frac{\frac{1025}{1024}}{\frac{3}{2}} \\ =\frac{1025}{1536}\end{array}$
View full question & answer→MCQ 71 Mark
The product of $(64)(64)^{1 / 7}(64)^{1 / 49} \ldots \infty$ is equal to-
Answer(c) 128
Explanation:
Here, (64)(64) $\frac{1}{7}(64) \frac{1}{49} \ldots \infty$
$\left(1+\frac{1}{7}+\frac{1}{49}+\ldots \infty\right)$
Since $1+\frac{1}{7}+\frac{1}{49}+\ldots \infty$ is in G.P.
Here, $a=1, r=\frac{1}{7} \div 1=\frac{1}{7}$
$S_{\infty}=\frac{a}{1-r}=\frac{1}{1-\frac{1}{7}}=\frac{7}{6}$
So, from equation (i),
$\begin{array}{l}64^{\left(1+\frac{1}{7}+\frac{1}{49}+\ldots \infty\right)}={ }_{64^{\frac{7}{6}}} \\ ={ }_{(2)}^{66 \frac{7}{6}} \\ =2^7 \\ =128\end{array}$
View full question & answer→MCQ 81 Mark
If the sum of the G.P. $1,3,9, \ldots$. is 1093 , then the number of terms in the G.P. is:
Answer(c) 7
Explanation:
Sum of G.P. 1, 3, 9, ... is
Let n be the number of terms.
Here, first term $(a)=1$, common ratio $(r)=\frac{3}{1}=3$
Sum of n terms, $\left( S _n\right)=1093$
$\begin{array}{l}\Rightarrow 1093=\frac{a\left(r^n-1\right)}{r-1} \\ \Rightarrow 1093=(0)\left(\frac{3^n-1}{3-1}\right) \\ \Rightarrow 1093 \times 2=3^n-1 \\ \Rightarrow 2186=3 n-1 \\ \Rightarrow 3^n=2186+1 \\ \Rightarrow 3^n=2187 \\ \Rightarrow 3^n=(3) 7 \\ \Rightarrow n=7\end{array}$
View full question & answer→MCQ 91 Mark
The sum of teh first 10 terms of the series $1+\sqrt{2}+2+\ldots$ is:
AnswerCorrect option: C. $31(\sqrt{2}+1)$
(c) $31(\sqrt{2}+1)$
Explanation
Here, first term (a) $=1$ common ratio $=\sqrt{2}+1-\sqrt{2}$
Number of the terms $( n )=10$
Sum of n terms $\left(S_n\right)=\frac{a\left(r^2-1\right)}{r-1}$
$\begin{array}{l}=\frac{\left[(\sqrt{2})^{10}-1\right]}{\sqrt{2}-1} \\ =\frac{(2)^{\frac{1}{2}}{ }^{10}-1}{\sqrt{2}-1}=\frac{2^5-1}{\sqrt{2}-1}=\frac{32-1}{\sqrt{2}-1} \\ =\frac{31(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}\end{array}$
View full question & answer→MCQ 101 Mark
The list of number $\frac{1}{64}, \frac{1}{16}, \frac{1}{4}, 1,-4, \ldots$ is $a$ :
AnswerCorrect option: B. P. with $r=4$
(b) P. with $r=4$
Explanation
Given, list is:
$\frac{1}{64}, \frac{1}{16}, \frac{1}{4}, 1,-4, \ldots$
Here, first term $(a)=\frac{1}{64}$
Common ratio $(r)=\frac{1}{16} \cdot \frac{1}{64}$
$=\frac{1}{16} \times 64=4$
View full question & answer→MCQ 111 Mark
The product of $(32)(32)^{1 / 6}(32)^{1 / 36} \ldots \infty$ is equal to:
Answer(a) 64
Explanation:
$(32)(32)^{1 / 6}(32)^{1 / 36} \ldots \infty$
$=32^{(1+1 / 6+1 / 36+\ldots \infty)}$.
Therefore, $1+\frac{1}{6}+\frac{1}{36}+\ldots \infty$
It is in G.P.
First term $(a)=1$,
Common ratio $(r)=\frac{1}{6}$
Sum of infinite series $\left(S_{\infty}\right)=\frac{a}{1-r}$
$=\frac{1}{1-\frac{1}{6}}=\frac{6}{5}$
So, from equation (i),
$\begin{aligned} & 32^{(1+1 / 6+1 / 36 \ldots \infty)}=32^{6 / 5} \\ = & (2)^{5 \times 6 / 5} \\ = & 2^6 \\ = & 64\end{aligned}$
View full question & answer→MCQ 121 Mark
If the sum of the G.P. $1,4,16, \ldots$ is 341 , then the number of terms in the G.P. is:
Answer(d) 5
Explanation:
Sum of G.P. $1,4,11, \ldots$ is 341 .
Let $n$ be the number of terms.
Here, First term $(a)=1$, Common ratio $(r)=\frac{4}{1}=4$, Sum of $n$ terms $\left(S_n\right)=341$
$\begin{array}{l}341=\frac{a\left(r^2-1\right)}{r-1} \\ 341=\frac{1\left(4^n-1\right)}{4-1}=\frac{4^n-1}{3} \\ 341 \times 3=4^n-1 \\ 1023=4^n-1 \\ 1023+1=4^n \\ 4^n=1024 \\ 4^n=4^5 \\ n=5 .\end{array}$
View full question & answer→MCQ 131 Mark
The sum of the first 8 terms of the series $1+\sqrt{3}+3+\ldots$ is:
- A
$40(\sqrt{3}-1)$
- ✓
$40(\sqrt{3}+1)$
- C
$80(\sqrt{3}-1)$
- D
$80(\sqrt{3}+1)$
AnswerCorrect option: B. $40(\sqrt{3}+1)$
(b)$40(\sqrt{3}+1)$
Explanation:
Here, First term (a) = 1
Common ratio $(r)=\frac{\sqrt{3}}{1}=\sqrt{3}$
Number of terms $(n)=8$
Sum of $n$ terms $\left( S _n\right)=\frac{a\left(r^2-1\right)}{r-1}$
$\begin{array}{l}=\frac{\left.1\left[(\sqrt{3})^8-1\right)\right]}{\sqrt{3}-1} \\ =\frac{81-1}{\sqrt{3}-1}=\frac{80}{\sqrt{3}-1}\end{array}$
Rationalising the denominator
$\begin{array}{l}=\frac{80(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \\ =\frac{80(\sqrt{3}+1)}{3-1}=\frac{80(\sqrt{3}+1)}{2} \\ =40(\sqrt{3}+1).\end{array}$
View full question & answer→MCQ 141 Mark
Which term of the G.P., $18,-12,8, \ldots$ is $\frac{512}{729} ?$
- A
$12^{\text {th }}$
- B
$11^{\text {th }}$
- C
$10^{\text {th }}$
- ✓
$9^{\text {th }}$
AnswerCorrect option: D. $9^{\text {th }}$
(d) $9^{\text {th }}$
Explanation :
Which term of the G.P. $18,-12,8, \ldots \frac{512}{729}$
First term $(a)=18$,
Common ratio $(r)=\frac{-12}{18}-\frac{-2}{3}$
$\begin{array}{l}\therefore a_n=a r^{n-1} \\ \frac{512}{729}=18\left(\frac{-2}{3}\right)^{n-1} \\ \frac{512}{729} \times \frac{1}{18}=\left(\frac{-2}{3}\right)^{n-1} \\ \frac{256}{729 \times 9}=\left(\frac{-2}{3}\right)^{n-1} \\ \left(\frac{-2}{3}\right)^8=\left(\frac{-2}{3}\right)^{n-1}\end{array}$
Comparing $n-1=8$
n=8+1
n=9
It is $9^{\text {th }}$ term.
View full question & answer→MCQ 151 Mark
The $11^{\text {th }}$ term of the G.P. $\frac{1}{8}, \frac{-1}{4}, 2,-1, \ldots$ is:
Answer(c) 128
Explanation:
$11^{\text {th }}$ term of the G.P. $\frac{1}{8}, \frac{-1}{4}, 2,-1, \ldots$ is
Here, First term $(a)=\frac{1}{8}$
Common ratio $(r)=\frac{-1}{4}+\frac{1}{8}$
$=\frac{-1}{4} \times \frac{8}{1}=-2$
$\begin{array}{l}11^{\text {th }} \text { term of G.P. }=a_{11} \\ a_{11}=a r^{n-1} \\ a_{11}=\frac{1}{8} \times(-2)^{11-1} \\ =\frac{1}{2^3} \times(-1)^{10} \times 2^{10} \\ =1 \times 2^{10-3}=2^7 \\ =128 .\end{array}$
View full question & answer→MCQ 161 Mark
If $k, 2(k+1), 3(k+1)$ are consecutive terms of a G.P., then the value of $k$ is:
Answer(b) -4
Explanation:
Since $k, 2(k+1), 3(k+1)$ are in G.P.
$\begin{array}{l}{[2(k+1)]^2=k \times 3(k+1)} \\ 4(k+1)^2=3 k(k+1) \\ 4(k+1)=3 k \\ 4 k+4=3 k \\ 4 k-3 k=-4 \\ k=-4\end{array}$
View full question & answer→MCQ 171 Mark
The list of number $\frac{1}{9}, \frac{1}{3}, 1,-3, \ldots$ is a:
- ✓
G.P. with $r=-3$
- B
G.P. with $r=-1 / 3$
- C
G.P. with $r=3$
- D
AnswerCorrect option: A. G.P. with $r=-3$
(a) G.P. with $r=-3$
Explanation:
$\frac{1}{9}, \frac{1}{3}, 1,-3, \ldots$
Here, First term $(a)=\frac{1}{9}$
Common ratio $(r)=\frac{1}{3}+\frac{1}{9}$
$=\frac{1}{3} \times 9=3$
It is a G.P. with $r=3$.
View full question & answer→