MCQ 11 Mark
Assertion $(A):$ If the sum of first two terms of an infinite $GP$ is $5$ and each term is three times the sum of the succeeding terms, then the common ratio is $\frac{1}{4}$.
Reason $(R):$ In an $AP\ 3, 6, 9, 12 .........$ the $10^{th}$ term is equal to $33.$
Reason $(R):$ In an $AP\ 3, 6, 9, 12 .........$ the $10^{th}$ term is equal to $33.$
- ABoth $A$ and $R$ are true and $R$ is the correct explanation of $A.$
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A.$
- ✓$A$ is true but $R$ is false
- D$A$ is false but $R$ is true.
Answer
View full question & answer→Correct option: C.
$A$ is true but $R$ is false
Assertion Let a be the first term and $r(|r| < 1)$ be the common ratio of the $GP.$
$\therefore$ The $GP$ is $a , ar ^2, ar ^2, \ldots$
According to the question,
$T_1+T_2=5$
$\Rightarrow a+a r=5$
$\Rightarrow a(1+r)=5$
and $T_n=3\left(T_{n+1}+T_{n+2}+T_{n+3}+\ldots\right)$
$\Rightarrow a r^{n-1}=3\left(a r^n+a r^{n+1}+a r^{n+2}+\ldots\right)$
$\Rightarrow a r^{n-1}=3 a r^n\left(1+r+r^2+\ldots\right)$
$\Rightarrow 1=3 r\left(\frac{1}{1-r}\right)$
$\Rightarrow 1=r=3 r$
$\Rightarrow r=\frac{1}{4}$
Reason: Given, $3, 6, 9, 12 ...$
Here, $a = 3, d = 6 - 3 = 3$
$\therefore T _{10}= a +(10-1) d$
$=3+9 \times 3$
$=3+27=30$
$\therefore$ The $GP$ is $a , ar ^2, ar ^2, \ldots$
According to the question,
$T_1+T_2=5$
$\Rightarrow a+a r=5$
$\Rightarrow a(1+r)=5$
and $T_n=3\left(T_{n+1}+T_{n+2}+T_{n+3}+\ldots\right)$
$\Rightarrow a r^{n-1}=3\left(a r^n+a r^{n+1}+a r^{n+2}+\ldots\right)$
$\Rightarrow a r^{n-1}=3 a r^n\left(1+r+r^2+\ldots\right)$
$\Rightarrow 1=3 r\left(\frac{1}{1-r}\right)$
$\Rightarrow 1=r=3 r$
$\Rightarrow r=\frac{1}{4}$
Reason: Given, $3, 6, 9, 12 ...$
Here, $a = 3, d = 6 - 3 = 3$
$\therefore T _{10}= a +(10-1) d$
$=3+9 \times 3$
$=3+27=30$