Question

Let $t_{n}$ be $n^{\text{th }}$ term of a geometric progression with common ratio $r$ . If $r < 0,\frac{t_{6}}{t_{8}}=\frac{1}{4}$ and $t_{3}+t_{7}=100$ and $t_{1}=\frac{a}{b}$ , then the value of $a+b$ is ( $a$ and $b$ are coprime numbers $)$ .

• A. $37$
• B. $83$
• C. $50$
• D. $42$

Answer 1

Answer: (D)

Let the geometric progression be
$n,nr,nr^{2},nr^{3},nr^{4},......nt^{n}$
Now,
$t_{1}=n=\frac{a}{b}$
Then,
$\frac{t_{6}}{t_{8}}=\frac{1}{4}$
$\Rightarrow \frac{n r^{5}}{n r^{7}}=\frac{1}{4}$
$\Rightarrow \frac{\left(\right. a / b \left.\right) r^{5}}{\left(\right. a / b \left.\right) r^{7}}=\frac{1}{4}$
$\Rightarrow r=-2$
Also,
$t_{3}+t_{7}=100$
$\Rightarrow nr^{2}+nr^{6}=100$
$\Rightarrow n\left(- 2\right)^{2}+n\left(- 2\right)^{6}=100$
$\Rightarrow 4n+64n=100$
$\Rightarrow 68n=100$
$\Rightarrow n=\frac{100}{68}=\frac{50}{34}=\frac{25}{17}$
Hence, $a=25,b=17$
$a+b=42$