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
$n,nr,n{r}^2,n{r}^3,n{r}^4,......n{t}^n$
Now,
$t_1=n=\frac{a}{b}$
Then,
$\frac{t_6}{t_8}=\frac{1}{4}$
$\Rightarrow \frac{n{r}^5}{n{r}^7}=\frac{1}{4}$
$\Rightarrow \frac{(a/b)r^5}{(a/b)r^7}=\frac{1}{4}$
$\Rightarrow r=-2$
Also,
$t_3+t_7=100$
$\Rightarrow n{r}^2+n{r}^6=100$
$\Rightarrow n{\left(-2\right)}^2+n{\left(-2\right)}^6=100$
$\Rightarrow 4n+64n=100$
$\Rightarrow 68n=100$
$\Rightarrow n=\frac{100}{68}=\frac{50}{34}=\frac{25}{17}$
Hence, $a=25,b=17$
$a+b=42$