Question

The fourth, seventh and the last term of a $G.P$. are $10,80$and $2560,$ respectively. Then the first term and number of terms in the G.P. are, respectively,

• A. $\frac{5}{4},10$
• B. $\frac{5}{4},12$
• C. $\frac{4}{5},10$
• D. $\frac{4}{5},12$

Answer 1

Answer: (B)

Let $a$ be the first term and $r$ be the common ratio of the given $G.P$. Then
$a_4=10,a_7=80$
$\Rightarrow a{r}^3=10$ and $a{r}^6=80$
$\Rightarrow \frac{a{r}^6}{a{r}^3}=\frac{80}{10}$
$\Rightarrow r^3=8$
$\Rightarrow r=2$
Putting $r=2$ in $a{r}^3=10,$ we get: $a=\frac{10}{8}=\frac{5}{4}$
Let there be $n$ terms in the given G.P. then, $a_n=2650$
$\Rightarrow a{r}^{n-1}=2560$
$\Rightarrow \frac{10}{8}\left(2^{n-1}\right)=2560\Rightarrow 2^{n-4}=256\Rightarrow 2^{n-4}=2^8\Rightarrow n-4=8$
$\Rightarrow n=12$