Question

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

• A. first term is $\frac{5}{2}$
• B. common ratio is $2$
• C. number of terms is $12$
• D. ${10}^{th}$ term is $640$

Answer 1

Answer: (B), (C), (D)

The correct options are
B common ratio is $2$
C number of terms is $12$
D ${10}^{th}$ term is $640$

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,
$a_n=2560$
$\Rightarrow a{r}^{n-1}=2560$
$\Rightarrow \frac{5}{4}\left(2^{n-1}\right)=2560$
$\Rightarrow 2^{n-1}=2048\Rightarrow 2^{n-1}=2^{11}$
$\Rightarrow n=12$

$a_{10}=a{r}^{10-1}=\frac{5}{4}\times 2^9=640$