Question

If $\frac{a_{k+1}}{a_k}$ is constant for every $k\ge 1$. If $n>m\Rightarrow a_n>a_m$ and $a_1+a_n=66,a_2{a}_{n-1}=128$ and $\sum _{i=1}^n{a}_i=126$, then

• A. $n=6$
• B. $n=5$
• C. $\frac{a_{k+1}}{a_k}=2$
• D. $\frac{a_{k+1}}{a_k}=4$

Answer 1

The correct option is A $n=6$

given

$\frac{a_{k+1}}{a_k}$ is constant implies it is a $GP$ series let the constant be $r$

and given

$a_1{+a}_n=66,a_2{a}_{n-1}=128$

if $a_1=a$ then $a_n=a{r}^{n-1}$

therefore

$a(1+r^{n-1})=66,a^2{r}^{n-1}=128$

also given

${\sum }_{i=1}^n{a}_i=126\Rightarrow \frac{a(r^n-1)}{r-1}=126$

$a+a{r}^{n-1}=66$

$\Rightarrow a+\frac{128}{a}=66$

$\Rightarrow a=2,r^{n-1}=32$

$\Rightarrow r=2,n=6$