Question

In an increasing $G.P$, the sum of the first and the last term is $66$, the product of the second term and the second last term is $128$, and the sum of all the term is $126$. How many terms are there in the progression?

Answer 1

G.P. is increasing, $\therefore r>1$,
${a+ar}^{n-1}=66$ ....................(1)
$ar.a{r}^{n-2}=a^2{r}^{n-1}=128$
$\therefore r^{n-1}=\frac{128}{a^2}$
Putting in $\left(1\right)$, we get $a+\frac{128}{a}=66$
$\therefore a^2-66a+128=0$
or $(a-2)(a-64)=0$
$\therefore a=2or64$

and $r^{n-1}=32or1/32$
We reject the second value as $r>1$

$\therefore r^{n-1}=32$
$Sum=\frac{a(r^n-1)}{r-1}=126$

or $\frac{2(32r-1)}{r-1}=126$ $[\because r^{n-1}=32]$
$\therefore 32r-1=63r-63$
$\therefore r=2$

and $r^{n-1}=32$
$\therefore 2^{n-1}=2^5$

$\Rightarrow n-1=5$

$Hence,n=6$