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Q. In a $G.P.$ the sum of the first and last terms is $66$ , the product of the second and the last but one is $128$ , and the sum of the terms is $126$.
If the decreasing $G.P. $ is considered, then the sum of infinite terms is

Sequences and Series

Solution:

Let a be the first term and $r$ the common ratio of the given $G.P. $
Further, let there be $n$ terms in the given G.P. Then
$a _{1}+ a _{ n }=66 $
$\Rightarrow a + ar ^{ n -1}=66\,\,\, ...(i)$
$a_{2} \times a_{n-1}=128$
$\Rightarrow a r \times a r^{n-2}=128 $
or $ a^{2} v^{n-1}=128$
or $a \times\left( ar ^{ n -1}\right)=128 $
or $ar ^{ n -1}=\frac{128}{ a }$
Putting this value of $a r^{n-1}$ in (i), we get
$a +\frac{128}{ a }=66$
or $ a^{2}-66 a+128=0$
or $(a-2)(a-64)=0$
or $a=2,64$
Putting $a =2$ in (i), we get
$2+2 \times 1^{n-1}=66 $
or $r^{n-1}=32$
Putting $a =64$ in (i), we get
$64+641^{ n -1}=66 $
or $ r ^{ n -1}=\frac{1}{32}$
For an increasing G.P. $r>1$.
Now, $S _{ n }=126$
$\Rightarrow 2\left(\frac{ r ^{ n }-1}{ r -1}\right)=126$
or $\frac{ r ^{ n }-1}{ r -1}=63$
or $\frac{ r ^{ n -1} \times r -1}{ r -1}=63$
or $\frac{32 r-1}{r-1}=63$
or $r=2$
$\therefore 1^{ n -1}=32 $
$\Rightarrow 2^{ n -1}=32=2^{5} $
$\Rightarrow n -1=5$
$\Rightarrow n =6$
For decreasing G.P., $a =64$ and $r =1 / 2$.
Hence, the sum of infinite terms is $64 /\{1-(1 / 2)\}=128 .$
For $a =2, r =2$, terms are $2,4,8,16,32,64$
For $a =64, r =1 / 2$ terms are $64,32,16,8,4,2$.
Hence, difference is $62$ .