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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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The sum of first three terms of a G.P. is $16$ and the sum of next three terms is $128$. Find the $1^{st}$ term, common ratio and the sum of $n$ terms of the G.P.

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Let the first $6$ terms of the G.P. be $a,ar,ar^2,ar^3,ar^4,ar^5$
It is given that sum of first 3 terms= 16 and
sum of next three terms =128
$i.e.,a+ar+ar^2=16\:\;and\:\:ar^3+ar^4+ar^5=128$
$\Rightarrow\:a(1+r+r^2)=16$...(i) and $ar^3(1+r+r^2)=128$...(ii)
$\Rightarrow\:1+r+r^2=\large\frac{16}{a}$
Step 2
Substituting the value of $1+r+r^2$ in (ii) we get
$\Rightarrow\:ar^3\times \large\frac{16}{a}$$=128$
$\Rightarrow\:r^3=\large\frac{128}{16}$$=8$
$\Rightarrow\:r=2$
Step 3
Substituting the value of $r$ in (i) we get
$1+2+2^2=\large\frac{16}{a}$
$\Rightarrow\:a=\large\frac{16}{7}$
Step 4
Substituting the value of $a$ and $r$ in $S_n$ we get
Sum of $n$ terms of the G.P.$ =S_n=a.\large\frac{r^n-1}{r-1}$
$=\large\frac{16}{7}.\frac{2^n-1}{2-1}=\large\frac{16}{7}$$.(2^n-1)$
answered Feb 26, 2014 by rvidyagovindarajan_1
 

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