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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of $n$ terms of a G.P. Prove that $P^2R^n=S^n$

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  • $n^{th}$ term of a G.P.$=a.r^{n-1}$
  • Sum of $n$ terms of a G.P. $=a.\large \frac{r^n-1}{r-1}$$=a.\large\frac{1-r^n}{1-r}$
Let the G.P. be $a,ar,ar^2,...............ar^{n-1}$
Given that $S=a+ar+ar^2+..............ar^{n-1}$
$\qquad\:P=a.(ar).(ar^2)..........(ar^{n-1})$ and
$\qquad\:R=\large\frac{1}{a}+\frac{1}{ar}+.........\frac{1}{a.r^{n-1}}$
$\Rightarrow\:S=a.\large\frac{r^n-1}{r-1}$..........(i)
$P=(a.a.a..........n\:times)\times (r.r^2.r^3..........r^{n-1})$
$\qquad = a^n.r^{1+2+.......(n-1)}=a^n.r^{\large\frac{n(n-1)}{2}}$.....(ii)
Since $1+2+......(n-1)=\large\frac{n(n-1)}{2}$
and
$R=\large\frac{1}{a}$$\big[1\large+\frac{1}{r}+\frac{1}{r^2}+.......\frac{1}{r^{n-1}}\big]$
$\qquad=\large\frac{1}{a}$$\large\frac{1-(1/r)^n}{1-1/r}$$=\large\frac{1}{a}.\large\frac{r^n-1}{r^n}\times \frac{r}{r-1}$.......(iii)
Step 2
Now
$P^2=\bigg[a^n.r^{\large\frac{n(n-1)}{2}}\bigg]^2$
$\qquad =a^{2n}.r^{\large\frac{n(n-1)}{2}\times 2}$
$\qquad =a^{2n}.r^{n(n-1)}$
Step 3
$R^n=\bigg[\large\frac{1}{a}.\large\frac{(r^n-1).r}{r^n(r-1)}\bigg]^n$
$\qquad =\large\frac{(r^n-1)^n.r^n}{a^n.r^{n^2}.(r-1)^n}$
Step 4
$S^n=\bigg[a.\large\frac{r^n-1}{r-1}\bigg]^n$$=a^n.\large\frac{(r^n-1)^n}{(r-1)^n}$
$\therefore\:P^2.R^n=\bigg[a^{2n}.r^{n(n-1)}\bigg]$$\times \bigg[\large\frac{(r^n-1)^n.r^n}{a^n.r^{n^2}.(r-1)^n}\bigg]$
$\qquad =a^n.\large\frac{(r^n-1)^n}{(r-1)^n}$$.r^{n-n^2+n(n-1)}$
$\qquad =a^n.\large\frac{(r^n-1)^n}{(r-1)^n}$$.r^{n-n^2+n^2-n)}$
$\qquad =a^n.\large\frac{(r^n-1)^n}{(r-1)^n}$$=S^n$
Hence proved.
answered Mar 31, 2014 by rvidyagovindarajan_1
 

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