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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 92 of 106
Marks: +1, -0
Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2RnP^{2}R^{n} = SnS^{n}
Solution:  
Let the G.P. with first term a and common ratio r be a , ar , ar2ar^{2} , ... , arn−1ar^{n-1}
Then, Sum (S) = a(1−rn)1−r\frac{a(1-r^{n})}{1-r}
Product (P) = a . ar ... arn−1ar^{n-1}
= a . a .... a (n times) (r1⋅r2⋯rn−1)(r^{1}\cdot r^{2}\cdots r^{n-1}) = anr1+2+⋯+(n−1)a^{n}r^{1+2+\cdots+(n-1)}
= anrn(n−1)2a^{n}r^{\frac{n(n-1)}{2}}
Since 1 + 2 + ... + n - 1 = n(n−1)2\frac{n(n-1)}{2}
Sum of reciprocals (R)
= 1a+1ar\frac{1}{a}+\frac{1}{ar} + ... + 1arn−1\frac{1}{ar^{n-1}} = 1a[1−(1r)n1−1r]\frac{1}{a}\left[\frac{1-\left(\frac{1}{r}\right)^{n}}{1-\frac{1}{r}}\right] = r(rn−1)a(r−1)rn\frac{r(r^{n}-1)}{a(r-1)r^{n}} = rn−1arn−1(r−1)\frac{r^{n}-1}{ar^{n-1}(r-1)}
Now, P2RnP^{2}R^{n} = [anrn(n−1)2]2[rn−1arn−1(r−1)]n\left[a^{n}r^{\frac{n(n-1)}{2}}\right]^{2}\left[\frac{r^{n}-1}{ar^{n-1}(r-1)}\right]^{n}
= [a2nrn(n−1)][(rn−1)nanrn(n−1)(r−1)n]\left[a^{2n}r^{n(n-1)}\right]\left[\frac{(r^{n}-1)^{n}}{a^{n}r^{n(n-1)}(r-1)^{n}}\right]
= an(rn−1)n(r−1)n\frac{a^{n}(r^{n}-1)^{n}}{(r-1)^{n}} = [a(rn−1)r−1]n\left[\frac{a(r^{n}-1)}{r-1}\right]^{n} = (a(1−rn)1−r)n\left(\frac{a(1-r^{n})}{1-r}\right)^{n} = SnS^{n}
Hence P2RnP^{2}R^{n} = SnS^{n}
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