Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P²Rⁿ = Sⁿ

Solution:

Let the G.P be a, ar, ar², ar³ ...., ar^{n - 1}

According to the given information sum of the n terms is,

S = a (rⁿ - 1)/(r - 1)

Product of the n terms

P = aⁿ x r^{1 + 2 + 3 + ... n - 1}

= aⁿr^{n(n - 1)/2}

[∵ sum of first n natural numbers = n (n + 1)/2]

Sum of reciprocals of n terms

R = 1/a + 1/ar + ... + 1/ar^{n - 1}

= (r^{n - 1} + r^{n - 2} + .... + r + 1)/ar^{n - 1}

= 1 (rⁿ - 1)/(r - 1) x 1/ar^{n - 1}

= [rⁿ - 1] /[ar^{n - 1} (r - 1)]

Now,

LHS = P²Rⁿ

= a²ⁿr^{n(n - 1)} x (rⁿ - 1)ⁿ/ [aⁿr^{n(n - 1)} (r - 1)ⁿ]

= aⁿ(rⁿ - 1)ⁿ/(r - 1)ⁿ

= [a(rⁿ - 1)/(r - 1)]ⁿ

= Sⁿ

= RHS

Hence, P²Rⁿ = Sⁿ proved

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NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 14

Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P²Rⁿ = Sⁿ

Summary:

We know that S, P and R are the sum, product and the reciprocal respectively of n terms in a G.P and using that we proved that P²Rⁿ = Sⁿ