If the sum to infinity of the series 3 + 5r + 7r2 + ...... ∞ is \(\f

1.	If the sum to infinity of the series 3 + 5r + 7r2 + ...... ∞ is \(\frac{44}{9}\), find the value of r.
Answer» 3 + 5r + 7r² + ...... ∞ is an infinite arithmetico-geometric series, where a = 3, d = 2, common ratio (r) = r. Sum to infinity of an A.G.P., with first term of A.P, as a, common difference d and common ratio r is S_∞= \(\frac{a}{1-r}\) + \(\frac{dr}{(1-r)^2}\) ∴ \(\frac{44}{9}\) = \(\frac{3}{1-r}\) + \(\frac{2r}{(1-r)^2}\) ⇒ \(\frac{44}{9}\) = \(\frac{3(1-r)+2r}{(1-r)^2}\) ⇒ 44 (1 – r)² = 9 (3 – r) ⇒ 44 (1 – 2r + r²) = 27 – 9r ⇒ 44 – 88r + 44r² = 27 – 9r ⇒ 44r² – 79r + 17 = 0 ⇒ (4r – 1) (11r – 17) = 0 ⇒ r = \(\frac{1}{4}\) or \(\frac{17}{11}\) r ≠ \(\frac{17}{11}\) as it is not possible to find the sum of an infinite G.P. with \| r \| > 1. So r = \(\frac{1}{4}\).

If the sum to infinity of the series 3 + 5r + 7r2 + ...... ∞ is \(\frac{44}{9}\), find the value of r.

Answer»

3 + 5r + 7r² + ...... ∞ is an infinite arithmetico-geometric series, where

a = 3, d = 2, common ratio (r) = r.

Sum to infinity of an A.G.P., with first term of A.P, as a, common difference d and common ratio r is

S_∞= \(\frac{a}{1-r}\) + \(\frac{dr}{(1-r)^2}\)

∴ \(\frac{44}{9}\) = \(\frac{3}{1-r}\) + \(\frac{2r}{(1-r)^2}\) ⇒ \(\frac{44}{9}\) = \(\frac{3(1-r)+2r}{(1-r)^2}\)

⇒ 44 (1 – r)² = 9 (3 – r) ⇒ 44 (1 – 2r + r²) = 27 – 9r

⇒ 44 – 88r + 44r² = 27 – 9r ⇒ 44r² – 79r + 17 = 0

⇒ (4r – 1) (11r – 17) = 0 ⇒ r = \(\frac{1}{4}\) or \(\frac{17}{11}\)

r ≠ \(\frac{17}{11}\) as it is not possible to find the sum of an infinite G.P. with | r | > 1. So r = \(\frac{1}{4}\).