Consider an infinite geometric series with first term a and common rat

1.	Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second term is \(\frac{3}{4}\) , then(a) a = 2, r = \(\frac{1}{2}\) (b) a = 2, r = \(\frac{3}{8}\) (c) a = 1, r = \(\frac{3}{4}\)(d) None of these
Answer» (c) a = 1, r = \(\frac{3}{4}\) Sum of an infinite G.P = \(\frac{a}{1-r},\) where first term = a, common ratio = r and \| r \| < 1 Given, \(\frac{a}{1-r}\) = 4 and ar = \(\frac{3}{4}\) ⇒ a = 4 – 4r and a = \(\frac{3}{4r}\) ⇒ \(\frac{3}{4r}\) = 4 - 4r ⇒ 3 = 16r - 16r² ⇒ 16r² – 16r + 3 = 0 ⇒ (4r – 3) (4r – 1) = 0 ⇒ 4r = 3 or 4r = 1 ⇒ r = \(\frac{3}{4}\) or \(\frac{1}{4}\) Now when r = \(\frac{3}{4}\), a = \(\frac{3}{4\times\frac{3}{4}}\) = 1 r = \(\frac{1}{4}\), a = \(\frac{3}{4\times\frac{1}{4}}\) = 3 ∴ (a, r) = \(\big(1,\frac{3}{4}\big)\) or \(\big(3,\frac{1}{4}\big)\).

Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second term is \(\frac{3}{4}\) , then(a) a = 2, r = \(\frac{1}{2}\) (b) a = 2, r = \(\frac{3}{8}\) (c) a = 1, r = \(\frac{3}{4}\)(d) None of these

Answer»

(c) a = 1, r = \(\frac{3}{4}\)

Sum of an infinite G.P = \(\frac{a}{1-r},\) where first term = a, common ratio = r and | r | < 1

Given, \(\frac{a}{1-r}\) = 4 and ar = \(\frac{3}{4}\)

⇒ a = 4 – 4r and a = \(\frac{3}{4r}\)

⇒ \(\frac{3}{4r}\) = 4 - 4r ⇒ 3 = 16r - 16r²

⇒ 16r² – 16r + 3 = 0

⇒ (4r – 3) (4r – 1) = 0

⇒ 4r = 3 or 4r = 1 ⇒ r = \(\frac{3}{4}\) or \(\frac{1}{4}\)

Now when r = \(\frac{3}{4}\), a = \(\frac{3}{4\times\frac{3}{4}}\) = 1

r = \(\frac{1}{4}\), a = \(\frac{3}{4\times\frac{1}{4}}\) = 3

∴ (a, r) = \(\big(1,\frac{3}{4}\big)\) or \(\big(3,\frac{1}{4}\big)\).