Sequences

[sukhman]

The sum to infinity of a GP is 27 and sum of squares of the terms is 243 , Find the common ratio
a)1/2 b)2 c)3 d)1/3 e) 1

[vishalwin]

OA: A

point 1: a + ar + ar^2.........infinity = a/ (1-r) = 27 -----(1)

point 2: a^2 + a^2 r^2......infinity = a^2/(1-r^2) = 243

point 2 can be written as a/(1-r) * a/(1+r) = 243

27 * a/(1+r) = 243

so, a/(1+r) = 9 ------(2)



divide (1) by (2)

(1+r)/(1-r) = 3

on solving this you will get r = 1/2 = common ratio


1+r = 3-3r

4r = 2

r = 1/2

[sukhman]

I thought point 2 be a^2/(1-r)^2 = 243
but it turned out a^2 + a^2 r^2......infinity = a^2/(1-r^2) = 243

[Matt@VeritasPrep]

Another option:

a + ar + ar² + ... = 27

a² + a²r² + a²r� + ... = 243

so ...

a * (1 + r + r² + ...) = 27

a² * (1 + r² + r� + ...) = 243

We know that if r < 1, we can write (1 + r + r² + ...) as 1 / (1 - r).

Assuming this to be true (otherwise the series diverges), we can write the first series as

a/(1 - r) = 27

In the second GP, since we have ((r²)� + (r²)¹ + (r²)² + ...), our ratio is r², so we have (1 + r² + r� + ...) = 1 / (1 - r²), and

a²/(1 - r²) = 243

Using the difference of squares, this gives us

a/(1 + r) = 9

So we have a/(1 + r) = 9 and a/(1 - r) = 27, or

9(1 + r) = 27(1 - r), or r = 1/2