A geometric sequence has a sum to infinity of –3 - AQA - A-Level Maths Mechanics - Question 3 - 2021 - Paper 1

To find the sum to infinity of a geometric sequence, we use the formula:

$S = \frac{a}{1 - r}$

where ( S ) is the sum to infinity, ( a ) is the first term, and ( r ) is the common ratio. Since the original geometric sequence has a sum to infinity of ( -3 ), we know:

1. The sum is given as ( S_1 = \frac{a}{1 - r} = -3 ).

When forming the new sequence, each term of the original sequence is multiplied by ( -2 ). Therefore, the first term becomes ( -2a ), and the common ratio becomes ( -2r ). The new sum to infinity ( S_2 ) is then calculated as:

$S_2 = \frac{-2a}{1 - (-2r)} = \frac{-2a}{1 + 2r}$

Using the original sum's formula, we can express ( a ) in terms of ( S_1 ):

Since ( a = -3(1 - r) ), substituting in gives:

$S_2 = \frac{-2(-3(1 - r))}{1 + 2r} = \frac{6(1 - r)}{1 + 2r}$

To find the result, we assess the limit of the new sequence. Assuming the original series converges, we see that the ratio ( r ) must satisfy ( |r| < 1 ). Given that the original sum to infinity is negative, the common ratio must be positive but less than 1. Therefore, when you multiply by ( -2 ), the entire sequence will flip and converge in the positive direction. Hence, solving for ( S_2 ):

Thus we conclude: ( S_2 = 6 ).