Question

Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P.
with common difference six. If first and last terms of this sequence are equal, then the last
term is :​

Answer 1

AnSwer:

Here it is given that 1st and last term is equal.

Let us consider the term be x .

The last 3 numbers are said to be in AP with common difference 6.

★So, 2nd and 3rd term would be (x - 12) and (x - 6) respectively.

Now since it is given that first 3 numbers are in GP,

Thus,

$\implies \sf \: \frac{x - 12}{x} = \frac{x - 6}{x - 12} \\ \\ \implies \sf \: {(x - 12)}^{2} = x(x - 6) \\ \\ \implies \sf \: {x}^{2} - 24x + 144 = {x}^{2} - 6x \\ \\ \implies \sf \:18x = 144 \\ \\ \implies \sf \:x = 8$

Thus, the numbers are 8,-4,2,8

Last number = 8 + 8 + 2 - 4 = 18 - 4 = 14