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JEE Mains · Maths · STD 11 - 8. sequence and series

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 in this sequence are equal, then the last term is

  1. A \(16\)
  2. B \(8\)
  3. C \(4\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(8\)

Step-by-step Solution

Detailed explanation

Let \(a,b,c,d\) be four number of the sequence. Now, according to the question \({b^2} = ac\) and \(c-b=6\) and \(a-c=6\) Also, given \(\boxed{a = d}\) \(\therefore {b^2} = ac \Rightarrow {b^2} = a\left[ {\frac{{a + b}}{2}} \right]\) (\(\because \) \(2c=a+b\))…