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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
- A \(16\)
- B \(8\)
- C \(4\)
- D \(2\)
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 \(\boxed{a = d}\) \(\therefore {b^2} = ac \Rightarrow {b^2} = a\left[ {\frac{{a + b}}{2}} \right]\) (\(\because \) \(2c=a+b\))…
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