JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(a_1=8, a_2, a_3, \ldots a_n\) be an \(A.P.\) If the sum of its first four terms is \(50\) and the sum of its last four terms is \(170\) , then the product of its middle two terms is
- A \(753\)
- B \(752\)
- C \(754\)
- D \(751\)
Answer & Solution
Correct Answer
(C) \(754\)
Step-by-step Solution
Detailed explanation
\(a_1+a_2+a_3+a_4=50\) \(\Rightarrow 32+6 d=50\) \(\Rightarrow d=3\) and, \(a_{n-3}+a_{n-2}+a_{n-1}+a_n=170\) \(\Rightarrow 32+(4 n -10) \cdot 3=170\) \(\Rightarrow n =14\) \(a _7=26, a _8=29\) \(\Rightarrow a _7 \cdot a _8=754\)
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