JEE Mains · Maths · STD 11 - 8. sequence and series
If \(\left\{a_{i}\right\}_{i=1}^{n}\) where \(n\) is an even integer, is an arithmetic progression with common difference \(1\) , and \(\sum \limits_{ i =1}^{ n } a _{ i }=192, \sum \limits_{ i =1}^{ n / 2} a _{2 i }=120\), then \(n\) is equal to
- A \(48\)
- B \(96\)
- C \(92\)
- D \(104\)
Answer & Solution
Correct Answer
(B) \(96\)
Step-by-step Solution
Detailed explanation
\(\sum \limits_{ i =1}^{ n } a _{ i }=\frac{ n }{2}\left\{2 a _{1}+( n +1)\right\}=192\) \(\Rightarrow 2 a _{1}+( n -1)=\frac{384}{ n } \ldots-(1)\) \(\sum \limits_{ i =1}^{ n / 2} a _{2 i }=\frac{ n }{4}\left[2 a _{1}+2+\left(\frac{ n }{2}-1\right) 2\right]=120\)…
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