JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(a _{ n }\) be the \(n ^{\text {th }}\) term of the series \(5+8+14+23\) \(+35+50+\ldots\) and \(S _{ n }=\sum \limits_{ k =1}^{ n } a _{ k }\). Then \(S _{30}- a _{40}\) is equal to
- A \(11310\)
- B \(11280\)
- C \(11290\)
- D \(11260\)
Answer & Solution
Correct Answer
(C) \(11290\)
Step-by-step Solution
Detailed explanation
\(S _{ n }=5+8+14+23+35+50+\ldots+a_n\) \(S _{ n }=5+8+14+23+35+\ldots+ a _{ n }\) \(O =5+3+6+9+12+15+\ldots . a _{ n }\) \(a _{ n }=5+(3+6+9+\ldots( n -1) \text { terms })\) \(a _{ n }=\frac{3 n ^2-3 n +10}{2}\) \(a _{40}=\frac{3(40)^2-3(40)+10}{2}=2345\)…
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