JEE Mains · Maths · STD 11 - 8. sequence and series
If \(S _{ n }=4+11+21+34+50+\ldots\) to \(n\) terms, then \(\frac{1}{60}\left( S _{29}- S _9\right)\) is equal to \(.......\).
- A \(226\)
- B \(220\)
- C \(223\)
- D \(227\)
Answer & Solution
Correct Answer
(C) \(223\)
Step-by-step Solution
Detailed explanation
\(S _{ n }=4+11+21+34+50+\ldots .+ n \text { terms }\) Difference are in \(A.P.\) Let \(T_n=a n^2+b n+c\) \(T _1= a + b + c =4\) \(T _2=4 a +2 b + c =11\) \(T _3=9 a +3 b + c =21\) By solving these \(3\) equations \(a =\frac{3}{2}, b =\frac{5}{2}, c =0\) So…
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