JEE Main 2023 (Online) 8th April Evening ShiftQuestion 1
Let $a_n$ be the $n^{\text{th}}$ term of the series $5+8+14+23+35+50+\ldots$ and $S_n=\sum_{k=1}^{n}a_k$. Then $S_{30}-a_{40}$ is equal to
Sequence and Series · Standard Sums (Σ formulas)
Solution
Formulas used. (1) $\sum_{n=1}^{N}n(n-1)=\frac{(N-1)N(N+1)}{3}$(2) Differences in A.P. $\Rightarrow$ general term is a quadraticStep 1 - General term. $5,8,14,23,35,50$ has differences $3,6,9,12,15$: $$a_n=5+3(1+2+\cdots+(n-1))=5+\frac{3n(n-1)}{2}.$$ Step 2 - $S_{30}$. $$S_{30}=5(30)+\frac32\sum_{n=1}^{30}n(n-1)=150+\frac32\cdot\frac{29\cdot30\cdot31}{3}=150+13485=13635 .$$ Step 3 - $a_{40}$. $$a_{40}=5+\frac{3\cdot40\cdot39}{2}=5+2340=2345 .$$ Step 4. $$S_{30}-a_{40}=13635-2345=11290 .$$ Answer: (C)