JEE Main 2026 (Online) 23rd January Evening ShiftQuestion 1
Let $\sum_{k=1}^{n}a_k=\alpha n^2+\beta n$. If $a_{10}=59$ and $a_6=7a_1$, then $\alpha+\beta$ is equal to :
Sequence and Series · Arithmetic Progression
Solution
Formulas used. (1) $a_n=S_n-S_{n-1}$ for $n\ge2$, and $a_1=S_1$Step 1 - General term. $$a_n=\alpha n^2+\beta n-\left[\alpha(n-1)^2+\beta(n-1)\right]=\alpha(2n-1)+\beta .$$ For $n=1$ this gives $\alpha+\beta=S_1$, so it holds for every $n$. Step 2 - First condition. $$a_{10}=19\alpha+\beta=59 .$$ Step 3 - Second condition. $$a_6=7a_1\ \Longrightarrow\ 11\alpha+\beta=7\alpha+7\beta\ \Longrightarrow\ 4\alpha=6\beta\ \Longrightarrow\ \alpha=\frac{3\beta}{2}.$$ Step 4 - Solve. $$19\cdot\frac{3\beta}{2}+\beta=59\ \Longrightarrow\ \frac{59\beta}{2}=59\ \Longrightarrow\ \beta=2,\ \alpha=3 .$$ $$\alpha+\beta=5 .$$ Answer: (D)