JEE Main 2018 (Offline)Question 1
Let $a_1,a_2,a_3,\ldots,a_{49}$ be in A.P. such that $\sum_{k=0}^{12}a_{4k+1}=416$ and $a_9+a_{43}=66$. If $a_1^2+a_2^2+\ldots+a_{17}^2=140m$, then $m$ is equal to:
Sequence and Series · Arithmetic Progression · Standard Sums (Σ formulas)
Solution
Formulas used. (1) An odd number of A.P. terms adds up to (number of terms) $\times$ (middle term)(2) $\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}$Step 1 - First condition. $a_1,a_5,\dots,a_{49}$ are $13$ terms with middle term $a_{25}$: $$13a_{25}=416\ \Longrightarrow\ a_{25}=32 .$$ Step 2 - Second condition. $9+43=2\times26$, so $a_9+a_{43}=2a_{26}=66$, $a_{26}=33$. Step 3 - The A.P. $d=1$ and $a_k=32+(k-25)=k+7$. Step 4 - Sum of squares. $$\sum_{k=1}^{17}(k+7)^2=\sum_{r=8}^{24}r^2=\frac{24\cdot25\cdot49}{6}-\frac{7\cdot8\cdot15}{6}=4900-140=4760=140\times34 .$$ Answer: (D)