JEE Main 2019 (Online) 10th April Morning ShiftQuestion 1
If $a_1,a_2,a_3,\ldots,a_n$ are in A.P. and $a_1+a_4+a_7+\ldots+a_{16}=114$, then $a_1+a_6+a_{11}+a_{16}$ is equal to :
Sequence and Series · Arithmetic Progression
Solution
Formulas used. (1) $a_p+a_q=a_1+a_{16}$ whenever $p+q=17$Step 1 - The given sum. $a_1,a_4,a_7,a_{10},a_{13},a_{16}$ pair up as $(a_1,a_{16}),(a_4,a_{13}),(a_7,a_{10})$; each pair has index sum $17$: $$3\left(a_1+a_{16}\right)=114\ \Longrightarrow\ a_1+a_{16}=38 .$$ Step 2 - The required sum. $6+11=17$ too, so $a_6+a_{11}=a_1+a_{16}$: $$a_1+a_6+a_{11}+a_{16}=2\times38=76 .$$ Answer: (D)