JEE PYQ: Sequence And Series Question 51
Question 51 - 2019 (10 Apr Shift 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:
(1) 98
(2) 76
(3) 38
(4) 64
Show Answer
Answer: (2)
Solution
$a_1 + a_4 + a_7 + \ldots + a_{16} = 114$
$\Rightarrow 3(a_1 + a_{16}) = 114$
$\Rightarrow a_1 + a_{16} = 38$
Now, $a_1 + a_6 + a_{11} + a_{16} = 2(a_1 + a_{16}) = 2 \times 38 = 76$
BETA
