Let \(a_{1}, a_{2}, a_{3}, \ldots ., a_{49}\) be in A.P. such that \(\sum_{k=0}^{12} a_{4 k+1}=416\) and \(a_{9}+a_{43}=66\). If \(\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\ldots . .+\mathrm{a}_{17}^{2}=140 \mathrm{~m}\), then m is equal to :

JEE Main · 2018 · 08 Apr 2018 · Offline · Sequences and Series

Let

a_{1}, a_{2}, a_{3}, \dots ., a_{49}

be in A.P. such that

\sum_{k = 0}^{12} a_{4 k + 1} = 416

and

a_{9} + a_{43} = 66

. If

a_{1}^{2} + a_{2}^{2} + \dots .. + a_{17}^{2} = 140 m

, then m is equal to :

Answer: (D)

Let the progression have first term

a_{1}

and common difference

d

. The condition

\sum_{k = 0}^{12} a_{4 k + 1} = 416

becomes

13 a_{1} + 4 d (0 + 1 + \dots + 12) = 13 a_{1} + 312 d = 416

, that is

a_{1} + 24 d = 32

. The condition

a_{9} + a_{43} = 66

gives

2 a_{1} + 50 d = 66

, that is

a_{1} + 25 d = 33

. Subtracting,

d = 1

and

a_{1} = 8

, so

a_{r} = 7 + r

and

\sum_{r = 1}^{17} a_{r}^{2} = \sum_{s = 8}^{24} s^{2} = \frac{24 \cdot 25 \cdot 49}{6} - \frac{7 \cdot 8 \cdot 15}{6} = 4900 - 140 = 4760

. Then

140 m = 4760

gives

m = 34

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option