JEE Main2018MathematicsSequences and SeriesActual
Let a 1 , a 2 , a 3 , … … , a 49 be in A . P . such that Σ k ⁡ = 0 1 2 a 4 k + 1 = 416 and a 9 + a 43 = 66 . If a 1 2 + a 2 2 + … + a 17 2 = 140 m , then m is equal to:
Options
- A33
- B66
- C68
- D34
Correct answer
D. 34
Step-by-step solution
a 1 + a 5 +   .   .   .   + a 49 = 416 Let common difference =   d 13 a 1   + 4 d 1 + 2 +   .   .   .   + 12 = 416 13 a 1   + 4 d 12 13 2 = 416 13 a 1   + 24 d 13 = 416 a 1 + 24 d = 32 .   .   .   ( i ) a 9 + a 43 = 66 a 1 + 8 d + a 1 + 42 d = 66 2 a 1 + 50 d = 66 a 1 + 25 d = 33 .   .   .   ( i i ) a 1 = 8       ;     d = 1 ∑ i = 1 17 a i 2 = 140 m ⇒ a 2 + a + d 2 +   . .   .