Let `a_1,a_2,a_3…., a_49` be in A.P . Such that `Sigma_(k=0)^(12) a_(4k+1)=416` and `a_9+a_(43)=66` .If `a_1^2+a_2^2 +…+ a_(17)` = 140 m then m is equal to

Options

A. 33
B. 66
C. 68
D. 34

Answer

D

Explanation

Let the first term of A.P be 'a ' and the common difference be d .
`underset(k=0)overset(12)Sigma a_(4k+1)= 416`
`rArr a_1+a_5+a_9+……+ a_(49)=46`
`rArr 13/2 xx (a_1-a_49)=416`
`rArr a+a++48 d = 64`
`therefore a+24d = 32`
`a_9+a_43=66`
`rArr a+8d+a+42d=66`
`rArr a+25d = 33`
Solving (i) and (ii) ,we get
d=1,a=8
Now 140 m `=a_1^2+a_^2+....+a_(17)^2`
`8^2+9^2+10^2+.....+24^2`
`(1^2+2^2+....+7^2+8^2+....+24^2)-(1^2+2^2 +...+7^2)`
`(24xx 25 xx 49)/(6)-(7xx 8xx 15)/(6)`
=4900-140
=140 (35-1)
`=140 xx 34`
`therefore m= 34`