JEE Main 2018Session, Shift 1MathematicsHardSequence and Series179× since 2002Let a1{a_1}a1, a2{a_2}a2, a3{a_3}a3, ......... ,a49{a_{49}}a49 be in A.P. such that ∑k=012a4k+1=416\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416k=0∑12a4k+1=416 and a9+a43=66{a_9} + {a_{43}} = 66a9+a43=66. a12+a22+.......+a172=140ma_1^2 + a_2^2 + ....... + a_{17}^2 = 140ma12+a22+.......+a172=140m, then m is equal toAOption A. 33BOption B. 66COption C. 68DOption D. 34CheckSkip · show solution