Let a1, a2, a3,....,a49 be in A.P. such that Σ_(k=0)^(12) a_(4k+1) = 416 and a9 + a43 = 66. If a21 + a22 +.....+ a217 = 140m, then m is equal to

⇒nthterman=a+(n𕒵)d⇒a9+a43=66∴a+8d+a+42d=66∴a+25d=33... (1)Now,∑12k=0a4k+1=416∴13a+312d=416(using sum of AP on the common difference parts)∴a+24d=32... (2)from (1) and (2), we getd=1anda=8∴∑17k=1a2k=82+92⋯+242=(12+22⋯+242)−(12+22⋯+72)=24×25×496𕒻×8×156(using sum of squares ofnnatural numbers isn(n+1)(2n+1)6=4760=140×34Hence, the required answer is34.