Let a1,a2,a3,…a49 be in A.P such that ∑k=012 a4k+1=416 and a9+a43=66. If a12+a22+..+a172=140m then m is

∑k=012 a4k+1=416⇒1322a1+48d=416⇒a1+24d=32  →1⇒a9+a43=66⇒2a1+50d=66   →21 and 2⇒a1=8,d1=1∑r=117 ar2=∑r=117 [8+(r−1)]2=140m⇒∑r=117 (r+7)2=140m⇒∑r=117 r2+49+14r=140m⇒17.18.356+14(17×18)2+49×17=140m⇒m=34