Let {a_1}, {a_2}, {a_3}, ......... ,{a_{49}} be in A.P. such that \sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416 and {a_9} + {a_{43}} = 66. a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m, then m is equal to

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Answer: D. 34 a1, a2, a3 . . . a43 are in AP So, a2 = a1 + d a3 = a1 + 2d . . . a49 =a1 + 48d Now given, {a_9} + {a_{43}} = 66 \Rightarrow \,\,\,\, a1 + 8d + a1 + 42d = 66 \Rightarrow \,\,\,\, 2a1 + 50d = 66 \Rightarrow \,\,\,\, a1 + 25d = 33 . . . . . (1) \sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416 \Rightarrow \,\,\,\, a1 + a5 + a9 + a13 +. . . . . 13 items = 416 \Rightarrow \,\,\,\, a1 + a1 + 4d + a1 + 8d + . . . . a1 + 48d = 416 \Rightarrow \,\,\,\, 13a1 + 4d +8d + 12d + . . . . . 48d = 416 \Rightarrow \,\,\,\, 13a1 + 4 (1+ 2 + 3 + . . . + 12) d = 416 \Rightarrow \,\,\,\,13\,\,a{}_1 + \,4\,\, 	imes \,{{12 	imes 13} \over 2} 	imes d = 416 \Rightarrow \,\,\,\, 13a1 + 24 	imes 13d = 416 \Rightarrow \,\,\,\, a1 + 24 d =32 . . . .(2) Solving (1) and (2) we get, d = 1 and {a_1} = 8 	herefore\,\,\, a1 = 8 a2 = 8 + 1 = 9 a3 = 8 + 2 = 10 . . . a17 = 8 + 16 = 24 Now, a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2\,\, = \,\,140m \Rightarrow \,\,\,\, a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2 = 140\,m \Rightarrow \,\,\,\,\,{8^2}\, + \,\,{9^2}\, + \,{10^2} + ......{(24)^2} = 140\,m We can write above series like this, \Rightarrow \,\,\,\,\, (12 +22 + . . . . +242) - (12 + 22 + . . . . .+ 72) = 140 m \Rightarrow {{24\left( {25} ight)\left( {49} ight)} \over 6} - {{7 	imes 8 	imes 15} \over 6} = 140\,m \Rightarrow \,\,\,\,\, 490 - 140 = 140 m \Rightarrow \,\,\,\,\,4760 = 140 m \Rightarrow \,\,\,\,\, m = 34

Let $${a_1}$$, $${a_2}$$, $${a_3}$$, ......... ,$${a_{49}}$$ be in A.P. such that $$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$$ and $${a_9} + {a_{43}} = 66$$. $$a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m$$, then m is equal to

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