If S_{n} = 4 + 11 + 21 + 34 + 50 + ldots to n | vedclass

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(C) The sequence is $4, 11, 21, 34, 50, \ldots$. The differences between consecutive terms are $7, 10, 13, 16, \ldots$,which form an $A.P.$ with commo

Vedclass · Accepted Answer

$223$

Vedclass · Answer

(C) The sequence is $4, 11, 21, 34, 50, \ldots$. The differences between consecutive terms are $7, 10, 13, 16, \ldots$,which form an $A.P.$ with common difference $3$.Let the $n^{th}$ term be $T_{n} = an^2 + bn + c$.For $n=1, 2, 3$: $a + b + c = 4$$4a + 2b + c = 11$$9a + 3b + c = 21$Solving these,we get $a = \frac{3}{2}, b = \frac{5}{2}, c = 0$.Thus,$T_{n} = \frac{3}{2}n^2 + \frac{5}{2}n = \frac{n(3n+5)}{2}$.$S_{n} = \sum_{k=1}^{n} T_{k} = \frac{3}{2} \sum k^2 + \frac{5}{2} \sum k = \frac{3}{2} \frac{n(n+1)(2n+1)}{6} + \frac{5}{2} \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1) + 5n(n+1)}{4} = \frac{n(n+1)(2n+6)}{4} = \frac{n(n+1)(n+3)}{2}$.Now,$S_{29} = \frac{29 	imes 30 	imes 32}{2} = 29 	imes 15 	imes 32 = 13920$.$S_{9} = \frac{9 	imes 10 	imes 12}{2} = 9 	imes 5 	imes 12 = 540$.$\frac{1}{60}(S_{29} - S_{9}) = \frac{1}{60}(13920 - 540) = \frac{13380}{60} = 223$.

If $S_{n} = 4 + 11 + 21 + 34 + 50 + \ldots$ to $n$ terms,then $\frac{1}{60}(S_{29} - S_{9})$ is equal to $.......$.

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