2.5 + 5.9 + 8.13 + 11.17 + ldots to 10 terms | vedclass

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(B) The given series is $S = 2.5 + 5.9 + 8.13 + 11.17 + \ldots$ to $10$ terms.The $n^{th}$ term $T_n$ is the product of the $n^{th}$ term of the arith

Vedclass · Accepted Answer

$4555$

Vedclass · Answer

(B) The given series is $S = 2.5 + 5.9 + 8.13 + 11.17 + \ldots$ to $10$ terms.The $n^{th}$ term $T_n$ is the product of the $n^{th}$ term of the arithmetic progression $(2, 5, 8, 11, \ldots)$ and $(5, 9, 13, 17, \ldots)$.$T_n = (3n - 1)(4n + 1) = 12n^2 + 3n - 4n - 1 = 12n^2 - n - 1$.Sum $S = \sum_{n=1}^{10} (12n^2 - n - 1) = 12 \sum_{n=1}^{10} n^2 - \sum_{n=1}^{10} n - \sum_{n=1}^{10} 1$.Using standard summation formulas:$\sum_{n=1}^{10} n^2 = \frac{10(11)(21)}{6} = 385$.$\sum_{n=1}^{10} n = \frac{10(11)}{2} = 55$.$\sum_{n=1}^{10} 1 = 10$.$S = 12(385) - 55 - 10 = 4620 - 65 = 4555$.

$2.5 + 5.9 + 8.13 + 11.17 + \ldots$ to $10$ terms $=$

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