Find the sum of the last ten terms of the AP: | vedclass

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(C) To find the sum of the last ten terms,we can write the given $AP$ in reverse order.The given $AP$ is $8, 10, 12, \ldots, 126$.The reverse $AP$ is

Vedclass · Accepted Answer

$1170$

Vedclass · Answer

(C) To find the sum of the last ten terms,we can write the given $AP$ in reverse order.The given $AP$ is $8, 10, 12, \ldots, 126$.The reverse $AP$ is $126, 124, 122, \ldots, 10, 8$.In this reversed $AP$,the first term $(a) = 126$ and the common difference $(d) = 124 - 126 = -2$.We need to find the sum of the first $10$ terms of this reversed $AP$ using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.For $n = 10$:$S_{10} = \frac{10}{2} [2(126) + (10 - 1)(-2)]$$S_{10} = 5 [252 + 9(-2)]$$S_{10} = 5 [252 - 18]$$S_{10} = 5 	imes 234 = 1170$.Thus,the sum of the last ten terms is $1170$.

Find the sum of the last ten terms of the $AP: 8, 10, 12, \ldots, 126$.

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