Let a_1=8, a_2, a_3, ldots, a_n be an A.P. If | vedclass

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(C) Given the first term $a_1 = 8$ and the sum of the first four terms $S_4 = 50$.Using the formula for the sum of an $A.P.$,$S_4 = \frac{4}{2}(2a_1 +

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$754$

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(C) Given the first term $a_1 = 8$ and the sum of the first four terms $S_4 = 50$.Using the formula for the sum of an $A.P.$,$S_4 = \frac{4}{2}(2a_1 + 3d) = 50$.$2(16 + 3d) = 50$ $\Rightarrow 16 + 3d = 25$ $\Rightarrow 3d = 9$ $\Rightarrow d = 3$.Now,the sum of the last four terms is $a_{n-3} + a_{n-2} + a_{n-1} + a_n = 170$.This can be written as $(a_1 + (n-4)d) + (a_1 + (n-3)d) + (a_1 + (n-2)d) + (a_1 + (n-1)d) = 170$.$4a_1 + (4n - 10)d = 170$.$4(8) + (4n - 10)(3) = 170 \Rightarrow 32 + 12n - 30 = 170$.$12n + 2 = 170$ $\Rightarrow 12n = 168$ $\Rightarrow n = 14$.The middle two terms of an $A.P.$ with $n=14$ are $a_7$ and $a_8$.$a_7 = a_1 + 6d = 8 + 6(3) = 8 + 18 = 26$.$a_8 = a_1 + 7d = 8 + 7(3) = 8 + 21 = 29$.The product of the middle two terms is $a_7 	imes a_8 = 26 	imes 29 = 754$.

Let $a_1=8, a_2, a_3, \ldots, a_n$ be an $A.P.$ If the sum of its first four terms is $50$ and the sum of its last four terms is $170$,then the product of its middle two terms is

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