How many terms of the A.P. -6, -frac{11}{2}, | vedclass

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(C) Let the sum of $n$ terms of the given $A.P.$ be $S_n = -25$.The formula for the sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.Here,the firs

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$20$ or $5$

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(C) Let the sum of $n$ terms of the given $A.P.$ be $S_n = -25$.The formula for the sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.Here,the first term $a = -6$ and the common difference $d = -\frac{11}{2} - (-6) = -\frac{11}{2} + 6 = \frac{1}{2}$.Substituting the values into the formula:$-25 = \frac{n}{2}[2(-6) + (n-1)(\frac{1}{2})]$$-50 = n[-12 + \frac{n}{2} - \frac{1}{2}]$$-50 = n[\frac{n-25}{2}]$$-100 = n^2 - 25n$$n^2 - 25n + 100 = 0$Factoring the quadratic equation:$n^2 - 20n - 5n + 100 = 0$$n(n-20) - 5(n-20) = 0$$(n-20)(n-5) = 0$Thus,$n = 20$ or $n = 5$.Both values are positive integers,so both are valid.

How many terms of the $A.P.$ $-6, -\frac{11}{2}, -5, \ldots$ are needed to give the sum $-25$?

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