How many terms of the AP : 9, 17, 25, ldots m | vedclass

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(B) Let there be $n$ terms of this $AP$.For this $AP$,the first term $a = 9$ and the common difference $d = 17 - 9 = 8$.The sum of $n$ terms is given

Vedclass · Accepted Answer

$12$

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(B) Let there be $n$ terms of this $AP$.For this $AP$,the first term $a = 9$ and the common difference $d = 17 - 9 = 8$.The sum of $n$ terms is given by the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.Substituting the given values: $636 = \frac{n}{2}[2(9) + (n - 1)8]$.$636 = \frac{n}{2}[18 + 8n - 8]$.$636 = \frac{n}{2}[10 + 8n]$.$636 = n(5 + 4n)$.$4n^2 + 5n - 636 = 0$.Solving the quadratic equation using the factorization method: $4n^2 + 53n - 48n - 636 = 0$.$n(4n + 53) - 12(4n + 53) = 0$.$(4n + 53)(n - 12) = 0$.This gives $n = -\frac{53}{4}$ or $n = 12$.Since the number of terms $n$ must be a positive integer,we discard the negative fractional value.Therefore,$n = 12$.

How many terms of the $AP : 9, 17, 25, \ldots$ must be taken to give a sum of $636$?

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