The sum of how many terms of the A.P. 9, 17, | vedclass

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(B) Given $A.P.$ is $9, 17, 25, \ldots$Here,the first term $a = 9$ and common difference $d = 17 - 9 = 8$.Let the sum of $n$ terms be $S_n = 636$.The

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$12$

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(B) Given $A.P.$ is $9, 17, 25, \ldots$Here,the first term $a = 9$ and common difference $d = 17 - 9 = 8$.Let the sum of $n$ terms be $S_n = 636$.The formula for the sum of $n$ terms is $S_n = \frac{n}{2} [2a + (n - 1)d]$.Substituting the values: $636 = \frac{n}{2} [2(9) + (n - 1)8]$.$636 = \frac{n}{2} [18 + 8n - 8]$.$636 = \frac{n}{2} [10 + 8n]$.$636 = n(5 + 4n)$.$4n^2 + 5n - 636 = 0$.Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{-5 \pm \sqrt{25 - 4(4)(-636)}}{8} = \frac{-5 \pm \sqrt{25 + 10176}}{8} = \frac{-5 \pm \sqrt{10201}}{8} = \frac{-5 \pm 101}{8}$.Since $n$ must be positive,$n = \frac{96}{8} = 12$.Thus,the sum of $12$ terms is $636$.

Part $I$	Part $II$
$1$. Sixth term of the $A.P.$ $1, 3, 5, 7, \ldots$	$a$. $105$
$2$. Eleventh term of the $A.P.$ $3, 6, 9, 12, \ldots$	$b$. $11$
$3$. Sixteenth term of the $A.P.$ $4, 6, 8, 10, \ldots$	$c$. $33$
$4$. Twenty-first term of the $A.P.$ $5, 10, 15, 20, \ldots$	$d$. $34$

The sum of how many terms of the $A.P.$ $9, 17, 25, \ldots$ is $636$?

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