Question

How many terms of AP : 9, 17, 25....must be taken to give a sum of 636??​

Answer 1

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▪Given :-

An A.P. :-

$\huge\mathfrak{9, 17, 25.\: .\: . \:.}$

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▪To Find :-

Number of Terms to Get Sum of 636

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▪Formula for Sum :-

$\bf S_n= \dfrac{ n}{2} \bigg \{2a +(n-1)d\bigg \}$

Where ,

a = First term

n = Number of terms

d = Common difference

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▪Solution :-

Here

a = 9

d = 8

$\sf S_n$ = 636

Using Formula For Sum :-

$\sf636 = \dfrac{n}{2} \bigg\{2(9) + (n - 1)8 \bigg \} \\ \\ \sf1272 = {n} (18 +8n - 8) \\ \\ \sf1272 = n(10 + 8n) \\ \\ \sf1272 = 10n +{8n}^{2} \\ \\ \sf {8n}^{2} + 10n - 1272 = 0$

Dividing whole equation by two :

$\bf {4n}^{2} + 5n - 636 = 0 \\ \\ \sf {4n}^{2} + 53n - 48n - 636 = 0 \\ \\ \sf n(4n + 53) - 12(4n + 53) = 0 \\ \\ \sf(4n + 53)(n - 12) = 0 \\ \\ \large\purple{ \implies  \underline {\boxed{{\bf n = 12 \: \: or \: \: n = - \frac{ 53}{4} } }}}$

As n is number of terms so it can't be Negative

$\Large\purple{ \implies  \underline {\boxed{{\bf n = 12} }}}$

Hence ,

12 terms of the given AP should be taken to get A Sum of 636.

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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