How Many Terms of The A P 9, 17, 25,..... Must Be Taken So That Their Sum is 636?

Given

Arithmetic Progression, AP= 9, 17, 25, ……

First term, a = 9

Second term = 17

Required Sum = 636

Solution

Common difference, d = Second term - first term =9 - 17=-8

Required sum = 636.

So, substituting the given values to find the value of n, i,e., number of terms.

S_{n} = \frac{n}{2}[2a+(n-1)d]

\Rightarrow 636 = \frac{n}{2}[2.9+(n-1)8]

\Rightarrow 1272 = n (18 + 8n - 8)

\Rightarrow 1272 = n (10 + 8n)

\Rightarrow 1272 = 8n^{2}+ 10 n

\Rightarrow 0 = 8n^{2} + 10n - 1272

\Rightarrow 0 = 8n^{2} + 106n - 96n - 1272

\Rightarrow 0 = 2n (4n + 53) - 24 (4n + 53)

\Rightarrow 0 = (2n - 24) (4n + 53)

\Rightarrow n = \frac{24}{2} = 12 or n = \frac{-53}{4}

For n = \frac{-53}{4} , as the value of n cannot be negative or fraction, this value of n is impossible.

Hence, 12 terms of the AP 9, 17, 25,..... needs to be added to obtain 636.

Conclusion

Hence, for the given AP 9,17,25….; 12 terms need to be added to get the sum of 636.