How Many Terms of The A P 9, 17, 25,..... Must Be Taken So That Their Sum is 636?
AP stands for Arithmetic progression. It is referred to as the sequence of numbers whose differences of consecutive numbers are equal. An arithmetic sequence is another name for Arithmetic Progression.
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.
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Conclusion
Hence, for the given AP 9,17,25….; 12 terms need to be added to get the sum of 636.
