Math, asked by Vedant1Gangurde, 1 year ago

How many term of the ap 9,17,25,.... must be taken to given a sum of 636

Answers

Answered by Anonymous
5

\bf\huge\boxed{\boxed{\bf\huge\:Hello\:Mate}}}



\bf\huge Let: first\: term\; be\: a \:and\: CD\: = 17 - 9 = 8



\bf\huge => S_{n} = 636



\bf\huge => \frac{N}{2}[2a + (n - 1)d] = 636



\bf\huge => \frac{N}{2}[2\times 9 + (n - 1)8] = 636



\bf\huge => \frac{N}{2} (8n - 10) = 636



\bf\huge => n(4n + 5) = 636



\bf\huge => 4n^2 + 5n + 636 = 0



\bf\huge => n = \frac{-5 + \sqrt{25 - 4\times 4\times -636}}{2\times 4}



\bf\huge = \frac{-5 + \sqrt{25 + 10176}}{8}



\bf\huge = \frac{- 5 + \sqrt{10201}}{8}



\bf\huge = \frac{-5 + 101}{8}



\bf\huge = \frac{96}{8} , \frac{-106}{8}



\bf\huge = 12 , \frac{-53}{4}



\bf\huge But\: n \:cannot\: be\: Negative



\bf\huge => n = 12



\bf\huge Hence\:Sum\: of\: 12\: terms\: is\: 636




\bf\huge\boxed{\boxed{\:Regards=\:Yash\:Raj}}}


Answered by subhrajena189
3

Answer:

HEY MATE...... .

PLZ REFER TO THE ATTACHMENT FOR ANSWER

HOPE IT HELPS YOU....... ❤️❤️

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