Math, asked by pinkybhuyan2323, 1 year ago

how many terms of the AP...
9, 17, 25 ....must be taken to give a sum of 636

Answers

Answered by Anonymous
11
heyy frnd...

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your answer is given in the picture⤴

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Answered by Anonymous
2

\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}}}


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