Math, asked by jay376, 1 year ago

How many terms of A. P. 63,60,57,.......taken so that their sum is 693?

Answers

Answered by Nikki57
18
Hey!

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Given A.P = 63 , 60 , 57 ....

Here,

a = 63
d = 60 - 63 = -3

We know,

S = n/2 [ 2a + (n - 1) d ]

693 = n/2 [ 2×63 + (n - 1) -3 ]

693 = n/2 ( 126 - 3n + 3 )

693 × 2 = n ( 129 - 3n )

1386 = n ( 129 - 3n)

1386 = 129n - 3n²

1386 - 129n +3n²

Factorising -:

3n² - 66n - 63n + 1386 = 0

(3n² - 66n) - (63n - 1386) = 0

3n (n - 22) - 63 ( n - 22) = 0

(3n -63) (n - 22) = 0

So,

3n - 63 = 0
n = 63/3
n = 21

Or

n - 22 = 0
n = 22

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Hope it helps...!!!

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Answered by jaideep88
9
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