Question

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

Answer 1

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...!!!

Answer 2

A.P is given by 63,60,57.......

So, a= 63 , d= -3 and Sn=693

using the formula,

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

693 = (n/2)*[2(63)+(n-1)(-3)]

⇒ 1386 = n[126 -3n+3]

⇒ 1386 = n[-3n+129]

⇒ 3n² -129n + 1386 = 0

⇒ n² - 43n + 462 = 0

⇒ n² - 21n - 22n + 462 = 0

⇒ n(n-21) -22(n-21) = 0

⇒ (n-21)(n-22) = 0

⇒ n=21 and n=22