Question

how many terms of AP 63,60 ,57 must be so that their sum is 693?

Answer 1

Given AP is 63,60,57...

a = 63, d = 60 - 63 = -3.

We know that sn = n/2(2a + (n-1)d)

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

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

                      1386 = n(129-3n)

                      3n^2 - 129n + 1386 = 0

                      3n^2 - 66n - 63n + 1386 = 0

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

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

                      n = 21 (or) 22.


Hope this helps! :-))

Answer 2

For given A.P,

first term,a=63
common diffrence, d= 60-63= -3

sum of n terms, Sn=693


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

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

1386= n[126-3n+3]

1386= n[129-3n]

1386=n[129-3n]

3n²-129n+1386=0

√D=√ [(-129)²-4*3*1386] =√9 =3

n=[-b±√D]÷2a
n= [129±3]÷6

n=(129+3)/6 OR n=(129-3)/6

n= 22,,

n=126/6=21


Hence, no of terms may be 22 or 21