Question

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

Answer 1

Answer:

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

=> 693 = n/2 [2.63+(n-1)(-3)]

=> 1386 = n(129-3n)

=> 3n^2-129n+1386 = 0

=> n2-43n+462=0

=> (n-21)(n-22) = 0

=> n = 21 or 22

Answer 2

Answer:

The required number of terms is 21 or 22.

Explanation:

a = 63, d = 60 - 63 = -3, S = 693

Sum of A.P. = n/2[2a + (n - 1)d]

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

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

693 = n/2[126 + (n - 1)(-3)]

693 x 2 = n[126 + (n - 1)(-3)]

1386 = n[126 - 3n + 3]

1386 = 126n - 3n² + 3n

1386 = 129n - 3n²

129n - 3n² - 1386

3n² - 129n + 1386

3n² - 66n - 63n + 1386

3n(n - 22) - 63(n - 22)

(n - 22)(3n - 63)

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

n = 22 or n = 21

So, the sum of 21 terms as well as that of 22 terms is 693. This is because the 22nd term of an A.P. is 0.

a$_{22}$ = 63 + (22 - 1)(-3) = 63 - 63 = 0

Hence, the required number of terms is 21 or 22.

I hope this answer is helpful for you.