Question

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

Answer 1

In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the number of terms.

A.P. 63, 60, 57

So here, let us find the number of terms whose sum is 693. For that, we will use the formula,

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

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

The first term (a) = 63

The sum of n terms (S_n) = 693

Common difference of the A.P. (d) = `a_2 - a_1`

= 60 - 63

= -3

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

`639 = n/2 [2(63) + (n -2)(-3)]`

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

`693 = (n/2) [129 - 3n`]

`693(2) = 129n - 3n^2`

So, we get the following quadratic equation,

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

`n^2 - 43n + 462 = 0`

On solving by splitting the middle term, we get,

`n^2 - 22n - 21n + 462 = 0`

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

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

Further,

n - 22 =0 

n = 22

or

n - 21 = 0

n = 21

Here 22 nd term wil be

`a_22 = a_1 + 21d`

= 63 + 21(-3)

= 63 - 63

= 0

So, the sum of 22 as well as 21 terms is 693. Therefore, the number of terms (n) is 21 or 22