Question

How many terms of the AP 63, 60, 57, 54, ... must be taken so that their sum is 693. Explain the double answer.

Answer 1

Given, AP 63, 60, 57
where, a = 63
and the difference (d) = 60 - 63 = -3
also given that $S_n$ = 693

∴ to find a,
we know
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

By substituting the values of a, d and $S_n$we get;

693 = $\frac{n}{2}$ [ 2 ×63 + (n - 1) - 3 ]

693 = $\frac{n}{2}$[ 126 - 3n + 3 ]

693 = $\frac{n}{2}$ [ 129 - 3n ]

$693=\frac{129n}{2}-\frac{3n^2}{2}$

693 × 2 = 129n - $3n^2$

1386 = 129n -$3n^2$

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

By dividing the whole equation by 3

we get,

$\frac{1386}{3}-\frac{129n}{3}+\frac{3n^2}{3}=\frac{0}{3}$

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

ie; $n^2$ - 43n + 462 = 0

using factorisation method :--

sum = - 43 and product = 462

∴ the numbers are -21 and -22

So by splitting the middle term we get;

( $n^2$ - 21n ) ( - 22n + 462 ) = 0

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

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

∴ n = 21 and n = 22

ie; We get the sum of the given AP as 693 when we take first 21 terms of it or 22 terms of the same AP.

Verification of the Answer

First take n as 21, the $S_{21}=\frac{21}{2}(a+a_{21})$

$a_{21}$ = a + ( 21 - 1 ) d

= 63 + [ 20 × ( - 3 ) ]

= 63 - 60

$a_{21}=3$

∴$S_{21}=\frac{21}{2}$[ 63 + 3 ]

= $\frac{21}{2}$ × 66

$S_{21}$ = 693

So the condition is satisfied for when n = 21

Now check for when n = 22

$S_{22}$ = $\frac{22}{2}$ ( a + $a_{22}$ )

$a_{22}$ = a + ( 22 - 1 ) d

= 63 + [ 21 × ( -3 ) ]

= 63 - 63

$a_{22}$ = 0

We know

$S_{22}$ = $S_{21}$ + $a_{22}$

= 693 + 0

= 693

∴ the condition is satisfied in both the cases

so n = 21 or n = 22