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

AP is $63,60,57,54,....$

Consider

$a=$ First term

$d=$ Common difference

$n=$ number of terms

Here, $a=63,d=60-63=-3$ and sum $=S_n=693$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

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

$693\times 2=n(126-3n+3)$

$1386=n(129-3n)$

$1386=129n-3n^2$

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

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

Which is a quadratic equation

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

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

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

Either, $n-21=0$, then $n=21$

or $n-22=0$, then $n=22$

Number of terms$=21$ or $22$

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

$=63+(22-1)(-3)$

$=63+21\times (-3)=63-63=0$

Which shows that, $22$th term of AP is zero.

Number of terms are $21$ or $22$. So there will be no effect on the sum.