Question

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

Answer 1

Given A.P is

$63,60,57,...\dots ..$

sum of n terms of A.P is $S_n=693$

first term of this A.P is $a_1=63$

second term of this A.P is $a_2=60$

common difference

$d=a_2-a_1=60-63=-3$

we know that sum of n term of an A.P is given by

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

$⟹S_n=\frac{n}{2}[2\times 63+(n-1)\times -3)]$

$⟹S_n=\frac{n}{2}[126-3n+3)]$.

$⟹S_n=\frac{n}{2}[129-3n]$

put value of $S_n=693$ in above equation

$⟹693=\frac{n}{2}[129-3n]$

$⟹[129n-3n^2]=1386$

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

by solving the above quadratic equation we get

$n=(129\pm \sqrt{(-129{)}^2-4\times 3\times 1386})/6$

$n=(129\pm 3)/6$

for minus sign

$n=\frac{126}{6}=21$

or

$n=\frac{132}{6}=22$

hence the sum of $21$ or $22$ term of this A.P is $693$.

the sum is equal for $21$ and $22$ term because $22$th term of this A.P is $0$

since addition of zero does not change the sum