Question

The sum of three numbers in A.P. is $21$ and the sum of their squares is $197$. Find the middle number.

Answer 1

Solution:-

Let the terms are $a-d,a,a+d$

The sum of three numbers in A.P. is 21

$\therefore sum=(a-d)+a+(a+d)=3a=21$

$\Rightarrow a=7$

Second condition:-

The sum of their squares is 197.

$\therefore {(a+d)}^2+a^2+{(a-d)}^2=197$

$a^2+d^2-2ad+a^2+a^2+d^2+2ad=197$

$\Rightarrow 3a^2+2d^2=197$

$3\times 7^2+2d^2=197$

$\Rightarrow 147+2d^2=197$

$\Rightarrow 2d^2=18$

$d=3$

so the series is-

$a-d=7-3=4$

$a=7$

$a+d=7+3=10$

Hence the series is $4,7,10$ and the middle no. is $7$.