Question

Find an increasing arithmetic progression in which the sum of the first three terms is $27$ and the sum of their squares is $275$.

Answer 1

Let the first $3$ terms be $a-d,a,a+d$.

$\therefore a-d+a+a+d=27\Rightarrow 3a=27\Rightarrow a=\frac{27}{3}=9$

Sum of squares $=275$.

$\Rightarrow {(a-d)}^2+a^2+{(a+d)}^2=275\Rightarrow a^2-2ad+d^2+a^2+a^2+2ad+d^2=275\Rightarrow {3a}^2+{2d}^2=275\Rightarrow 3\times 9^2+{2d}^2=275\Rightarrow 243+{2d}^2=275\Rightarrow {2d}^2=275-243=32\Rightarrow d^2=16\Rightarrow d=\pm 4$

As it is an increasing $AP$,so the value of $d$ is $4$.

$\therefore$The series is $9,13,14,21,........$