Find five numbers in an AP , whose sum is 25 and the sum of whose squares is 135

Dear Student,
Solution) Five terms be (a-2d ),(a-d),a,(a+d) and (a+2d)
Therefore, acc. to question, 
Sum of terms = 25.
⇒ (a-2d ) + (a-d) + a + (a+d) + (a+2d) = 25
⇒ a-2d + a-d +a + a+d +a +2d =25 
⇒ 5a = 25
​​​​​​⇒ a= 5 = first term
It is also given that,
sum of squares of terms of AP  = 135
⇒ (a-2d )²+(a-d)²+a²+(a+d)²+(a+2d)² = 135
Put a = 5, we get,
⇒  (5-2d )²+(5-d)²+5²+(5+d)²+(5+2d)² = 135
$$\begin{gathered} \{ because {(a-b)}^2=a^2+b^2-2ab and {(a+b)}^2=a^2+b^2+2ab\} \\ \Rightarrow 25+4d^2-20d+25+d^2-10d+25+25+4d^2+20d+25+d^2+10d=135 \\ \Rightarrow 10d^2+125-20d-10d+20d+10d=135 \\ \Rightarrow 10d^2+125=135 \\ \Rightarrow 10d^2=135-125 \\ \Rightarrow 10d^2=10 \\ \Rightarrow d^2=1 \\ \Rightarrow d=\pm 1=common difference \\ Case 1) a=5 and d=1 \\ then, \\ a-2d=5-2=3 \\ a-d=5-1=4 \\ a=5 \\ a+d=5+1=6 \\ a+2d=5+2=7 \\ Five terms are 3,4,5,6 and 7. \\ Case 2) a=5,d=-1 \\ then, \\ a-2d=5+2=7 \\ a-d=5+1=6 \\ a=5 \\ a+d=5-1=4 \\ a+2d=5-2=3 \\ Five terms are 3,4,5,6 and 7. \\ Regards! \end{gathered}$$