Question

Find five numbers in A.P whose sum is 25 and the sum of whose square is 135

Answer 1

terms of AP are 3, 4, 5, 6 and 7

the 5 terms be (a-2d ), (a-d) , a , (a+d) (a+2d)

Sum of terms is 25.

⇒ (a-2d ) + (a-d) + a + (a+d) + (a+2d) = 25

⇒ 5a = 25

⇒ a= 5

the sum of squares of terms of AP is 135 (given)

(a-2d )^2+(a-d)^2+a^2+(a+d)^2+(a+2d)^2=135

Put a = 5

25+4d^2-20d+25+d^2-10d+5^2+25+d^2+10d+25+4d^2+20d = 135

4d^2+d^2+d^2+4d^2= 135-125

10d^2=10

d^2=1

d=± 1

Case 1) when a = 5 and d= 1

(a-2d ) = 5-2 = 3

(a-d) = 5-1 = 4

 a = 5

(a+d) = 5+1 = 6

and (a+2d) = 5 +2 = 7

Case 2) when a = 5 and d= -1

(a-2d ) = 5+2 = 7

(a-d) = 5+1 = 6

 a = 5

(a+d) = 5-1 = 4

and (a+2d) = 5 - 2 = 3

Thus, terms of AP are 3, 4, 5, 6 and 7