Question

The sum of 55 consecutive integers a1,a2,a3,a4,......a55 is 2750.what is the sum of their squares?

Answer 1

let the consecutives numbers are x,x+1,x+2 ... x+54

sum of these numbers are x+x+1+x+2...x+54 = 2750

55x+ (1+2+3+...+54) = 2750

it is an AP 1+2+3+4...54

a =1 ,d = 1, l= 54, n= 54

S = n/2 (a+l)

S = 54/2(1+54) =   1485

55x +1485 = 2750

55 x = 2750-1485

x = 1265/55 = 23

So,the numbers are 23,24,25...78

Now calculate the sum of their squares are x² + (x+1)²+(x+2)²...+(x+54)²

= x²+x²+2x+1+x²+4x+4+x²+6x+9...x²+108x+2916

= 55 x²+ x ( 2 +4 +6 + ...+108) +( 1+4+9+16 ...+2916)

= 55 x² + x [( 54/2)(2+108)]+ (1² + 2²+3²+4²...+54²)

(1² + 2²+3²+4²...+54²): sum of this series is  [n(n+1)(2n+1)/6]

= 55x²+ 2970 x + [n(n+1)(2n+1)/6]

= 55 (23)²+ 2970×23 +[54(54+1)(108+1)/6]

= 290,095+68,310 + 53,955

= 412,360