The sum of 55 consecutive integers a1,a2,a3,a4,......a55 is 2750.What is
the sum of their squares?
Answers
Dear Student,
Answer: 412,360
Solution:
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
is the required answer
Hope it helps you
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