Latest Huawei Aptitude Question SOLUTION: Sum of the 66 consecutive integers is 5181, then sum of the squares is?

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Sum of the 66 consecutive integers is 5181, then sum of the squares is?

Asked In Huawei Dilipan M (5 years ago)

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arun nachiappan  (4 years ago)

sum of 66 consecutive integers is 5181,
Let consecutive integers are : x , (x + 1) , ( x + 2) , ( x + 3) ........ (x + 65)

now, sum of consecutive integers = x + (x + 1) + (x + 2) + (x + 3) + (x + 4) ....... + (x + 65)
= 66x + (1 + 2 + 3 + .... + 65)
= 66x + 65(65 + 1)/2
= 66x + 65 × 33

so, 66x + 65 × 33 = 5181
11(6x + 65 × 3) = 471 × 11
(6x + 65 × 3) = 471
6x = 471 - 195 = 276
x = 46

hence, consecutive integers are 46, 47, 48, ... 111

now, sum of square of consecutive integers
= 46² + 47² + 48² + .... 111²
=
use formula

=

= 430,661

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