Twisting question on AP

Algebra Level 4

Calculate the sum of the squares of the first 100 terms of an arithmetic progression, given that the sum of the first 100 terms is −1 and that the sum of the second, fourth, sixth, ..., and the hundredth terms is 1.


The answer is 299.98.

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Ravi Dwivedi by Ravi Dwivedi , 24, India

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1 solution

Ravi Dwivedi
Jul 17, 2015

Let the arithmetic progression be x 1 , x 2 , . . . , x 100 x_1,x_2,...,x_{100} having common difference d d

x 1 + x 2 + . . . + x 100 = 1 2 ( x 1 + x 100 ) ⋅ 100 = − 1 x_1+x_2+...+x_{100}=\frac{1}{2}(x_1+x_{100})\cdot 100 =-1

x 1 + x 100 = − 1 50 x_1+x_{100}=-\frac{1}{50}

Also x 2 + x 4 + . . . + x 100 = 1 2 ( ( x 1 + d ) + 100 ) x_2+x_4+...+x_{100}=\frac{1}{2}((x_1+d)+100)

x 1 + x 100 + d = 1 25 x_1+x_{100}+d=\frac{1}{25}

d = 3 50 d=\frac{3}{50}

x 1 + x 100 = x 1 + ( x 1 + 99 d ) = − 1 50 x_1+x_{100}=x_1+(x_1+99d)=-\frac{1}{50}

x 1 = − 149 50 x_1=-\frac{149}{50}

Desired quantity is x 1 2 + x 2 2 + . . . x 100 2 = 100 x 1 2 + 2 x 1 d ( 1 + 2 + . . . + 99 ) + 1 2 + . . . + 9 9 2 = 14999 50 x^2_1+x^2_2+...x^2_{100}=100x^2_1+2x_1d(1+2+...+99)+1^2+...+99^2\\=\large \boxed{\frac{14999}{50}}

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Moderator note:

Simple standard approach with Arithmetic Progressions.

I think this question came in NMTC 2015 ( Senior level).

Vishwak Srinivasan - 5 years, 11 months ago

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I don't know

Ravi Dwivedi - 5 years, 11 months ago

Nice question , overrated though.

Nihar Mahajan - 5 years, 11 months ago

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