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CBSE Class 11-science Answered

Q. Calculate the sum of the squares of first 100 terms of an AP , given that the sum of the first 100 terms is -1. and that the sum of 2nd , 4th, ......., 100th term is 1.
Asked by Shrivatsa | 22 Nov, 2018, 05:36: PM
answered-by-expert Expert Answer
In A.P, sum of 100 terms is -1, (100/2) [ 2a +99d ] = 50 [ 2a+99d ] = -1 .................(1)
where a is the first term and d is common difference. eqn.(1) can be rewritten as  [ 2a + 99d ] = (-1/50) ...................(2)
 
Similarly if sum of 2nd, 4th, 6th .......100th terms are -1 , then (50/2) [ 2(a+d) + 49×2d ] = 1  or 2a+100d = (1/25) ...............(3)
 
if we solve eqns. (2) and (3), we get d = 3/50  and a = -149/50
 
we can write the nth term of A.P. as, tn = (-149/50)+(n-1)(3/50)
 
Hence the required sum = begin mathsize 12px style sum from k equals 1 to 100 of t subscript k superscript 2 space equals space sum from k equals 1 to 100 of open parentheses fraction numerator negative 149 over denominator 50 end fraction plus left parenthesis k minus 1 right parenthesis 3 over 50 close parentheses squared space equals space 1 over 2500 sum from k equals 1 to 100 of open parentheses negative 152 plus 3 k close parentheses squared space equals space 1 over 2500 sum from k equals 1 to 100 of open parentheses 152 squared space space minus space 912 k space plus 9 k squared close parentheses end style .....................(4)
In eqn.(4) we get sum of integers  and sum of squares of integers from 1 to 100.
 
Hence the required sum as given in eqn.(4) is written as begin mathsize 12px style 1 over 2500 open curly brackets space 152 cross times 152 cross times 100 space minus space fraction numerator 912 cross times 100 cross times 101 over denominator 2 end fraction space plus space fraction numerator 9 cross times 100 cross times 101 cross times 201 over denominator 6 end fraction close curly brackets end style
further simplification can done by the student
Answered by Thiyagarajan K | 23 Nov, 2018, 10:02: PM

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