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Let Sbe the sum of the first n terms of the series 1+ 2.52 + 32 + 2.72 + 52 + 2.9+... if S20 = 20A then A is equal to 

a) 2001

b) 2019

c) 1851

d) 1951

e) None of these 

Please provide an explanation as well. 

 Apr 19, 2021
 #1
avatar+114818 
+3

Firstly, this is not really the sum of 1 sequence, it is the sum of 2 sequences

 

\(1^2+3^2+5^2+ .......\\ 2.5^2+2.7^2+2.9^2......\\ S_{20}=1^2+3^2+5^2+ .......+19^2\quad+ \quad 2.5^2+2.7^2+2.9^2......+4.3^2\)

 

I suppose I could just calculate the answer.

\(S_{20}=1330+118.9 = 1448.9\\ A=1448.9/20=72.445\)

 

Doesn't look like any of those answers to me.

 Apr 19, 2021
edited by Melody  Apr 19, 2021
edited by Melody  Apr 19, 2021
 #2
avatar+32633 
+3

Another way is to notice that every term in the series is less than 20^2. Since there are 20 terms in all we must have S20 < 20*20^2, which means that 20A < 20*20^2, or A < 20^2  or A < 400.  Hence e) is the answer.

 Apr 19, 2021
 #3
avatar+114818 
0

Thanks Alan, that is better :)

Melody  Apr 19, 2021

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