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sums

by Viper83 » Tue Nov 23, 2010 6:57 am
the sum of the first 50 positive even integers is 2,550. what is the sum of the even integers from 102 to 200, inclusive?

a) 5100
b) 7550
c) 10,100
d) 15,500
e) 20,100

b
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by abhishekg21 » Tue Nov 23, 2010 7:24 am
Given is sum (2+4+6...+100)=2550

to find sum of (102+104+106=...200)=?

102+104+106...+200 ncan be written as

(100+2)+(100+4)+(100+6)...+(100+100)=100*50 +(2+4+6..+100)=100*50+2550=7550
so answer is b
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by rishab1988 » Tue Nov 23, 2010 7:37 am
Although abhishek solved this question.Here is a more universal approach that you can apply to all arithmetic sequence questions

The first term : 102
The last term : 200

(because the first even term is 102 and last even term is 200.Buut if the question were for odd, the first term would be 103 and the last term would be 199).

To find number of terms -> subtract last first term from last term -> 200-102=98
Divide by common difference ( here it is 2 because all even terms are 2 spaces apart,if it were consecutive terms the difference would be 1) -> 98/2=49.

Add 1 if the question includes the world inclusive -> 49+1=50

Avg in an any arithmetic progression is : average of first and last term in the sequence .here it is: (102+200)/2=151.

Since Average = sum/no of terms, sum = average * no of terms.here -> 151 (avg)* 50= 7550 or B
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