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# ap

Asked by MANOJKUMAR 16th May 2010, 8:57 PM

Let's explicitly note such numbers,

1, 3, 7, 9, 11....

So we can identify two AP,

1, 3, 5, 7, 9, 11, ...

with a0 = 1, d = 2,

n = 5x10 = 50 since between every ten numbers we have 5 numbers of our series like 1, 3, 5, 7, 9 from 1 to 10

Therefore,

S55 = 55(1 + 1 + 54x2)/2 =   3025.

But we have included 5, 15, 25 in our series, so for this AP, and to get correct result we must subtract this sum from S55,  we have,

a0 = 5, d = 10 and n = 1x10 = 10, since we have just one number for every 10 numbers.

S10 = 10(5 + 5 + 9x10)/2 = 500

Required Sum = S55 - S10 = 3025 - 500 = 2525.

Regards,

Team,

TopperLearning.

Answered by Expert 17th May 2010, 7:22 AM
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