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by mrkylee » Fri Dec 21, 2007 12:43 am
the sum of the first 50 positve even integer is 2,550. What is the sum of the even integers from 102 to 200, inclusive? If anyone can help, I'd really appreciate it...
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by camitava » Fri Dec 21, 2007 3:44 am
sum of the even integers from 102 to 200, inclusive means we can use the basic formula of AP. Sn = sum of n digits = n/2(2a + (n - 1)d)
Where a = first term of the series, d = common diff and n = no of terms.
Tn = nth term = a + (n - 1)d = 102 + (n - 1) * 2 = 200 or n = 50
So Sn = 50/2(102 + 200). Hope u get me, mrkylee!
Correct me If I am wrong


Regards,

Amitava
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by TSonam » Fri Dec 21, 2007 11:01 am
camitava,

you are correct and the answer comes out to 7550
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by Bschool08 » Sun Dec 23, 2007 9:03 pm
camitava/Tsonam,

why is it that the nth term equation is equated to 200? is this because of the inclusive aspect?


thanks
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by camitava » Sun Dec 23, 2007 9:12 pm
Bschool08, u r correct! As the Qs is saying that 200 is the max limit - inclusive. So we will take 200 as the nth term. Hope u understand what I want to mean!
Correct me If I am wrong


Regards,

Amitava
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by Bschool08 » Sun Dec 23, 2007 9:18 pm
yes i do- thanks Camitava!!
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by samirpandeyit62 » Sun Dec 23, 2007 9:28 pm
100+ 102 .... 200

= 2+ 4+6.... +100 + 50(100)

= 2550 + 50*100 = 7550
Regards
Samir
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