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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


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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


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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
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Samir
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