Problem Solving
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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!
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
Regards,
Amitava

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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
Regards,
Amitava

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