Part (a): Find the sum a + (a + 1) + (a + 2) + ... + (a + n - 1) in terms of a and n. Part (b): Find all pairs of positive integers (a,n) such that n \ge 2 and \[a + (a + 1) + (a + 2) +... + (a + n - 1) = 100.
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a + (a+1) + (a+2)+............+ (a+n-1)
The given series is an arithmetic progression series, with 'n' terms starting with 'a' as the first term and a common difference of 1 between any two successive terms.
The first term of the series, a1 =a
and the nth term of the series, an = a+n-1
The sum of such an Arithmetic progression series will be given as:
sum = `n/2 (a1+an) = n/2 (a+a+n-1) = n/2 (2a+n-1)`
(b) n`>=` 2 and sum of the series = 100.
i.e. (n/2)(2a+n-1) = 100.
We can start by substituting values of n starting from 2 and determining if a is a positive integer or not.
Lets try with n = 5
i.e. (5/2) (2a+5-1) =100
or, 2a+4 = 100*2/5 = 40 or a = 36/2 =18.
This way we can find the values of pair (a,n) that satisfy this equation and are positive.
In fact, only 2 pairs of (a,n): (18,5) and (9,8) satisfy the relationship.
Hope this helps.
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