Latest Others Aptitude Question SOLUTION: how to find the sum of series whose difference is in AP? eg. 1, 3, 6, 10, 15, 21.....

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(#M40019358) OTHERS question Keep an EYE

how to find the sum of series whose difference is in AP?
eg. 1, 3, 6, 10, 15, 21.....

Asked In Others mohan lal garg (10 years ago)

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Chandni Maurya  (10 years ago)

let s is the summation of the series and Tn is the nth term.
S = 1 + 3 + 6 + 10 + 15 + 21 ...+ Tn ---(1).
S = 1 + 3 + 6 + 10 + 15 ...+ T(n-1) + Tn ---(2).
subtracting eqation (2). from (1). we have...
0 = 1 + 2 + 3 + 4 + 5 + 6 ... +(Tn- T(n-1)) -Tn
=> Tn= 1 + 2 + 3 + 4 + 5 + 6 ... +(Tn- T(n-1))
=> Tn= 1 + 2 + 3 + 4 + 5 + 6 ... + n terms
(according to the series apply the formula... eg. A.P. or G.P. series
here we can use,sum of n natural no, i.e. sum=n(n+1)/2)
=> Tn= n(n+1)/2
now,
S = ∑Tn
=> S = ∑ (n^2+n)/2
=> S = 1/2(∑n^2 + ∑n)
=> S = 1/2( (n(n+1)(2n+1))/6 + (n(n+1))/2 )
=> S = 1/2 (n/6)[ (n+1)(2n+1) + 3n +3]
=> S = n/12(2n^2+ 3n +1+ 3n +3)
=> S =n/12(2n^2+ 6n +4)
=> S=n/12[(2n +2) (n+2)]
=> S=n/6[(n +1) (n+2)]

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