Is it in A.P?

Algebra Level pending

What is the sum of the series: 1 + 5 + 12 + 22 + 35 + . . . . . + 14950 1+5+12+22+35+.....+14950 ?

Note that the second order difference of the consecutive terms of this series is constant.


The answer is 505000.

Ayush G Rai by Ayush G Rai , 20, India

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1 solution

Ayush G Rai
Aug 7, 2016

We get n t h nth term to be 1 2 ( 3 n 2 − n ) . \dfrac{1}{2}(3n^2-n). We know the n t h nth term to 14950. 14950. So 3 n 2 − n − 29900 = 0. 3n^2-n-29900=0. After solving this quadratic equation,we get n = 100. n=100.
We can also generalize the sum of the series = n 2 ( n + 1 ) 2 . =\dfrac{n^2(n+1)}{2}. Putting n = 100 , n=100, we get the sum to be 505000 . \boxed{505000}.

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