Question

find the sum of the first 90 terms of the sequence -4,-1,2,5,8...

a. 11654
b.11657
c.11656
d.11655

Answer 1

Since it is moving 3 by 3 we can assume \[u _{n}=3n + c _{1}\]\[u _{1}=-4 \rightarrow c _{1}=-7\] Becoming,\[u _{n}=3n-7\]
\[\sum_{n=1}^{90}3n-7=(-7)\times90+3\sum_{n=1}^{90}n=-630+3\times90\times(90+1)\div2=11655\]

Some other time I might explain why \[\sum_{k=1}^{n}k=n \times(n+1)\div2\]
Anyway answer is d. :)

Cheers

PS: FIRST POST, BANZAI!!!!