drraju's problem please solve it

[munimM]

1. Given a series a1 a2.....an. If a1=-9, a2=-4 and an=a(n-1)-a(n-2) then find the sum of first 100 numbers in the series.
NOTE: 1, 2, n, n-1, n-2 are subscripts.

[munimM]

solve it with details explanation

[omnara]

a1 = -9
a2 = -4
a3 = a2- a1
a4 = a3-a2 = -a1
a5 = a4 - a3 = -a2
a6 = a5 - a4 = -a2 + a1
a7 = a6 - a5 = a1
a8 = a7 - a6 = a2
a9 = a8 - a7 = a2 - a1

repeats.
Sum of a3 through a8 = 0 => sum of a1-a8.
So would be a9 through a14 and so on...

=> sum of a1 through a100 = a1 + a2 + 0 + 0 + .... a99+a100
=> a1 + a2 + a2-a1+ -a1
2a2-a1 = 2* -4 + -9 => -17

Thanks,
Omnara.

[munimM]

if a1=-4, a2=5 what is the value of an

[Kwan]

My answer is 1.
Here is my way: Write the series and find the repeatability
-9,-4, 5, 9, 4, -5, -9, -4, 5, 9, 4, -5, -9, -4, ...
Except a1 = -9 and a2 = -4, every six number sub-series (from a3 to a8) has a sum of 0, so 16 sub-series like that occupy 16x6 = 96 slots, plus the first 2 slots for a1 and a2 we already had 98 numbers. The two next numbers are 5 and 9. Therefore, the sum of the first 100 numbers in this series is: -9 + (-4) + 5 + 9 = 1
Am I correct?

[munimM]

thanks kawn