Answer to Question #258323 in Math for Bura

Question

Find the 100th partial sum of 2+10+8+14+14+18

Answer 1

$2+10+8+14+14+18+...$

$=(2+8+14+...)+(10+14+18+...)$

Arithmetic progression $2, 8, 14, ...$

$a=2, d=8-2=6$

$S_{50}=\dfrac{2(2)+6(50-1)}{2}\cdot50=7450$

Arithmetic progression $10, 14, 18, ...$

$a=10, d=14-10=4$

$S_{50}=\dfrac{2(10)+4(50-1)}{2}\cdot50=5400$

The 100th partial sum is

$7450+5400=12850$

Or

$2+10+8+14+14+18+...$

$=12+22+32+...$

Arithmetic progression $10, 14, 18, ...$

$a=12, d=22-12=10$

$S_{50}=\dfrac{2(12)+10(50-1)}{2}\cdot50=12850$

The 100th partial sum is $12850.$