Sequence | Frankie's Blog

S= $t_{1}+ t_{2}+ t_{3}+ t_{4}$ + …..

$t_1$ = 20

Common difference: + 10

What’s the $t_{250}$ and What’s the $S_250$

Solution:

$t_n$ = $t_1$ +(n-1)d

$t_{250}$ =20+(250-1)10=2510

$t_{1}$ =20

$t_{2}$ =20+20+(10)=50

$t_{3}$ =20+20+(10)+20+(10)+(10)=90

$t_{4}$ =20+20+(10)+20+(10)+(10)+20+(10)+(10)+(10)=140

$t_{250}$ = 20+20+(10)+20+(10)+(10)+20+(10)+(10)+(10)+***

= $\frac{n}-2$ x (20 + 510)

= $\frac{n}-2$ x 530

=56250

In conclusion, $t_250$ = 2510 and $S_250$ = 56250