Question

PLEASE HELP ME !ASAP!!

Answer 1

Solution :

First term, a₀ = -10 .

Common difference of A.P. , d = -8 - (-10) = 2 .

Last term, aₙ = 24 .

We know, last term is given by :

aₙ = a₀ + (n-1)d

24 = -10 + (n-1)×2

2n - 2 = 34

n = 18

Now, sum of series is given by :

$S_n = \dfrac{n}{2} \times ( 2a+(n-1)d)\\\\S_n = \dfrac{18}{2} \times ( 2\times -10 + ( 18 -1) \times 2)\\\\S_n = 126$

Hence, this is the required solution.