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 :
[tex]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[/tex]
Hence, this is the required solution.
