A8 = 40 and a13 = - 10, find a1 and d

A term in an arithmetic progression is given by the formula a1 + (n-1) d

Thus; a8 = a1 + 7d = 40

a13 = a1 + 12 d = - 10

Solving them simultaneously;

a1 + 7d = 40

a1 + 12d = - 10 subtracting the two equations

-5d = 50

d = - 10

a1 = 40 - (7 * - 10)

= 110

Therefore; a1 = 110 and d = - 10