Find t12 term in an arithmetic progression having t3 = 10 and t10 = - 4.

t12 = - 8

Step-by-step explanation:

t3

a+2d=10 (1)

t10

a+9d=-4. (2)

From (1)

a=10-2d. (3)

Sub into (3) into equ (2)

a+9d=-4

10-2d+9d=-4

10+7d=-4

7d=-4-10

7d=-14

Divide both sides by 7

d=-14/7

d = - 2

10=a+2d

10=a+2 (-2)

10=a-4

10+4=a

14=a

a=14

t12=a+11d

= 14+11 (-2)

=14-22

= - 8

t12 = - 8