Question 399540
T3=-8, T17=48 and T20=?

Tn=a+(n-1)d
T3=a+(3-1)d
-8=a+2d......(1)

Tn=a+(n-1)d
T17=a+(17-1)d
48=a+16d.....(2)

subtracting (2) from (1)
48=a+16d
-8=a+ 2d
-  -  - 
____________
56=  14d
d=4

-8=a+2d  putting d=4
-8=a+2(4)
-8=a+8
a=8+8
a=16

now a=16, d=4 ,T20=?
Tn=a+(n-1)d
T20=16+(20-1)(4)
   =16+(19)(4)
   =16+76
   =92
T20=92