Question

find the first 51 term of an ap whose second and third terms are 14and 18

Answer 1

T2=14
T3=18
You get d as 4(do it by subtracting T2 from T3)

T2=a+d
14=a+4
14-4=a
a=10

T51=a+50d
T51=10+50(4)
T51=10+200
T51=210

Therefore the 51 term is 210

Answer 2

t2 can be expressed as a+d
and t3 can be expressed as a+2d
so we get the equations
a + d =14
a + 2d=18
on solving we get d=4 and a=10
now t51 can be expressed as a + 50d = 10 + 50×4=210