Question

If p +2, 4p – 6 and 3p -2 are three consecutive terms of an AP, find p​

Answer 1

Answer:

a=p+2

d=a2-a1=a3-a2

=4p-6-(p+2)=3p-2-(4p-6)

=4p-6-p-2=3p-2-4p+6

3p-8=-p+4

3p+p=4+8

4p=12

p=12/4

p=3

Answer 2

First term (a) = p + 2

Common difference (d) = a2 - a1 = a3 - a2 = a4 - a3 .......

=> a2 - a1 = a3 - a2

=> 4p - 6 - ( p + 2) = 3p - 2 - ( 4p - 6 )

=> 3p - 8 = -p +4

=> 4p = 12 => p = 3

∴ P = 3