Question

for what value of p ,are 2p+1, 13 and 5p -3 ,three consecutive term of ap

Answer 1

AP = 2p + 1 , 13 , 5p - 3.




T1 = 2p + 1 , T2 = 13 and T3 = 5p - 3.


Common difference = T2 - T1 = 13 - ( 2p + 1 ) = 13 - 2p - 1 = 12 - 2p.


Common difference = T3 - T2 = 5p - 3 - 13 = 5p - 16.



T2 - T1 = T3 - T2


12- 2p = 5p - 16.



5p + 2p = 12 + 16


7p = 28


p = 28/7


p = 4.

Answer 2

If 2p+1 , 13 and 5p-3 are consecutive terms of AP
Let us consider this AP is a1, a2, a3
Here, a1 = 2p+1
a2 = 13
a3 = 5p-3
We know that , d = a2 - a1
= 13 - (2p+1)
= 13 -2p -1
a3 = a1 + 2d (n-1 = 3-1 =2)
5p-3 = 2p + 1 + 2 ( 13-2p - 1)
5p-3 = 2p +1 +26 -4p -2
5p = (-2p) + 28
7p = 28
p = 28÷7 = 4