Question

Find the value of p for which the numbers 2p - 1, 3p + 1, 11 are in AP. Hence, find the numbers.

Answer 1

AP Series :  2p - 1 , 3p + 1 , 11

In an AP, the common difference is consistent

⇒ a2 - a1 = a3 - a2

Find P:

(3p + 1) - (2p - 1) = 11 - (3p + 1)

3p + 1 - 2p + 1 = 11 - 3p - 1

p + 2 = 10 - 3p

4p = 8

p = 8 ÷ 4 = 2

Find the numbers:

1st term = 2p - 1 = 2(2) - 1 = 3

2nd term = 3p + 1= 3(2) + 1 = 7

3rd term = 11

Answer: The numbers in the AP are 3, 7 and 11

Answer 2

Answer:

=> The numbers are 3, 7, and 11.

Step-by-step explanation:

As per the data given in the question, we have to find the numbers.

Given numbers:-

$(2p-1),(3p+1),11$

The numbers $a$, $b$, and $c$ are in $AP$,

Then, $b-a=c-b$

$=>2b=a+c$

Therefore,

$=>2(3p+1)=(2p-1)+11$

$=>6p+2=2p-1+11$

$=>6p+2=2p+10$

$=>6p-2p=10-2$

$=>4p=8$

Divide both sides by 4, we get

$=>p=2$

So, the value of p is $2.$

Hence, the numbers are:-

$=>(2p-1)=(2\times 2-1)=3$

$=>(3p+1)=(3\times 2+1)=7$

$=>11$