Question

For what value of P, are 2p-1, 7 and 11/2p three consecutive terms of an A.P.

Answer 1

Answer:

let's say

a(n)=2p-1

a(n+1)=7

a(n+2)=11/2p

we know,

a(n+1)-a(n)=d.............(¡)

a(n+2)-a(n+1)=d...........(¡¡)

from ¡ and ¡¡

7-2p+1=11/2p-7

7+7=11/2p+2p

14=11/2p+4/2p

14=15/2p

p=14×2/15

p=28/15

Answer..!

Answer 2

The value of p is $\frac{19}{4}$.

Step-by-step explanation:

Given : $2p-1$ , 7 and $\frac{11}{2}p$  three consecutive terms of an A.P.

To find : What is the value of p ?

Solution :

The A.P is $2p-1$ , 7 and $\frac{11}{2}p$ .

The common difference is the difference between the successive term and its preceding term.

So, second term - first term = third term - second term

$7-(2p-1)=\frac{11}{2}p-7$

$7-2p+1=\frac{11}{2}p-7$

$-2p-\frac{11}{2}p=-7-8$

$\frac{-4p-11p}{2}=-15$

$-4p-11=-30$

$-4p=-30+11$

$-4p=-19$

$p=\frac{19}{4}$

Therefore, The value of p is $\frac{19}{4}$.

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