Question

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

Answer 1

The answer is given below :

FORMULA :

If the numbers a, b and c are in AP, then

b - a = c - b

=> 2b = a + c

SOLUTION :

The given three numbers (2p - 1), (3p + 1) and 11 are in AP.

Then,

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

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

=> 6p + 2 = 2p + 10

=> 6p - 2p = 10 - 2

=> 4p = 8

=> p = 8/4

=> p = 2

So, the value of p is 2.

Then, the given numbers be

(2×2 - 1), (3×2 + 1) and 11

i.e., 3, 7 and 11.

So, we get an Arithmetic progression

3, 7, 11.

Thank you for your question.

Answer 2

Hey mate your answer is

We know that,

a3-a2=a2-a1

So,

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

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

1p+2=10-3p

1p+3p=10-2

4p=8

p=2

Now,put in the given numbers

2p-1=2*2-1=4-3=1

3p+1=3*2+1=6+1=7

11

HOPE this helps ❤️

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