Question

If 3, 4+p^2 , 6−p  are in A.P.,then p must be equal to

Answer 1

Solution:-

Given A.P:-

3 , ( 4 + p²) , ( 6 - p) , ............

To Find :-

p = ?

Find:-

We know that,

In A.P. series the Common Difference is always same.

For Ex:-

A.P. - 1 , 4 , 7, ........

So, D = 3 [ 4 -1 = 3 and 7-4 = 3 ].

So, In the Given Question. The Common Difference must be equal if the series is in A.P.

=) a2 - a1 = a3 - a2

=) ( 4 + p²) - 3 = ( 6 - p) - ( 4 + p²)

=) 4 + p² - 3 = 6 - p - 4 - p²

=) 1 + p² -6 + p + 4 + p² = 0

=) 2p² + p - 1 = 0

=) 2p² + ( 2 - 1)p - 1 = 0

=) 2p² + 2p - p - 1 = 0

=) 2p ( p + 1) -1 ( p + 1) = 0

=) ( p +1) and ( 2p - 1)

=) [ p = -1 ] and [ p = 1/2 ]

Hence Solved!

Answer 2

3, 4+p^2 , 6−p are in A.P

It means d should be equal in both cases.

Let a1 = 3

a2 = 4 + p^2

a3 = 6-p

D = a2 - a1

= 4 + p^2 - 3 = 1 + p^2

D = a3 - a2

= 6 -p - 4-p^2 = 2- p - p^2

-p^2 -p +2 = 1 + p^2

2p^2 + p -1 = 0

2p^2 + 2p - p -1 = 0

2p(p + 1) -1(p+1) = 0

(2p - 1)(p + 1)

p = 1/2 or -1.