Question

The first three of four given numbers are in G.P. and their last three are in A.P., with common difference 6. If first and fourth numbers are equal, then the first number is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is D

8

The first and the last numbers are equal

Let the four given numbers be p, q, r and p

The first three of four given numbers are in G.P.

$\therefore q^2=p.r$ ...... (i)

And, the last three numbers are in A.P. with common difference 6.

We have :

First term = q

Second term = r = q + 6

Third term = p = q + 12

Also, 2r = q + p

Now, putting the values of p and r in (i) :

$q^2=(q+12)(q+6)$

$\Rightarrow q^2=q^2+18q+72$

$\Rightarrow q+4=0$

$\Rightarrow q=-4$

Now, putting the value of q in p = q + 12 :

$p=-4+12=8$