Question

Find out four numbers such that, first three numbers are in G.P., last three numbers are in A.P. having common difference $6$, first and last numbers are same.

• A. $8,4,2,8$
• B. $-8,4,-2,-8$
• C. $8,-4,2,8$
• D. $-8,-4,-2,-8$

Answer 1

The correct option is C

$8,-4,2,8$

Step 1. Find the four numbers:

Given: $d=6$

and the Last $3$ of the $4$ numbers are in AP.

Let the number are,$a-d,a,a+d$.

Also, the first number is the same as the 4th,$a+d$.

$\therefore$the 4 numbers are $a+d,a-d,a,a+d$

The first 3 numbers of these are in G.P.

$\Rightarrow$ ${(a-d)}^2=a(a+d)$

Step 2. Put the value of $d$

$\Rightarrow$ ${(a-6)}^2=a(a+6)$

$\Rightarrow$ $a^2+36-12a=a^2+6a$

$\Rightarrow$ $18a=36$

$\Rightarrow$ $a=\frac{36}{18}$

$\Rightarrow$ $a=2$

Step 3. Find all four terms of the GP:

Now, First term $=a+d$

$=2+6$

$=8$

Second term $=a-d$

$=2-6$

$=-4$

Third term $=a$

$=2$

Fourth term $=a+d$

$=2+6$

$=8$

$\therefore$ the required series is $\mathbf{8}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{,}\mathbf{}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{8}$

Hence, Option ‘C’ is Correct.