Question

Let a,b be two real numbers between $3$ and $81$such that the resulting sequence $3,a,b,81$ is in a geometric progression. The value of $a+b$ is

• A. $36$
• B. $29$
• C. $90$
• D. $27$
• E. $81$

Answer 1

Answer: (A)

$36$

Answer 2

Given that

The G.P series is: $3,a,b,81$

means here first term is $=3$

last term $=81$

So, let

        $a=3.r$

       $ b=3.r=3r^2$

Similarly, $81=3r^3$$$

          $⇒ 27=r^3$$$

          $⇒ r=3$

Therefore, $a=3×3=9$

            $ b=3×3^2=27$

So, $a+b=9+27=36$ (_Ans)