Question

Let $a,b,c,d$ be an increasing sequence of real numbers,which are in geometric progression .If $a+d=112$ and $b+c=48$ ,then the value of $\dfrac{a+c+8}{b}$ is

• A. $1$
• B. $5$
• C. $4$
• D. $3$
• E. $2$

Answer 1

Answer: (C)

$4$

Answer 2

Given that:

$a + d = 112$

$b + c = 48$

$a + ar^3 = 112$

$⇒a (1 + r^3) = 112$-------(1)

$ar + ar2 = 48$

$⇒a ( r + r^2 ) = 48$--------(2)

Now comparing $a$ from above cases (1) and (2)we get:

$3r^3-7r^2-7r+3=0$

on solving we get :

$r= -1,3,0.3334$

Therefore from parent equation we get $a=\dfrac{48}{12}=4$

Then, sequence becomes $4, 12, 36, 108$$$

Therefore , $\dfrac{a + c +8}{ b} = \dfrac{4+36+8}{12} = 4$ (_Ans)