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

The Correct Option is C

\(a + d = 112\)

\(b + c = 48\)

\(a + ar^3 = 112\)

\(⇒a (1 + r^3) = 112\)-------(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 , \(\dfrac{a + c +8}{ b} = \dfrac{4+36+8}{12} = 4\) 

Answer 2

The terms can be expressed as:

\(a = a\)

\(b = ar\)

\(c = ar^2\)

\(d = ar^3\)

We are given the following information:

1. \(a + d = 112\)

2. \(b + c = 48\)

Substitute the GP terms into these equations:

1. \(a + ar^3 = 112 \implies a(1 + r^3) = 112\)

2. \(ar + ar^2 = 48 \implies ar(1 + r) = 48\)

Now, divide equation (1) by equation (2) to eliminate \(a\):

\[ \frac{a(1 + r^3)}{ar(1 + r)} = \frac{112}{48} \]

First, simplify the fraction on the right side: \(\frac{112}{48} = \frac{16 \times 7}{16 \times 3} = \frac{7}{3}\). Next, use the sum of cubes factorization for the term \(1 + r^3\): \(1 + r^3 = (1 + r)(1 - r + r^2)\). Substitute this into the equation:

\[ \frac{a(1 + r)(1 - r + r^2)}{ar(1 + r)} = \frac{7}{3} \]

Assuming \(a \neq 0\) and \(r \neq -1\) (which is true since the sequence is increasing), we can cancel \(a\) and \((1+r)\):

\[ \frac{1 - r + r^2}{r} = \frac{7}{3} \]

Cross-multiply to solve for \(r\):

\[ 3(1 - r + r^2) = 7r \]

\[ 3 - 3r + 3r^2 = 7r \]

Rearrange into a standard quadratic form:

\[ 3r^2 - 3r - 7r + 3 = 0 \]

\[ 3r^2 - 10r + 3 = 0 \]

Factor the quadratic equation:

\[ (3r - 1)(r - 3) = 0 \]

This gives two possible values for the common ratio: \(r = \frac{1}{3}\) or \(r = 3\). Since the sequence \(a, b, c, d\) is stated to be increasing, the common ratio \(r\) must be greater than 1. Thus, we must have:

\(r = 3\)

Now substitute \(r=3\) back into equation (2) to find \(a\):

\(ar(1 + r) = 48\)

\(a(3)(1 + 3) = 48\)

\(a(3)(4) = 48\)

\(12a = 48\)

\(a = \frac{48}{12} = 4\)

So, the first term is \(a=4\) and the common ratio is \(r=3\). The terms of the sequence are:

\(a = 4\)

\(b = ar = 4(3) = 12\)

\(c = ar^2 = 4(3^2) = 4(9) = 36\)

\(d = ar^3 = 4(3^3) = 4(27) = 108\)

The sequence is \(4, 12, 36, 108\). This is indeed an increasing sequence. Check conditions: \(a+d = 4+108=112\) (Correct), \(b+c = 12+36=48\) (Correct).

Finally, we need to calculate the value of the expression \(\frac{a+c+8}{b}\):

\[ \frac{a+c+8}{b} = \frac{4 + 36 + 8}{12} \]

\[ = \frac{40 + 8}{12} \]

\[ = \frac{48}{12} \]

\[ = 4 \]

The value of the expression is 4.