Question

If sum of four numbers in GP is 60 and AM of first and last is 18, then the numbers are:

• A. \(3, 9, 27, 81\)
• B. \(4, 8, 16, 32\)
• C. \(2, 6, 18, 54\)
• D. None of these

Hint:

For GP problems, use the ratio \(r\) and solve using sum and product/AM conditions.

Answer 1

The Correct Option is B

Concept: Let the four numbers in GP be \(a, ar, ar^2, ar^3\).

Step 1: Sum condition. \[ a(1 + r + r^2 + r^3) = 60 \]

Step 2: AM condition. \[ \frac{a + ar^3}{2} = 18 \quad \Rightarrow \quad a(1 + r^3) = 36 \]

Step 3: Divide the two equations. \[ \frac{a(1 + r + r^2 + r^3)}{a(1 + r^3)} = \frac{60}{36} = \frac{5}{3} \] \[ \frac{1 + r + r^2 + r^3}{1 + r^3} = \frac{5}{3} \] Note that \(1 + r + r^2 + r^3 = (1+r)(1+r^2)\) and \(1 + r^3 = (1+r)(1 - r + r^2)\). Cancelling \(1+r\) (since \(r \neq -1\)): \[ \frac{1 + r^2}{1 - r + r^2} = \frac{5}{3} \] \[ 3(1 + r^2) = 5(1 - r + r^2) \quad \Rightarrow \quad 3 + 3r^2 = 5 - 5r + 5r^2 \] \[ 0 = 2 - 5r + 2r^2 \quad \Rightarrow \quad 2r^2 - 5r + 2 = 0 \] \[ (2r - 1)(r - 2) = 0 \quad \Rightarrow \quad r = \frac{1}{2} \quad \text{or} \quad r = 2 \]

Step 4: Find \(a\). For \(r = 2\): \(a(1 + 8) = 36 \Rightarrow 9a = 36 \Rightarrow a = 4\).
GP: \(4, 8, 16, 32\).
Sum = \(4+8+16+32 = 60\),
AM of first and last = \((4+32)/2 = 18\).