Question

If the first term of a G.P. is 3 and the sum of second and third terms is 60, then the common ratio of the G.P. is

• A. 4 or -3
• B. 4 only
• C. 4 or 5
• D. 4 or -5
• E. -5 only

Answer 1

The Correct Option is D

Let the first term of the geometric progression (G.P.) be \( a = 3 \) and the common ratio be \( r \). The second term is \( ar = 3r \), and the third term is \( ar^2 = 3r^2 \). We are given that the sum of the second and third terms is 60, i.e., \[ 3r + 3r^2 = 60. \] Factor out the common factor 3: \[ 3(r + r^2) = 60 \quad \Rightarrow \quad r + r^2 = 20. \] Rearrange the equation: \[ r^2 + r - 20 = 0. \] Now solve this quadratic equation using the quadratic formula: \[ r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-20)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm \sqrt{81}}{2}. \] Thus, \[ r = \frac{-1 \pm 9}{2}. \] The two possible values for \( r \) are: \[ r = \frac{-1 + 9}{2} = 4 \quad \text{or} \quad r = \frac{-1 - 9}{2} = -5. \]

The correct option is (D) : \(4\) or \(-5\)

Answer 2

Let the first term of the geometric progression (G.P.) be \(a = 3\) and the common ratio be \(r\). The second term of the G.P. is \(ar = 3r\), and the third term is \(ar^2 = 3r^2\).

We are given that the sum of the second and third terms is 60:

\[ 3r + 3r^2 = 60 \] \[ 3(r + r^2) = 60 \] \[ r + r^2 = 20 \]

Rearrange this equation into a quadratic form:

\[ r^2 + r - 20 = 0 \]

Now, solve this quadratic equation using the quadratic formula:

\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For \(r^2 + r - 20 = 0\), \(a = 1\), \(b = 1\), and \(c = -20\). Substituting these values into the quadratic formula:

 

\[ r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-20)}}{2(1)} \] \[ r = \frac{-1 \pm \sqrt{1 + 80}}{2} \] \[ r = \frac{-1 \pm \sqrt{81}}{2} \] \[ r = \frac{-1 \pm 9}{2} \]

Therefore, we get two possible values for \(r\):

\[ r = \frac{-1 + 9}{2} = 4 \quad \text{or} \quad r = \frac{-1 - 9}{2} = -5 \]

Thus, the common ratio of the G.P. is 4 or -5.