Question

In an increasing geometric progression of positive terms, the sum of the second and sixth terms is \[ \frac{70}{3} \] and the product of the third and fifth terms is 49. Then the sum of the \(4^\text{th}, 6^\text{th}\), and \(8^\text{th}\) terms is:

• A. 96
• B. 78
• C. 91
• D. 84

Answer 1

The Correct Option is C

To address this problem, we will utilize the properties of a geometric progression (G.P.). Let the first term of the G.P. be denoted by \(a\) and the common ratio by \(r\). The terms of the sequence can be represented as:

• 1st term: \(a\)
• 2nd term: \(ar\)
• 3rd term: \(ar^2\)
• 4th term: \(ar^3\)
• 5th term: \(ar^4\)
• 6th term: \(ar^5\)
• 8th term: \(ar^7\)

The problem states that the sum of the second term (\(ar\)) and the sixth term (\(ar^5\)) is:

\(ar + ar^5 = \frac{70}{3}\) ... (1)

The product of the third term (\(ar^2\)) and the fifth term (\(ar^4\)) is:

\(ar^2 \times ar^4 = 49\)

This simplifies to:

\(a^2r^6 = 49\) ... (2)

Factoring equation (1) yields:

\(ar(1 + r^4) = \frac{70}{3}\)

From equation (2), we can express \(a\) in terms of \(r\): \(a = \frac{7}{r^3}\), which implies \(ar = \frac{7}{r^2}\).

Substituting this into equation (1):

\(\frac{7}{r^2}(1 + r^4) = \frac{70}{3}\)

Multiplying both sides by \(r^2\):

\(7(1 + r^4) = \frac{70r^2}{3}\)

Simplifying this equation:

\(21 + 21r^4 = 70r^2\)

Rearranging the terms results in:

\(21r^4 - 70r^2 + 21 = 0\)

Let \(x = r^2\). The equation becomes a quadratic in \(x\):

\(21x^2 - 70x + 21 = 0\)

Dividing the entire equation by 7 simplifies it to:

\(3x^2 - 10x + 3 = 0\)

Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find the roots:

\(x = \frac{10 \pm \sqrt{(10)^2 - 4 \times 3 \times 3}}{6}\)

\(x = \frac{10 \pm \sqrt{64}}{6}\)

\(x = \frac{10 \pm 8}{6}\)

The possible values for \(x\) are \(x = 3\) or \(x = \frac{1}{3}\).

Since \(x = r^2\) and \(r\) is positive, \(r^2 = 3\) implies \(r = \sqrt{3}\).

Substituting \(r = \sqrt{3}\) into the expression for \(a\):

\(a = \frac{7}{(\sqrt{3})^3} = \frac{7}{3\sqrt{3}} = \frac{7\sqrt{3}}{9}\)

Now, we calculate the sum of the 4th (\(ar^3\)), 6th (\(ar^5\)), and 8th (\(ar^7\)) terms:

The values of these terms are:

\(ar^3 = \frac{7\sqrt{3}}{9} \cdot 3(\sqrt{3}) = 7\)

\(ar^5 = \frac{7\sqrt{3}}{9} \cdot (\sqrt{3})^5 = 21\)

\(ar^7 = \frac{7\sqrt{3}}{9} \cdot (\sqrt{3})^7 = 63\)

The required sum is:

\(7 + 21 + 63 = 91\)

Therefore, the sum of the 4th, 6th, and 8th terms is 91.