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 solve this problem, we need to work with the properties of a geometric progression (G.P.). Let's denote the first term of the G.P. as \(a\) and the common ratio as \(r\). Therefore, the terms of the sequence can be expressed as follows:

• First term: \(a\)
• Second term: \(ar\)
• Third term: \(ar^2\)
• Fourth term: \(ar^3\)
• Fifth term: \(ar^4\)
• Sixth term: \(ar^5\)
• Eighth term: \(ar^7\)

According to the problem, the sum of the second (i.e., \(ar\)) and sixth (i.e., \(ar^5\)) terms is given by:

\(ar + ar^5 = \frac{70}{3}\) ... (1)

The product of the third (i.e., \(ar^2\)) and fifth (i.e., \(ar^4\)) terms is:

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

Which simplifies to:

\(a^2r^6 = 49\) ... (2)

From equation (1), factor out \(ar\):

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

Substitute the value of \(ar\) from equation (2):

\(a = \frac{7}{r^3}, \; \text{so} \; ar = \frac{7}{r^2}\)

Plugging it into equation (1):

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

Multiplying through by \(r^2\) gives:

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

Simplifying, this turns into:

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

Or:

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

Let \(x = r^2\). Then, the equation becomes a quadratic:

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

Dividing throughout by 7 gives:

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

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

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

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

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

Thus, \(x = 3\) or \(x = \frac{1}{3}\).

\(r^2 = 3 \; \Rightarrow \; r = \sqrt{3}\) (since \(r\) is positive).

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

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

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

The terms are:

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

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

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

Therefore, the required sum is:

\(7 + 21 + 63 = 91\)

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

Answer 2

Given: 
\( T_2 + T_6 = \frac{70}{3} \quad \text{and} \quad T_3 \cdot T_5 = 49. \) Let the first term of the geometric progression be \( a \) and the common ratio be \( r \). 
The second term is: 
\[ T_2 = ar, \] and the sixth term is: 
\[ T_6 = ar^5. \] 

Given: 
\[ ar + ar^5 = \frac{70}{3}. \] Factoring out \( ar \): 
\(ar(1 + r^4) = \frac{70}{3}.\)              (1)

The third term is: 
\[ T_3 = ar^2, \]and the fifth term is: 
\[ T_5 = ar^4.\] 

Given: 
\[ T_3 \times T_5 = ar^2 \times ar^4 = (ar^3)^2 = 49. \] 

Taking the square root: 
\(ar^3 = 7 \implies a = \frac{7}{r^3}.\)               (2)

Substituting the value of \( a \) from equation (2) into equation (1): 
\[ \frac{7}{r^3} \times r \times (1 + r^4) = \frac{70}{3}.\] 

Simplifying: 
\[ \frac{7}{r^2} (1 + r^4) = \frac{70}{3}. \] 

Multiplying both sides by \( 3r^2 \): 
\[ 21(1 + r^4) = 70r^2. \] 

Rearranging terms: 
\[ 21 + 21r^4 = 70r^2. \] 

Dividing by 7: 
\[ 3 + 3r^4 = 10r^2. \] 

Letting \( t = r^2 \), we have: 
\[ 3 + 3t^2 = 10t. \] 

Rearranging: 
\[ 3t^2 - 10t + 3 = 0. \] 

Solving this quadratic equation using the quadratic formula: 
\[ t = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6}.\] This gives: 
\[t = \frac{18}{6} = 3 \quad \text{or} \quad t = \frac{2}{6} = \frac{1}{3}.\] 

Since the GP is increasing, we take \( t = 3 \), 

so: 
\[ r^2 = 3 \implies r = \sqrt{3}. \] Using \( r = \sqrt{3} \) in equation (2): 
\[ a = \frac{7}{(\sqrt{3})^3} = \frac{7}{3\sqrt{3}} = \frac{7\sqrt{3}}{9}. \] Now, we find the sum of the 4th, 6th, and 8th terms: 
\[ T_4 = ar^3, \quad T_6 = ar^5, \quad T_8 = ar^7. \] Calculating: 
\[ T_4 + T_6 + T_8 = ar^3 + ar^5 + ar^7 = ar^3 (1 + r^2 + r^4). \] Substituting values: 
\[ ar^3 = 7, \quad 1 + r^2 + r^4 = 1 + 3 + 9 = 13. \] Thus: 
\[ T_4 + T_6 + T_8 = 7 \times 13 = 91. \] Therefore: 
\[ 91. \]