Question

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is:

• A. 760
• B. 755
• C. 750
• D. 757

Hint:

When dealing with G.P.s, use the sum of terms formula and the given relationships to solve for the first term and common ratio, then calculate the sum of the desired terms.

Answer 1

The Correct Option is D

To find the sum of the first nine terms of the given geometric progression (G.P.), we use the information provided in the question and the properties of G.P.

Let the first term of the G.P. be \(a\) and the common ratio be \(r\). Then, the terms of the G.P. can be expressed as \(a, ar, ar^2, ar^3, \ldots\)

The second, fourth, and sixth terms of the G.P. are:

• Second term: \(ar\)
• Fourth term: \(ar^3\)
• Sixth term: \(ar^5\)

According to the problem, the sum of these terms is 21:

\(ar + ar^3 + ar^5 = 21\)  (Equation 1)

The eighth, tenth, and twelfth terms of the G.P. are:

• Eighth term: \(ar^7\)
• Tenth term: \(ar^9\)
• Twelfth term: \(ar^{11}\)

According to the problem, the sum of these terms is 15309:

\(ar^7 + ar^9 + ar^{11} = 15309\)  (Equation 2)

Now, divide Equation 2 by Equation 1:

\(\frac{ar^7 + ar^9 + ar^{11}}{ar + ar^3 + ar^5} = \frac{15309}{21}\)

\(r^6 \cdot \frac{1 + r^2 + r^4}{1 + r^2 + r^4} = 729\)

Thus, \(r^6 = 729\)

\(r^6 = 3^6\), hence \(r = 3\)

Now, substitute \(r = 3\) back into Equation 1 to find \(a\):

\(a(3) + a(3^3) + a(3^5) = 21\)

\(a(3 + 27 + 243) = 21\)

\(a \cdot 273 = 21\)

\(a = \frac{21}{273} = \frac{1}{13}\)

Finally, we calculate the sum of the first nine terms of the G.P.:

The sum of the first \(n\) terms of a G.P. is given by:

\(S_n = a \frac{r^n - 1}{r - 1}\)

Substituting the known values \(a = \frac{1}{13}\), \(r = 3\), and \(n = 9\):

\(S_9 = \frac{1}{13} \cdot \frac{3^9 - 1}{3 - 1}\)

\(S_9 = \frac{1}{13} \cdot \frac{19683 - 1}{2}\)

\(S_9 = \frac{1}{13} \cdot \frac{19682}{2}\)

\(S_9 = \frac{1}{13} \cdot 9841\)

\(S_9 = 757\)

Hence, the sum of the first nine terms is 757.

Answer 2

Let the first term of the geometric progression (G.P.) be \( a \) and the common ratio be \( r \). The general term of a G.P. is given by: \[ T_n = a r^{n-1} \] We are given the following conditions:
Step 1: Sum of the second, fourth, and sixth terms
The second, fourth, and sixth terms are: \[ T_2 = a r^1, \quad T_4 = a r^3, \quad T_6 = a r^5 \] Their sum is: \[ T_2 + T_4 + T_6 = a r + a r^3 + a r^5 = 21 \] Factor out \( a r \): \[ a r (1 + r^2 + r^4) = 21 \] This equation is (1).
Step 2: Sum of the eighth, tenth, and twelfth terms
The eighth, tenth, and twelfth terms are: \[ T_8 = a r^7, \quad T_{10} = a r^9, \quad T_{12} = a r^{11} \] Their sum is: \[ T_8 + T_{10} + T_{12} = a r^7 + a r^9 + a r^{11} = 15309 \] Factor out \( a r^7 \): \[ a r^7 (1 + r^2 + r^4) = 15309 \] This equation is (2).
Step 3: Solve the system of equations
From equation (1), we have: \[ a r (1 + r^2 + r^4) = 21 \quad \text{(3)} \] From equation (2), we have: \[ a r^7 (1 + r^2 + r^4) = 15309 \quad \text{(4)} \] Divide equation (4) by equation (3): \[ \frac{a r^7 (1 + r^2 + r^4)}{a r (1 + r^2 + r^4)} = \frac{15309}{21} \] Simplifying: \[ r^6 = \frac{15309}{21} = 729 \] Taking the sixth root of both sides: \[ r = 3 \]
Step 4: Find the first term \( a \)
Substitute \( r = 3 \) into equation (3): \[ a \cdot 3 \cdot (1 + 3^2 + 3^4) = 21 \] \[ a \cdot 3 \cdot (1 + 9 + 81) = 21 \] \[ a \cdot 3 \cdot 91 = 21 \] \[ a = \frac{21}{273} = \frac{1}{13} \]
Step 5: Find the sum of the first nine terms
The sum of the first \( n \) terms of a G.P. is given by: \[ S_n = a \frac{1 - r^n}{1 - r} \] For the first nine terms: \[ S_9 = \frac{1}{13} \cdot \frac{1 - 3^9}{1 - 3} \] \[ S_9 = \frac{1}{13} \cdot \frac{1 - 19683}{-2} \] \[ S_9 = \frac{1}{13} \cdot \frac{-19682}{-2} = \frac{1}{13} \cdot 9841 = 757 \] Thus, the sum of the first nine terms is \( 757 \).