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:

Question

If \( A = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix} \), then \( A^{50} \) is:

Answer 1

The objective is to determine the sum of the initial nine terms of a specified geometric progression (G.P.).

Let the first term of the G.P. be denoted by \(a\) and the common ratio by \(r\). The sequence can be represented as \(a, ar, ar^2, ar^3, \ldots\).

The second, fourth, and sixth terms are:

The sum of these terms is given as 21:

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

The eighth, tenth, and twelfth terms are:

The sum of these terms is given as 15309:

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

Dividing Equation 2 by Equation 1 yields:

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

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

Therefore, \(r^6 = 729\).

\(r^6 = 3^6\), which implies \(r = 3\).

Substituting \(r = 3\) into Equation 1 to find \(a\):

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

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

\(a \cdot 273 = 21\)

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

The sum of the first nine terms of the G.P. is calculated using the formula:

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

With \(a = \frac{1}{13}\), \(r = 3\), and \(n = 9\):

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

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

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

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

\(S_9 = 757\).

The sum of the first nine terms is 757.

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