The 7th term of a gp is 27 and the 2nd term is 1by 9 find the common ratio and the sum of the Ist 9 terms
I did not see the answer to the question I ask for
Fist term: 81
Common ratio: 1/3
To find the common ratio (r) of a geometric progression (GP), we can use the formula:
\[ a_n = a_1 \times r^{(n-1)} \]
Given that the 7th term (a_7) is 27 and the 2nd term (a_2) is 1/9, we can set up the following equations:
\[ a_7 = a_1 \times r^{(7-1)} \]
\[ 27 = a_1 \times r^6 \]
\[ a_2 = a_1 \times r^{(2-1)} \]
\[ \frac{1}{9} = a_1 \times r^1 \]
From the second equation, we can rewrite it as:
\[ a_1 = \frac{1}{9r} \]
Substituting this value of a_1 into the first equation:
\[ 27 = \frac{1}{9r} \times r^6 \]
\[ 27 = \frac{r^6}{9r} \]
\[ 27 = \frac{r^5}{9} \]
Now, we can simplify the equation by multiplying both sides by 9 to get rid of the fraction:
\[ 27 \times 9 = r^5 \]
\[ 243 = r^5 \]
To solve for r, we take the fifth root of both sides of the equation:
\[ r = \sqrt[5]{243} \]
\[ r = 3 \]
Therefore, the common ratio (r) of the GP is 3.
To find the sum of the first 9 terms of the GP, we can use the formula for the sum of a geometric series:
\[ S_n = \frac{a_1 \times (1 - r^n)}{(1 - r)} \]
Plugging in the values, we have:
\[ S_9 = \frac{\frac{1}{9r} \times (1 - r^9)}{(1 - r)} \]
\[ S_9 = \frac{\frac{1}{9 \times 3} \times (1 - 3^9)}{(1 - 3)} \]
\[ S_9 = \frac{1}{27} \times \frac{1 - 19683}{-2} \]
\[ S_9 = \frac{1}{27} \times (-19682) \]
\[ S_9 = -728 \]
Therefore, the sum of the first 9 terms of the GP is -728.