8th term of a G.P. is 640.if the first term is 5 find the common ratio and 10th term
5r^7 = 640
r = 2
a_10 = 640r^2 = ___
Well, the 8th term of a G.P. is 640 and the first term is 5. So, in order to find the common ratio, we need to find the ratio between the 8th term and the first term.
640 / 5 = 128
So, the common ratio is 128.
To find the 10th term, we can use the formula
T(n) = a*r^(n-1)
where T(n) is the nth term, a is the first term, r is the common ratio, and n is the term number.
Plugging in the values we know, we have:
T(10) = 5 * 128^(10-1)
T(10) = 5 * 128^9
And I'm afraid I don't have a calculator with me to compute the actual value. But hey, since we're on the topic of G.P., maybe you can calculate it yourself and have some fun with numbers!
To find the common ratio of a geometric progression (G.P.), we can use the formula:
\[a_n = a_1 \cdot r^{(n-1)}\]
where \(a_n\) is the \(n\)th term of the G.P., \(a_1\) is the first term, and \(r\) is the common ratio.
Given that the 8th term of the G.P. is 640 and the first term is 5, we can substitute these values into the formula:
\[640 = 5 \cdot r^{(8-1)}\]
Simplifying the equation:
\[640 = 5 \cdot r^7\]
Dividing both sides by 5:
\[128 = r^7\]
To solve for \(r\), we can take the 7th root of both sides:
\[r = \sqrt[7]{128}\]
Using a calculator, we find that \(r\) is approximately 2.
To find the 10th term of the G.P., we can use the formula:
\[a_{10} = a_1 \cdot r^{(10-1)}\]
Substituting the given values:
\[a_{10} = 5 \cdot 2^{(10-1)}\]
Evaluating the expression:
\[a_{10} = 5 \cdot 2^9\]
Simplifying:
\[a_{10} = 5 \cdot 512\]
\[a_{10} = 2560\]
Therefore, the common ratio for the G.P. is 2, and the 10th term is 2560.
To find the common ratio in a geometric progression (G.P.), we can use the formula for the nth term of a G.P:
\[ T_n = a \cdot r^{(n-1)} \]
Where:
T_n is the nth term,
a is the first term, and
r is the common ratio.
According to the given information, the 8th term (T_8) is 640, and the first term (a) is 5. We can plug these values into the formula:
\[ T_8 = 5 \cdot r^{(8-1)} \]
\[ 640 = 5 \cdot r^7 \]
To find the common ratio (r), we can rearrange the equation and solve for r:
\[ r^7 = \frac{640}{5} \]
\[ r^7 = 128 \]
\[ r = \sqrt[7]{128} \]
\[ r = 2 \]
So, the common ratio (r) is 2.
To find the 10th term (T_10), we can again use the formula:
\[ T_n = a \cdot r^{(n-1)} \]
Plugging in the values:
\[ T_{10} = 5 \cdot 2^{(10-1)} \]
\[ T_{10} = 5 \cdot 2^9 \]
\[ T_{10} = 5 \cdot 512 \]
\[ T_{10} = 2560 \]
So, the 10th term (T_10) is 2560.