The 6th term of a GP is 2000.find its first term if it common ratio is 10.
ar^5 = 2000 , but r = 10
a(10^5) = 2000
100000a = 2000
a = 2000/100000 = 1/50 <------ first term
To find the first term of a geometric progression (GP) when the common ratio and the 6th term are given, we can use the formula for the nth term of a GP:
\[ T_n = a \times r^{(n-1)} \]
where:
- \( T_n \) is the nth term,
- \( a \) is the first term,
- \( r \) is the common ratio,
- and \( n \) is the position of the term in the sequence.
In this case, we are given:
- The 6th term (\( T_6 \)) is 2000.
- The common ratio (\( r \)) is 10.
To find the first term (\( a \)), we substitute the given values into the formula and solve for \( a \):
\[ T_6 = a \times r^{(6-1)} \]
Substituting the values and simplifying:
\[ 2000 = a \times 10^5 \]
Dividing both sides by \( 10^5 \):
\[ \frac{2000}{10^5} = \frac{a \times 10^5}{10^5} \]
Simplifying:
\[ \frac{2000}{10^5} = a \]
Calculating:
\[ a = 0.02 \]
Therefore, the first term of the GP is 0.02.
To find the first term of a geometric progression (GP) with a given common ratio and 6th term, you can use the formula:
\[
a_n = a_1 \cdot r^{(n-1)}
\]
where \(a_n\) is the nth term of the GP, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.
In this case, the 6th term is 2000, and the common ratio is 10. Let's solve for the first term (\(a_1\)).
Plugging the values into the formula:
\[
2000 = a_1 \cdot 10^{(6-1)}
\]
Simplifying further:
\[
2000 = a_1 \cdot 10^5
\]
To isolate \(a_1\), divide both sides of the equation by \(10^5\):
\[
\frac{2000}{10^5} = a_1
\]
The first term, \(a_1\), is equal to 0.02.
Therefore, the first term of the GP is 0.02.