In a geomatic sequence, the 7th term exceeds the 5th term by 48. Find the first term if the common ratio is 2.
Why did the geomatic sequence get a promotion?
Because it's always "exceeding" expectations!
Now, let's solve this problem. We know that in a geometric sequence, the nth term can be found using the formula:
Tn = a * r^(n - 1)
Where Tn is the nth term, a is the first term, r is the common ratio, and n is the term number.
In this case, we are given that the 7th term exceeds the 5th term by 48. So we can set up the following equation:
a * r^(7 - 1) = a * r^(5 - 1) + 48
Simplifying this equation gives us:
a * r^6 = a * r^4 + 48
Now, we can substitute the given common ratio of 2 into the equation:
a * 2^6 = a * 2^4 + 48
64a = 16a + 48
Combining like terms, we get:
48a = 48
Dividing both sides by 48, we find that:
a = 1
So, the first term of the geometric sequence is 1.
To find the first term in a geometric sequence, we need to know either the common ratio or the terms of the sequence. In this case, we are given that the common ratio is 2.
Let's assume that the first term of the geometric sequence is "a" and the common ratio is 2. We can write the terms of the sequence as follows:
1st term: a
2nd term: ar (where r is the common ratio)
3rd term: ar^2
4th term: ar^3
5th term: ar^4
6th term: ar^5
7th term: ar^6
We are also given that the 7th term exceeds the 5th term by 48.
So, ar^6 - ar^4 = 48
Now, substitute the value of the common ratio (which is 2) into the equation:
a(2^6) - a(2^4) = 48
64a - 16a = 48
48a = 48
Divide both sides of the equation by 48 to solve for a:
a = 1
Therefore, the first term of the geometric sequence is 1.
To find the first term of a geometric sequence, we can use the formula:
\[T_n = T_1 \times r^{(n-1)}\]
where:
- \(T_n\) is the \(n\)th term of the sequence
- \(T_1\) is the first term of the sequence
- \(r\) is the common ratio
- \(n\) is the position of the term in the sequence
In this case, we are given:
- \(T_7\) (7th term) exceeds \(T_5\) (5th term) by 48, which means \(T_7 = T_5 + 48\)
- The common ratio \(r\) is 2
Let's use the formula to find the terms:
\[T_7 = T_1 \times r^{(7-1)}\]
\[T_7 = T_1 \times 2^6\]
\[T_5 = T_1 \times r^{(5-1)}\]
\[T_5 = T_1 \times 2^4\]
We know from the given information that \(T_7 = T_5 + 48\). Substituting the expressions for \(T_7\) and \(T_5\) in terms of \(T_1\) and solving for \(T_1\):
\[T_1 \times 2^6 = T_1 \times 2^4 + 48\]
\[64T_1 = 16T_1 + 48\]
\[48T_1 = 48\]
\[T_1 = 1\]
Therefore, the first term of the geometric sequence is 1.
still can't spell "geometric"?
64a - 16a = 48
a = 1