What is the sum of the geometric sequence 4/3, 2/3, 1/3, 1/6, ....?
I got 8/3, am I right?
No.
You need to change the fractions to equivalent fractions with the common denominator of 6.
4/3 = 8/6
2/3 = 4/6
1/3 = 2/6
This would equal 2 1/2 or 2.50
You want the sum of "all " the terms
an infinite GS
you have a GS, where a = 4/3, r = 1/2
sum(of all the terms) = a/(1-r)
= (4/3)/(1-1/2)
= (4/3)/(1/2)
= 8/3 or 2 2/3
To find the sum of a geometric sequence, you can use the formula:
S = a * (1 - r^n) / (1 - r)
Where:
S is the sum of the sequence,
a is the first term of the sequence,
r is the common ratio, and
n is the number of terms.
In this case, the first term (a) is 4/3, and the common ratio (r) is 1/2. The terms go on indefinitely, so we can assume n approaches infinity.
Using the formula, the sum (S) would be:
S = (4/3) * (1 - (1/2)^infinity) / (1 - 1/2)
Now, let's simplify the expression.
Since (1/2)^infinity approaches zero as n approaches infinity, we can rewrite the expression as:
S = (4/3) * (1 - 0) / (1 - 1/2)
Simplifying further, we have:
S = (4/3) * (1) / (1/2)
Multiplying the numerator and denominator by the reciprocal of 1/2 (which is 2/1) gives us:
S = (4/3) * (2/1)
Now, multiply the numerators and denominators:
S = (4 * 2) / (3 * 1)
Simplifying the expression, we have:
S = 8/3
Therefore, according to the calculations, the sum of the given geometric sequence is indeed 8/3. Your answer is correct.