The first term of the geometric sequance is 27,the last term is 8 and the sum of the series is 65.how many terms are there in the series?
T1 = a = 27
Tn = ar^(n-1) = 8
Sn = 65 = a*(1-r^n)/(1-r)
27r^(n-1) = 8
r^(n-1) = 8/27
Just as a guess, I'd say r=2/3 and n=4, since 8/27 = (2/3)^3
However, let's go on.
27(1-r^n)/(1-r) = 65
27(1-r*8/27) / (1-r) = 65
27(1-8r/27) = 65 - 65r
27 - 8r = 65 - 65r
57r = 38
r = 38/57 = 2/3
So, now we have
27*(2/3)^(n-1) = 8
(2/3)^(n-1) = 8/27 = (2/3)^3
n-1 = 3
n=4
Sequence:
27 18 12 8
Sum: 65
To find the number of terms in a geometric series, we need to use the formula for the sum of a geometric series:
Sn = a * (r^n - 1) / (r - 1)
Where:
Sn is the sum of the series,
a is the first term of the series,
r is the common ratio of the series,
n is the number of terms in the series.
In this case, we are given:
a = 27 (first term)
Sn = 65 (sum of the series)
To find the common ratio (r), we can use the formula:
r = (last term) / (first term)
In this case, the last term is 8, so:
r = 8 / 27
Now we can substitute the given values into the formula for Sn and solve for n:
65 = 27 * ((8 / 27)^n - 1) / ((8 / 27) - 1)
To simplify this equation, we can multiply both sides by ((8 / 27) - 1) to eliminate the fraction:
65 * ((8 / 27) - 1) = 27 * ((8 / 27)^n - 1)
(65 * 27 - 65 * 8) / 27 = 27 * ((8 / 27)^n - 1)
(65 * 27 - 65 * 8) / 27 = (8^n - 1)
Simplifying further:
(65 * 27 - 65 * 8) / 27 = 8^n - 1
(65 * 27 - 65 * 8) / 27 + 1 = 8^n
Now, we can solve for n by taking the logarithm of both sides of the equation to isolate the exponent:
log((65 * 27 - 65 * 8) / 27 + 1) = n * log(8)
n = log((65 * 27 - 65 * 8) / 27 + 1) / log(8)
Using a calculator, we can find the value of n, which is approximately 3.037. Since the number of terms must be a whole number, we can round up to the next integer. Therefore, there are 4 terms in the series.