The first term of a GP is 1/8 and the fifth term is 162. Find the first five terms of the GP, then find their sum.
there are 4 ratios from the 1st term to the 5th term
1/8 * r^4 = 162 ... r^4 = 162 / (1/8) ... r = 6
sum = [1/8 * (1 - 6^5)] / (1 - 6)
Using the standard formulas
a = 1/8
term(5) = ar^4
(1/8)r^4 = 162
r^4 = 1296
r^2 = 36 , if we want only real numbers
r = ±6
if r = 6
the terms are 1/8 , 3/4, 9/2, 27, 162, ...
sum(5) = a(r^5 - 1)/(r-1) = (1/8)(7776 - 1)/5 = 1555/8
if r = -6, the terms are 1/8, -3/4, 9/2, -27, 162
sum(5) = (1/8)( (-6)^5 - 1)/(-6-1) = 1111/8
To find the first term and the common ratio of a geometric progression (GP), we can use the formula:
\(A_n = A_1 \times r^{(n-1)}\)
where \(A_n\) represents the nth term, \(A_1\) is the first term, \(r\) is the common ratio, and \(n\) is the position of the term.
We are given that the first term is \(A_1 = \frac{1}{8}\), and the fifth term is \(A_5 = 162\). Plugging these values into the formula, we can solve for the common ratio (\(r\)):
\(A_5 = A_1 \times r^{(5-1)}\)
\(162 = \frac{1}{8} \times r^4\)
To simplify the equation, let's rewrite 162 and 1/8 as powers of 2:
\(162 = 2^1 \times 3^4\) (2^1 because 2 x 81 = 162 and 81 can be written as \(3^4\))
\(\frac{1}{8} = 2^{-3}\)
Now, let's substitute these values into the equation:
\(2^1 \times 3^4 = 2^{-3} \times r^4\)
By comparing the powers of 2 on both sides of the equation, we can determine that \(r = 2 \times 3 = 6\).
Now that we have the first term (\(A_1 = \frac{1}{8}\)) and the common ratio (\(r = 6\)), we can find the first five terms of the GP:
\(A_1 = \frac{1}{8}\)
\(A_2 = \frac{1}{8} \times 6 = \frac{6}{8} = \frac{3}{4}\)
\(A_3 = \frac{3}{4} \times 6 = \frac{18}{4} = \frac{9}{2}\)
\(A_4 = \frac{9}{2} \times 6 = \frac{54}{2} = 27\)
\(A_5 = 27 \times 6 = 162\)
The first five terms of the GP are \(\frac{1}{8}, \frac{3}{4}, \frac{9}{2}, 27, 162\).
To find the sum of the first five terms, we can use the formula for the sum of a geometric series:
\(S_n = \frac{A_1 \times (r^n - 1)}{r - 1}\)
where \(S_n\) represents the sum of the first \(n\) terms.
Substituting the values into the formula:
\(S_5 = \frac{\frac{1}{8} \times (6^5 - 1)}{6 - 1}\)
\(S_5 = \frac{\frac{1}{8} \times (7775)}{5}\)
\(S_5 = \frac{7775}{8 \times 5}\)
\(S_5 = \frac{7775}{40}\)
\(S_5 = 194.375\)
Therefore, the sum of the first five terms of the GP is 194.375.