The product of the first five terms of a geometric progression is 32. If the fourth term is 17, compute the second term.
a(ar)(ar^2)(ar^3)(ar^4) = 32
a^5 r^10 = 32
take 5th root
a r^2 = 2 ---- #1
ar^3 = 17 ----#2
divide #2 by #1
r = 17/2
sub into #1
a(17/2)^2 = 2
a(289/4) = 2
a = 8/289
term(2) = ar = (8/289)(17/2) =4/17
To find the second term of a geometric progression, we need to use the formula for the Nth term of a geometric progression:
\(a_n = a_1 \times r^{(n-1)}\)
Where:
\(a_n\) is the Nth term,
\(a_1\) is the first term,
\(r\) is the common ratio,
\(n\) is the term number.
Given that the product of the first five terms is 32, we have:
\(a_1 \times a_2 \times a_3 \times a_4 \times a_5 = 32\)
And we are given that the fourth term is 17:
\(a_4 = 17\)
From here, we can deduce the value of the first term, \(a_1\), using the formula for the Nth term of a geometric progression:
\(a_4 = a_1 \times r^{(4-1)}\)
Substituting the given value for \(a_4\):
\(17 = a_1 \times r^3\)
Now we can solve for \(a_1\) by isolating it:
\(a_1 = \frac{17}{r^3}\)
Next, we substitute this value of \(a_1\) into the expression for the product of the first five terms:
\(a_1 \times a_2 \times a_3 \times a_4 \times a_5 = 32\)
\(\frac{17}{r^3} \times a_2 \times a_3 \times 17 \times a_5 = 32\)
Simplifying:
\(289 \times a_2 \times a_3 \times a_5 = 32 \times r^3\)
We're almost there! We're left with one unknown, the second term \(a_2\). To find its value, we divide both sides of the equation by \(289 \times a_3 \times a_5\):
\(a_2 = \frac{32 \times r^3}{289 \times a_3 \times a_5}\)
And there you have it – the formula to compute the second term of the geometric progression. To obtain the actual value of the second term, you would need to know the values of the common ratio, \(r\), as well as the third and fifth terms, \(a_3\) and \(a_5\), respectively.