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The third and seventh terms of a geometric sequence are43 and 12 respectively.(a) Find the 10th term.(b) What term is equal to 3072?

Question

The third and seventh terms of a geometric sequence are43 and 12 respectively.(a) Find the 10th term.(b) What term is equal to 3072?

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Solution

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A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.

Given that the third term (a3) is 43 and the seventh term (a7) is 12, we can use the formula for the nth term of a geometric sequence:

an = ar^(n-1)

where a is the first term, r is the common ratio, and n is the term number.

We have two equations from the problem:

43 = ar^2 (equation 1, from the third term)

12 = ar^6 (equation 2, from the seventh term)

We can divide equation 2 by equation 1 to solve for r:

(12/43) = r^4

Taking the fourth root of both sides, we find that r = (12/43)^(1/4) = 0.6436 (approximately).

Substitute r = 0.6436 into equation 1 to find a:

43 = a*(0.6436)^2

Solving for a, we get a = 43 / (0.6436)^2 = 103.97 (approximately).

(a) To find the 10th term, we substitute a = 103.97, r = 0.6436, and n = 10 into the formula:

a10 = 103.97 * (0.6436)^9 = 5.38 (approximately).

(b) To find which term is equal to 3072, we set an = 3072 in the formula and solve for n:

3072 = 103.97 * (0.6436)^(n-1)

Taking the natural logarithm of both sides and solving for n, we get n = log(3072/103.97) / log(0.6436) + 1 = 15.99.

Since n must be an integer, the 16th term is the first one to exceed 3072. So, no term is exactly equal to 3072.

This problem has been solved

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