Make up a geometric series ∑a rⁿ⁻¹ that converges to the numbe...

Question

Make up a geometric series ∑a rⁿ⁻¹ that converges to the number 5 if

b. a = 13/2

Accepted Answer

Welcome back everyone. Find the common ratio r for an infinite geometric series sigma from n equals 1 up to infinity of a multiplied by r to the power of n minus 1 that converges to 4, given that the first term is a equals 14/3, a 16, b 1/6, C 1/3, and D 2/3. For this problem we're going to use the sum formula for the infinite geometric series. Let's remember that sums is equal to our first term a divided by 1 minus R. And specifically we have to ensure that the absolute value of R is less than 1, right? Because our geometric theories converges. What we're going to do is solve for R, so a divided by s is equal to 1 minus r. And therefore if we subtract one from both sides, we're going to get a divided by s minus 1 is equal to r. Multiplying both sides by -1, we get 1 minus a divided by s is equal to r. Substituting the values, we get 1 minus a, which is 14/3 divided by s, which is our sum, it says converges to 4, right? This is going to be equal to r. And now let's simplify. So we get 1 minus 14 divided by 3 multiplied by 4 that's divided by 12 is equal to r. And now what we can do is simplify, right? So that would be 1 minus 7 divided by 6 equals R, or basically R is equal to -16. Which satisfies the required condition, the absolute value of -16 is less than 1, and the correct answer is option B. Thank you for watching.

Make up a geometric series ∑a rⁿ⁻¹ that converges to the number 5 ifb. a = 13/2

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Make up a geometric series ∑a rⁿ⁻¹ that converges to the number 5 if
b. a = 13/2