SOLUTION: The sum of the first two terms of an exponential sequence is 135 and the sum of the third and the fourth terms is 60.given that the common ratio is positive.calculate:the limit of

Question 1191149: The sum of the first two terms of an exponential sequence is 135 and the sum of the third and the fourth terms is 60.given that the common ratio is positive.calculate:the limit of the first n terms as n becomes larger and the least number of terms for which the 7um exceeds 240.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Let the sequence be a, ar, ar^2, ar^3. Then

a+ar=a(1+r)=135 [1]
a^2+ar^3=ar^2(1+r)=60 [2]

Divide [2] by [1] to determine the common ratio r:

Use r to determine the first term a.

The first term is 81; the common ratio is 2/3.

The limit of the sum is (first term) divided by (1 minus the common ratio).

ANSWER 1: 243

For the second question we need to use the formula for the sum of a finite number of terms of a geometric sequence.

We want the number of terms n for which the sum is greater than 240. Solve the equation for a sum of 240 and round up.

The variable is in an exponent -- solve using logarithms.

= 10.838 to 3 decimal places

That decimal number means with 10 terms the sum will still be less than 240 and with 11 terms it will be more than 240.

ANSWER 2: 11 terms

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