The sum of the two first two terms of an exponential sequence is 135 and the sum of the third and the fourth terms is 6. Given that the common ratio is positive, calculate:

Question

(I)the common ratio and the first tern
(II)the limit of the sum of the first n termsas n becomes large;
(III)the least number of terms for which the sum exceed 240

Answer 1

Let's call the first term of the exponential sequence "a" and the common ratio "r".

(I) We know that the sum of the first two terms is 135, so we can write the equation:

a + ar = 135 (equation 1)

We also know that the sum of the third and fourth terms is 6, so we can write the equation:

ar^2 + ar^3 = 6 (equation 2)

To solve these equations, we can use substitution. From equation 1, we can solve for "a" in terms of "r":

a = 135 - ar

Substitute this expression for "a" into equation 2:

(135 - ar)r^2 + (135 - ar)r^3 = 6

Expand and simplify:

135r^2 - ar^3 - 135r^3 + ar^4 = 6

Rearrange the terms:

ar^4 - ar^3 + 135r^3 - 135r^2 = 6

Factor out the common factor "r^2":

r^2(ar^2 - ar + 135r - 135) = 6

Since the common ratio is positive, we can divide both sides by "r^2":

ar^2 - ar + 135r - 135 = 6/r^2

We know that "r" cannot be zero, so we can divide both sides by "r":

ar - a + 135 - 135/r = 6/r^3

Rearrange the terms:

ar - a = 6/r^3 - 135 + 135/r

Factor out "a" on the left side:

a(r - 1) = 6/r^3 - 135 + 135/r

Divide both sides by "r - 1":

a = (6/r^3 - 135 + 135/r)/(r - 1)

Now we have an expression for "a" in terms of "r". We can substitute this expression into equation 1 to solve for "r":

(6/r^3 - 135 + 135/r)/(r - 1) + (6/r^3 - 135 + 135/r)/(r - 1)r = 135

Simplify:

(6/r^3 - 135 + 135/r + (6/r^3 - 135 + 135/r)r)/(r - 1) = 135

Combine like terms:

(6/r^3 - 135 + 135r^2/r^2)/(r - 1) = 135

Simplify:

(6/r^3 - 135 + 135r^2)/(r - 1) = 135

Multiply both sides by (r - 1):

6 - 135(r - 1) + 135r^2 = 135(r - 1)

Expand and simplify:

6 - 135r + 135 + 135r^2 = 135r - 135

Rearrange the terms:

135r^2 - 135r - 135 = 0

Divide both sides by 135:

r^2 - r - 1 = 0

This is a quadratic equation. We can solve it using the quadratic formula:

r = (-(-1) ± sqrt((-1)^2 - 4*1*(-1)))/(2*1)

Simplify:

r = (1 ± sqrt(1 + 4))/2

r = (1 ± sqrt(5))/2

Since the common ratio is positive, we take the positive solution:

r = (1 + sqrt(5))/2

Substitute this value of "r" back into equation 1 to solve for "a":

a + a((1 + sqrt(5))/2) = 135

Simplify:

a + (a + sqrt(5)a)/2 = 135

Multiply both sides by 2:

2a + a + sqrt(5)a = 270

Combine like terms:

3a + sqrt(5)a = 270

Factor out "a":

a(3 + sqrt(5)) = 270

Divide both sides by (3 + sqrt(5)):

a = 270/(3 + sqrt(5))

Simplify:

a = 270(3 - sqrt(5))/(3^2 - (sqrt(5))^2)

a = 270(3 - sqrt(5))/(9 - 5)

a = 270(3 - sqrt(5))/4

So the common ratio "r" is (1 + sqrt(5))/2 and the first term "a" is 270(3 - sqrt(5))/4.

(II) To find the limit of the sum of the first n terms as n becomes large, we need to find the limit of the sum of the geometric series. The formula for the sum of a geometric series is:

Sn = a(1 - r^n)/(1 - r)

As n becomes large, the term r^n approaches 0 if 0 < r < 1. In this case, 0 < (1 + sqrt(5))/2 < 1, so the sum of the series approaches:

S = a/(1 - r)

Substitute the values of "a" and "r" into the formula:

S = (270(3 - sqrt(5))/4)/(1 - (1 + sqrt(5))/2)

Simplify:

S = (270(3 - sqrt(5))/4)/((2 - sqrt(5))/2)

Multiply the numerator and denominator by 2:

S = (540(3 - sqrt(5)))/(2(2 - sqrt(5)))

Simplify:

S = (540(3 - sqrt(5)))/(4 - 2sqrt(5))

S = (540(3 - sqrt(5)))/(4 - 2sqrt(5))

Simplify:

S = (1350 - 540sqrt(5))/(4 - 2sqrt(5))

Therefore, the limit of the sum of the first n terms as n becomes large is (1350 - 540sqrt(5))/(4 - 2sqrt(5)).

(III) To find the least number of terms for which the sum exceeds 240, we need to find the value of n where the sum of the first n terms exceeds 240. We can use the formula for the sum of a geometric series:

Sn = a(1 - r^n)/(1 - r)

We need to find the smallest n such that:

Sn > 240

Substitute the values of "a" and "r" into the formula:

a(1 - r^n)/(1 - r) > 240

(270(3 - sqrt(5))/4)(1 - ((1 + sqrt(5))/2)^n)/(1 - (1 + sqrt(5))/2) > 240

Multiply both sides by 4:

270(3 - sqrt(5))(1 - ((1 + sqrt(5))/2)^n)/(2 - sqrt(5)) > 960

Multiply both sides by (2 - sqrt(5)):

270(3 - sqrt(5))(1 - ((1 + sqrt(5))/2)^n) > 960(2 - sqrt(5))

Divide both sides by 270(3 - sqrt(5)):

1 - ((1 + sqrt(5))/2)^n > (960(2 - sqrt(5)))/(270(3 - sqrt(5)))

Simplify:

1 - ((1 + sqrt(5))/2)^n > (640 - 320sqrt(5))/(270 - 135sqrt(5))

We can see that the right side is positive, so we can remove the negative sign on the left side:

((1 + sqrt(5))/2)^n - 1 < (320sqrt(5) - 640)/(270 - 135sqrt(5))

(((1 + sqrt(5))/2)^n - 1)(270 - 135sqrt(5)) < (320sqrt(5) - 640)

Expand:

((135 + 135sqrt(5))/2)^n - (270 - 135sqrt(5))/2^n - (270 - 135sqrt(5))/2^n + 1 < (320sqrt(5) - 640)

Combine like terms on the left side:

((135 + 135sqrt(5))/2)^n - 2*(270 - 135sqrt(5))/2^n + 1 < (320sqrt(5) - 640)

Multiply both sides by 2^n:

((135 + 135sqrt(5))/2)^n - 2(270 - 135sqrt(5)) + 2^n < 2^n(320sqrt(5) - 640)

We can see that the left side is increasing with n, while the right side is fixed. Therefore, the inequality will eventually be true for some value of n. Since we want to find the least number of terms, we can start with n = 1 and keep increasing n until the inequality is true.

For n = 1:

((135 + 135sqrt(5))/2) - 2(270 - 135sqrt(5)) + 2 < 2(320sqrt(5) - 640)

Rearrange the terms:

((135 + 135sqrt(5))/2) < 2(270 - 135sqrt(5)) + 638

Simplify:

((135 + 135sqrt(5))/2) < 2(908 - 135sqrt(5))

Divide both sides by 2:

135 + 135sqrt(5) < 2(908 - 135sqrt(5))

Simplify:

135 + 135sqrt(5) < 1816 - 270sqrt(5)

Combine like terms:

405sqrt(5) < 1681

Divide both sides by 405:

sqrt(5) < 1681/405

Simplify:

sqrt(5) < 1681/405

Take the square of both sides:

5 < (1681/405)^2

Simplify:

5 < 2.7625

This is true, so the least number of terms for which the sum exceeds 240 is n = 1.