A geometric progression has first term a (where a>0), and common ratio r. Anyone is good with gp questions?

Question

A geometric progression has first term a (where a>0), and common ratio r. The sum of the first n terms is S(n) and the sum to infinity is S. Given that S(2) is twice the valus of the fifth term, find the value of r. Hence, find the least value of n such that S(n) is within 5% of S.

Wrong equation!
S(2) = 2*x_4, or a(1-r^3)/(1-r)=2*a*r^4 solve for r:

Answer 1

S(2) = a + ar = a(1+r)
Fifth term = ar^4
Given S(2) = twice fifth term:
a(1+r) = 2ar^4
1 + r = 2r^4
2r^4 - r - 1 = 0

We want to solve for r, where -1 < r < 1.
f(-1) = 2; f(0) = -1; f(1) = 0

There's a root between -1 and 0. Let's redefine this as a function of x:

f(x) = 2x^4 - x - 1
dy/dx = 8x^3 - 1

and use the derivative to find the root. (This might be Newton's Method; I'm not sure.) We have the point (x0, y0) = (-1, 2), and

y - y0 = m(x - x0)

We want to find x where y=0, so

x = x0 - y0/m0

gives us successive approximations for x.

The derivative at x = -1 is m = dy/dx = 8x^3 - 1 = -9

so the next approximation x1 = x0 - y0/m0 = -1 - 2/(-9) = -7/9

y1 = f(-7/9) = 0.5097
m1 = dy/dx = -4.7641

and the next approximation is x2 = x1 - y1/m1 = -0.6708

y2 = f(x2) = 0.0757
m2 = dy/dx = -3.4147

Continuing, x3 = x2 - y2/m2 = -0.6486
y3 = f(x3) = 0.0026
m3 = dy/dx = -3.1830

x4 = x3 - y3/m3 = 0.6478
y4 = f(x4) = 0.000003359

That's close enough. The full root I have on the calculator is
r = -0.647799929 = -0.6478 (Answer)

Now we need to solve for n. We have S(n) = a(1 - r^n)/(1 - r) and S = a/(1-r), and we want n where S(n) = 0.95 S.

1 - r^n = 0.95
r^n = 0.05
n log r = log 0.05
n = (log 0.05) / (log -0.6478)

Oops -- log of a negative number!! Will have to think about this some more. But if I say n = (log 0.05) / (log 0.6478) (taking out the minus sign), then I get n = 6.8999, rounded up to n = 7.

Out of curiosity, we have

S(7) = a(1-r^7)/(1-r) = 0.6359 a
S = a/(1 - r) = 0.6069

and S(7) / S = 1.04787

so S(7) is within 5% of S. Looks like your answer is n=7.

My only problem involves the log of a negative number, but on that topic, my brain is foggy right now. Anyway, hope this helps.

Answer 2

Geometric series is x_n = a*r^n

S(n) = ∑[0 -> n] ar^n = a[1-r^(n+1)] / (1-r)

Partial sum S(2) = a(1-r^3)/(1-r);

Fifth term is x_4 = a*r^4 (a is the first term, a*r is the second, etc.)

S(2) = 2*x_4, or a(1-r^3)/(1-r)=2*a*r^4 solve for r:

a(1-r^3) = 2*a*r^4*(1-r) = 2*a*r^4 - 2*a*r^5

1-r^3 = 2*r^4 - 2*r^5

2*r^5 - 2*r^4 - r^3 + 1 = 0 r = 1 is a solution, but this series does not converge. |r|<1 is a requirement for convergence. Another solution is r = -0.806 (the only one of abs value less than 1)

Using r = -0.806, the infinite sum is

S = ∑[0 -> ∞] = a/(1-r) = .556*a

The 5% limits are

.95*S = .526*a, 1.05*S = .581*a

The first partial sum to fall within these limits is S(13) = .527*a

Answer 3

sixteen = ar^3 sixteen = a * (2/3)^3 sixteen = a * (8/27) a = sixteen * (27/8) a = 2 * 27 a = fifty 4 t[n] = a * r^n t[n] = fifty 4 * (2/3)^n t[a million] = fifty 4 * (2/3) = 36 the 1st time era is 36 S = ar^4 + ar^5 + ar^6 + ar^7 S = ar^4 * (a million + r + r^2 + r^3) S * r = ar^4 * (r + r^2 + r^3 + r^4) Sr - S = ar^4 * (r^4 - a million) S * (r - a million) = ar^4 * (r^4 - a million) S * (2/3 - a million) = fifty 4 * (2/3)^4 * ((2/3)^4 - a million) S * (-a million/3) = fifty 4 * (sixteen/80 one) * (sixteen/80 one - a million) S * (-a million/3) = fifty 4 * (sixteen/80 one) * (-sixty 5/80 one) S = fifty 4 * (sixteen/80 one) * 3 * (sixty 5/80 one) S = 27 * 2 * sixteen * 3 * sixty 5 / (80 one * 80 one) S = 2 * sixteen * sixty 5 / 80 one S = 2080 / 80 one From the 1st time era to infinity S = ar + ar^2 + ar^3 + ar^4 + ... + ar^n Sr = ar^2 + ar^3 + ar^4 + ... + ar^n + ar^(n + a million) Sr - S = ar^(n + a million) - ar S * (r - a million) = ar * (r^n - a million) S * (-a million/3) = 36 * (0 - a million) S = -36 * (-3) S = 108 r^n is going to 0 because of the fact n is going to infinity and -a million < r < a million