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Consider the infinite geometric sequence that has a third term equal to -1/6 and a sixth term equal to -1/1296. What is the sum to infinity of the terms of this sequence?

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MrRoyal

The sum to infinity of the terms of this sequence is -36/5

What is the sum to infinity of the terms of this sequence?

The given parameters are

T3 = -1/6

T6 = -1/1296

The nth term of a geometric sequence is

Tn = a * r^(n-1)

So, we have

T3 = ar^2

T6 = ar^5

Divide both equations

T6/T3 = ar^5/ar^2

Evaluate the quotient

T6/T3 = r^3

Substitute T3 = -1/6 and T6 = -1/1296 in T6/T3 = r^3

(-1/1296)/(-1/6) = r^3

Evaluate the quotient

1/216 = r^3

Take the cube root of both sides

r =1/6

Substitute r =1/6 in T3 = ar^2

T3 = a(1/6)^2

This gives

T3 = a/36

Substitute T3 = -1/6 in T3 = a/36

a/36 = -1/6

This gives

a = -6

The sum to infinity of the terms of this sequence is

S = a/(1 - r)

This gives

S = -6/(1 - 1/6)

This gives

S = -6/(5/6)

Evaluate

S = -36/5

Hence, the sum to infinity of the terms of this sequence is -36/5

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