The sum of all the terms in an infinite geometric sequence is 81 and the sum
of all terms starting from the third is 9. What is the second term in the sequence?

Step (1)

We know that the sum of all the terms in an infinite geometric sequence finding from following equation whenever |q| < 1 .

q = magnitude

[tex]s( \infty ) = \frac{t(1)}{1 - q} \\ [/tex]

So :

[tex]s( \infty ) = \frac{t(1)}{1 - q} \\ [/tex]

[tex]81 = \frac{t(1)}{1 - q} \\ [/tex]

[tex]81(1 - q) = t(1)[/tex]

Remember it I'll use it again.

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Step (2)

We know that the sum of the n first terms of a geometric sequence finding from following equation.

[tex]s(n) = \frac{t(1) \times (1 - {q})^{n} }{1 - q} \\ [/tex]

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Sum of all terms starting from the third is 9.

So :

[tex]s(2) + s(3 - \infty ) = 81[/tex]

[tex]s(2) + 9 = 81[/tex]

Sides minus 9

[tex]s(2) = 81 - 9[/tex]

[tex]s(2) = 72[/tex]

So :

[tex]t(1) + t(2) = 72[/tex]

[tex]t(1) + t(1) \times q = 72[/tex]

Factoring t(1)

[tex]t(1) \times (1 + q) = 72[/tex]

We have found t(1) = 81 ( 1 - q ) in step (1).

So :

[tex]81(1 - q)(1 + q) = 72[/tex]

Divided sides by 81

[tex](1 - q)(1 + q) = \frac{72}{81} \\ [/tex]

[tex](1 - q)( 1 + q) = \frac{8}{9} \\ [/tex]

[tex]1 - {q}^{2} = \frac{8}{9} \\ [/tex]

Subtract sides minus -1

[tex] - {q}^{2} = \frac{8}{9} - 1 \\ [/tex]

[tex] - {q}^{2} = \frac{8}{9} - \frac{9}{9} \\ [/tex]

[tex] - {q}^{2} = - \frac{1}{9} \\ [/tex]

Negatives simplifies

[tex] {q}^{2} = \frac{1}{9} \\ [/tex]

Radical sides

[tex]q = + \frac{1}{3} \\ q = - \frac{1}{3} [/tex]

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Step (4)

If q = 1/3 :

[tex]81(1 - q) = t(1)[/tex]

[tex]81(1 - \frac{1}{3}) = t(1) \\ [/tex]

[tex]81( \frac{2}{3}) = t(1) \\ [/tex]

[tex]t(1) = 27 \times 2[/tex]

[tex]t(1) = 54[/tex]

So :

[tex]t(2) = t(1) \times q[/tex]

[tex]t(2) = 54 \times \frac{1}{3} \\ [/tex]

[tex]t(2) = 18[/tex]

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

If q = - 1/3 :

[tex]81(1 - q) = t(1)[/tex]

[tex]81(1 - ( - \frac{1}{3})) = t(1) \\ [/tex]

[tex]81(1 + \frac{1}{3}) = t(1) \\ [/tex]

[tex]81( \frac{4}{3}) = t(1) \\ [/tex]

[tex]t(1) = 27 \times 4[/tex]

[tex]t(1) = 108[/tex]

So :

[tex]t(2) = t(1) \times q[/tex]

[tex]t(2) = 108 \times - \frac{1}{3} \\ [/tex]

[tex]t(2) = - 36[/tex]

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And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

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