Question

*Please help*For a given geometric sequence, the first term is 29/81. The sixth term is -87. What is the 10th term?

Answer 1

a₁₀ = -7047

Explanation:

The explicit form of the equation of a Geometric Sequence (used to find any term) is aₙ=a₁(r)ⁿ⁻¹, where a₁ is the first term in the sequence, aₙ is the nth term (ie, 2nd, 3rd, 4th, etc) of the sequence, n is the number of the term (1st, 2nd, 3rd, ... = 1, 2, 3, ...), and r is the common ratio (the number you multiply by to get the next, then the next, etc, term). For this sequence,
a₁ = 29/81
a₆ = -87
a₁₀ = ??

1. Use a₁ and a₆ to Find r:
-87 = (29/81)(r)⁶⁻¹
(81/29)(-87) = (81/29)(29/81)(r)⁵
-243 = r⁵

`\sqrt[5]{-243} =\sqrt[5]{r^5}`
-3 = r

2. The explicit equation used to find terms in this geometric sequence is
aₙ = (29/81)(-3)ⁿ⁻¹. Use it to find the 10th term (a₁₀):
a₁₀ = (29/81)(-3)¹⁰⁻¹
= (29/81)(-3)⁹
= (29/81)(-19683)
= -7047

Answer 2

a₁₀ = -7047

Explanation:

The general formula for the nth term of a geometric sequence is:

`\large\boxed{a_n=ar^(n-1)}`

where:

• 
  `a_n` is the nth term.
• a is the first term.
• r is the common ratio.
• n is the position of the term

Given that the first term is 29/81, then a = 29/81:

`a_n=\left((29)/(81)\right)r^(n-1)`

To find the common ratio (r), substitute a₆ = -87 into the equation:

`\begin{aligned}a_6=\left((29)/(81)\right)r^(6-1)&=-87\\\\\left((29)/(81)\right)r^(5)&=-87\\\\r^(5)&=-87\cdot (81)/(29)\\\\r^(5)&=-243\\\\\sqrt[5]{r^5}&=\sqrt[n]{-243}\\\\r&=-3\end{aligned}`

Therefore, the equation for the nth term of the given geometric sequence is:

`a_n=\left((29)/(81)\right)(-3)^(n-1)`

To find the 10th term, substitute n = 10 into the nth term equation:

`a_(10)=\left((29)/(81)\right)(-3)^(10-1)`

`a_(10)=\left((29)/(81)\right)(-3)^(9)`

`a_(10)=\left((29)/(81)\right)(-19683)`

`a_(10)=-(570807)/(81)`

`a_(10)=-7047`

Therefore, the 10th term of the given geometric sequence is -7047.