Question

The second term of a geometric sequence is and the sixth term is 2 and 32

The first term, a and the common ratio, r.

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Answer 1

There are two sets of solutions:

• either
  `a = 1` and
  `r = 2`, or
• 
  `a = -1` and
  `r = -2`.

Explanation:

The question states that
`a` and
`r` should represent the first term and the common ratio of this geometric sequence, respectively. Therefore:

• First term of this sequence:
  `a`.
• Second term of this sequence:
  `r\, a`.
• Third term of this sequence:
  `r^2\, a`.
• And so on so forth. Each term of this sequence after the first term is equal to
  `r` times the previous term.

Let
`n` denote a positive whole number. In general, the
`n\!`-th term of this geometric sequence would be
`r^(n-1)\, a`.

The sixth term of this sequence would thus be
`r^(6-1)\, a = r^5\, a`.

The fact that the second term is
`2` and the sixth term is
`32` gives two equations about
`a` and
`r`:

`\displaystyle \left\lbrace\begin{aligned}& r\, a = 2 \\ & r^5\, a= 32\end{aligned}\right.`.

Take the quotient of these two equations to eliminate
`a`:

`\displaystyle (r^5\, a)/(r\, a) = (32)/(2)`.

`r^4 = 16`.

There are two possible roots: either
`r = 2` or
`r = -2`.

When
`r = 2`, substitute back to the first equation and solve for
`a`:
`a = 1`.

On the other hand, when
`r = -2`, substituting back and solving for
`a` would give
`a = -1`.

Substitute these values into the second equation. Either set of values will work.