randRangeNonZero(-8, 8) randFromArray([-1, 1]) * randRange(1, 4) randRange(1, 4) randRange(2, 5) randRange(-N - 1, 0) _.map(_.range(N), function(i) { if (i + OFFSET >= 0) { return reduce(A * pow(RN, i + OFFSET), pow(RD, i + OFFSET)); } else { return reduce(A * pow(RD, -i - OFFSET), pow(RN, -i - OFFSET)); } }) fractionReduce(RN, RD) _.map(TERMS, function(f) { return fractionReduce.apply(KhanUtil, f); })

The geometric sequence `(a_i)` is defined by the formula:

`a_i = TERMS_TEX[0] \left(R_TEX\right)^{i - 1}`

What is `a_{N}`, the ordinalThrough20(N) term in the sequence?

A * pow(RN / RD, N - 1 + OFFSET)

From the given formula, we can see that the first term of the sequence is `TERMS_TEX[0]` and the common ratio is `R_TEX`.

The second term is simply the first term times the common ratio.

Therefore, the second term is equal to `a_2 = TERMS_TEX[0] \cdot R_TEX = TERMS_TEX[1]`.

To find `a_{N}`, we can simply substitute `i = N` into the given formula.

Therefore, the ordinalThrough20(N) term is equal to `a_{N} = TERMS_TEX[0] \left(R_TEX\right)^{N - 1} = TERMS_TEX[N-1]`.

`a_1 = TERMS_TEX[0]`
`a_i = R_TEXa_{i-1}`

From the given formula, we can see that the first term of the sequence is `TERMS_TEX[0]` and the common ratio is `R_TEX`.

The second term is simply the first term times the common ratio.

Therefore, the second term is equal to `a_2 = TERMS_TEX[0] \cdot R_TEX = TERMS_TEX[1]`.

To find the ordinalThrough20(N) term, we can rewrite the given recurrence as an explicit formula.

The general form for a geometric sequence is `a_i = a_1 r^{i - 1}`. In this case, we have `a_i = TERMS_TEX[0] \left(R_TEX\right)^{i - 1}`.

To find `a_{N}`, we can simply substitute `i = N` into the formula.

Therefore, the ordinalThrough20(N) term is equal to `a_{N} = TERMS_TEX[0] \left(R_TEX\right)^{N - 1} = TERMS_TEX[N-1]`.