Question

if you are given a3=2 a5=16, find a100.

Answer 1

I suppose $a_n$ denotes the n-th term of some sequence, and we're given the 3rd and 5th terms $a_3=2$ and $a_5=16$. On this information alone, it's impossible to determine the 100th term $a_{100}$ because there are infinitely many sequences where 2 and 16 are the 3rd and 5th terms.

To get around that, I'll offer two plausible solutions based on different assumptions. So bear in mind that this is not a complete answer, and indeed may not even be applicable.

• Assumption 1: the sequence is arithmetic (a.k.a. linear)

In this case, consecutive terms differ by a constant d, or

$a_n = a_{n-1} + d$

By this relation,

$a_{n-1} = a_{n-2} + d$

and by substitution,

$a_n = (a_{n-2} + d) + d = a_{n-2} + 2d$

We can continue in this fashion to get

$a_n = a_{n-3} + 3d$

$a_n = a_{n-4} + 4d$

and so on, down to writing the n-th term in terms of the first as

$a_n = a_1 + (n-1)d$

Now, with the given known values, we have

$a_3 = a_1 + 2d = 2$

$a_5 = a_1 + 4d = 16$

Eliminate $a_1$ to solve for d :

$(a_1 + 4d) - (a_1 + 2d) = 16 - 2 \implies 2d = 14 \implies d = 7$

Find the first term $a_1$ :

$a_1 + 2\times7 = 2 \implies a_1 = 2 - 14 = -12$

Then the 100th term in the sequence is

$a_{100} = a_1 + 99d = -12 + 99\times7 = \boxed{681}$

• Assumption 2: the sequence is geometric

In this case, the ratio of consecutive terms is a constant r such that

$a_n = r a_{n-1}$

We can solve for $a_n$ in terms of $a_1$ like we did in the arithmetic case.

$a_{n-1} = ra_{n-2} \implies a_n = r\left(ra_{n-2}\right) = r^2 a_{n-2}$

and so on down to

$a_n = r^{n-1} a_1$

Now,

$a_3 = r^2 a_1 = 2$

$a_5 = r^4 a_1 = 16$

Eliminate $a_1$ and solve for r by dividing

$\dfrac{a_5}{a_3} = \dfrac{r^4a_1}{r^2a_1} = \dfrac{16}2 \implies r^2 = 8 \implies r = 2\sqrt2$

Solve for $a_1$ :

$r^2 a_1 = 8a_1 = 2 \implies a_1 = \dfrac14$

Then the 100th term is

$a_{100} = \dfrac{(2\sqrt2)^{99}}4 = \boxed{\dfrac{\sqrt{8^{99}}}4}$

The arithmetic case seems more likely since the final answer is a simple integer, but that's just my opinion...