Question

A line of geometry, the 3rd term is `a^(-4)` and the 4th term is `a^x`. The 10th term is `a^(52)`, then, x is?

Answer 1

See below

Explanation:

For a geometric sequence, let the `n^th` term be `a*r^(n-1)`
where r is the fixed ratio and a is the first term.

`a_3 = a*r^(3-1)`
`a^-4 = a*r^2` ----(1)

`a_10 = a*r^(10-1)`
`a^52 = a*r^9` ----(2)

Divide equation (2) by (1)

`a^52/a^-4 = (ar^9)/(ar^2)`

`a^56 = r^7`
`(a^8)^7 = r^7`
`a^8 = r`

Put this value in (2)

`a^52 = a*(a^8)^9`
`a^52 = a*a^72`
`a^52/a^72 = a`
`a^-20 = a`

Now we got both a and r.

We can put these values in `a_4` to get the answer.

`a_4 = a*r^(4-1)`
`a^x = a^-20*(a^8)^3`
`a^x = a^-20*a^24`
`a^x = a^4`

Thus, x = 4

Answer 2

`x = 4`

Explanation:

`n^(th) term is Ar^(n-1)`
where A is the first term and r is the common ratio.

`A_3 = A r^2 = a^-4` Eqn (1)#

`A_(10) = Ar^9 = a^(52)` Eqn (2)

Dividing Eqn (2) by (1),

`r^7 = a^(56), r = a^8`

Substituting value of r in Eqn (1),

`A a^(16) = a^(-4)`

`A = a^(-20)`

`A_4 = A r^3 = a^(-20) * a^(24) = a^x`

`a^(4) = a^x, x = 4`