Question

Three numbers form a arithmetic sequence whose common difference is 3. if the numbers are increased by 2,5,& 9, respectively, the resulting numbers form a geometric sequence. which of the following gives the common ratio of the resulting geometric sequen?

Answer 1

`7/6`

Explanation:

the first sequence (the arithmetic sequence, common difference `3`) can be written as

`n`, `n + 3` and `n + 6`
where `n` is the starting term of the sequence.

if the numbers are increased by `2`, `5` and `9`:

`(n) + 2 = n + 2`
`(n+3) + 5 = n + 8`
`(n+6) + 9 = n + 15`

the geometric sequence, then, is `(n+2)`, `(n+8)` and `(n+15)`.

since the sequence is geometric with a common ratio, it is known that the same common ratio is multiplied by a term each time, to get the next term.

`r(n+2) = (n+8)`
`r(n+8) = (n+15)`

if these are rearranged:

`r = (n+8)/(n+2)`
`r = (n+15)/(n+8)`

since the two left-hand expressions are the same, we know that `(n+8)/(n+2) = (n+15)/(n+8)`.

since `(n+2)` and `(n+8)` have no common factors, the common denominator of `(n+2)` and `(n+8)` is `(n+2)(n+8)`.

`(n+8)/(n+2) = ((n+8)(n+8))/((n+8)(n+2))`

`(n+15)/(n+8) = ((n+15)(n+2))/((n+8)(n+2))`

hence, `(n+8)(n+8) = (n+15)(n+2)`

expanding brackets on either side gives

`n^2+16n+64 = n^2+17n+30`

subtracting `n^2+17n+30` from both sides gives
`-n+34 = 0`
subtracting `34` gives `-n = -34`
and multiplying by `-1` gives `n = 34`.

using this, we can find that the geometric sequence, with the three terms `(n+2)`, `(n+8)` and `(n+15)`, starts with the three terms `36, 42` and `49`.

using this, `r = 42/36 = 49/42`

`42/36 = 7/6` and `49/42 = 7/6`

hence, the common ratio is `7/6`.