Question

How do you find the formula for `a_n` for the arithmetic sequence `a_3=94, a_6=85`?

Answer 1

`n` th term is `a_n= 100 -3(n-1)`

Explanation:

`3` rd term `a_3= a_1+2 d =94; (1); a_1 and d` are first term

and common difference of A.P sequence.

`6` th term #a_6= a_1+5 d =85 ; (2); . Subtracting equation (1)

from equation (2) we get, `3 d = -9 :. d= -9/3 =-3` Putting

`d=-3` in equation (1) we get, `a_1+2 * (-3) =94` or

`a_1=94+6=100 :.` first term `a_1=100` Therefore,

`n` th term is `a_n= a_1+(n-1) d or a_n= 100 -3(n-1)` [Ans]

Answer 2

`a_n=103-3n`

Explanation:

Since this is an arithmetic sequence:
for some constant `k`
`color(white)("XXX")a_6=color(blue)(a_5)+k`
`color(white)("XXX"a_6)=color(blue)(""(color(lime)(a_4)+k))+k`
`color(white)("XXX"a_6)=color(blue)(""(color(lime)(""(a_3+k))+k))+k`
`color(white)("XXX"a_6)=a_3+3k`

since `a_6=85` and `a_3=64`
`color(white)("XXX")3k = 85-94=-9`
`color(white)("XXX")rArr k=-3`

Similarly
`color(white)("XXX")a_1=a_0+k`
`color(white)("XXX")a_2=a_0+2k`
`color(white)("XXX")a_3=a_0+3k`
and
since `a_3=94` and `k=-3`
`color(white)("XXX")94=a_0+3 * (-3)=a_0-9`

`color(white)("XXX")rArr a_0=103`

The general arithmetic formula is
`color(white)("XXX")a_n=a_0+ k * n`
So, for this specific case,
`color(white)("XXX")a_n=103+(-3)n`