Identify the 42nd term of an arithmetic sequence where a_1=-12 and a_27=66
2 answers:
Answer:
a₄₂ = 107
Step-by-step explanation:
The nth term of an arithmetic sequence is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Given a₂₇ = 66 , then
a₁ + 26d = 66 , that is
- 12 + 26d = 66 ( add 12 to both sides )
26d = 78 ( divide both sides by 26 )
d = 3
Then
a₄₂ = - 16 + (41 × 3) = - 12 + 123 = 111
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which indeed gives the recurrence you found,
but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that
, and substituting this into the recurrence, you find that 
for all 
.
Next, the linear term tells you that
, or 
.
Now, if
is the first term in the sequence, then by the recurrence you have
and so on, such that
for all 
.
Finally, the quadratic term gives
, or 
. Then by the recurrence,
and so on, such that
for all.
Now, the solution was proposed to be
so the general solution would be
which indeed gives the recurrence you found,
but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that
Next, the linear term tells you that
Now, if
and so on, such that
Finally, the quadratic term gives
and so on, such that
for all
Now, the solution was proposed to be
so the general solution would be