The 12th term of a sequence is 41 and the 15th term is 140. What is the nth term?
1 answer:
An= mth term.
an=a₁+(n-1)*d
a₁₂=41
a₁₅=140
a₁₂=41
41=a₁+(12-1)*d
41=a₁+11d
a₁+11d=41 (1)
a₁₅=140
140=a₁+(15-1)*d
140=a₁+14d
a₁+14d=140 (2)
With the equiations (1) and (2) build a system of equations
a₁+11d=41
a₁+14d=140
we solve it.
-(a₁+11d=41)
a₁+14d=140
--------------------
3d=99 ⇒d=99/3=33
a₁+11d=41
a₁+(11*33)=41
a₁+363=41
a₁=41-363=-322
an=a₁+(n-1)*d
an=-322+(n-1)*33
an=-322+33n-33
an=-355+33n
an=-355+33n
To check:
a₁₂=-355+33*12=-355+396=41
a₁₅=-355+33*15=-355+495=140.
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