What is a3 of the arithmetic sequence for which a10 is 41 and a15 is 61

1 answer:

3 0

For the arithmetic sequence
a₁, a₂, a₃, ...,
The n-th term is
$a_{n}=a_{1}+(n-1)d$
where d = common difference.

Because a₁₀ = 41,
a₁ + 9d = 41 (1)

Because a₁₅ = 61
a₁ + 14d = 61 (2)

Subtract (1) from (2).
5d = 20
d = 4
From (1), obtain
a₁ = 41 - 9*4 = 5

Therefore
a₃ = 5 + 2*4 = 13

Answer: a₃ = 13

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