Question

What is a1
of the arithmetic sequence for which a3=126
and a64=3,725
a
64
=
3
,
725
?

Answer 1

In an arithmetic sequence, every pair of consecutive terms differs by a fixed number c, so that the n-th term $a_n$ is given recursively by

$a_n=a_{n-1}+c$

Then for n ≥ 2, we have

$a_2=a_1+c$

$a_3=a_2+c = (a_1+c)+c = a_1 + 2c$

$a_4=a_3+c = (a_1 + 2c) + c = a_1 + 3c$

and so on, up to

$a_n=a_1+(n-1)c$

Given that $a_3=126$ and $a_{64}=3725$, we can solve for $a_1$:

$\begin{cases}a_1+2c=126\\a_1+63c=3725\end{cases}$

$\implies(a_1+63c)-(a_1+2c)=3725-126$

$\implies 61c = 3599$

$\implies c=59$

$\implies a_1+2\times59=126$

$\implies a_1+118 = 126$

$\implies \boxed{a_1=8}$