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*59=126`

`\implies a_1+118 = 126`

`\implies \boxed{a_1=8}`