Question

Determine the number of terms n in each arithmetic series.

a₁=14
d=10
Sn= 1260
i know the answer is 15, I'm just not sure how to get there.

Answer 1

We're given an arithmetic sequence
`\{a_n\}_(n\ge1)` that starts with
`a_1=14` and a common difference between terms of
`d=10`. Recursively, this sequence is given by

`\begin{cases}a_n=a_(n-1)+10&\text{for }n>1\\a_1=14\end{cases}`

We can find an explicit formula for the
`n`-th term
`a_n`:

`a_2=a_1+10`

`a_3=a_2+10=a_1+2(10)`

`a_4=a_3+10=a_1+3(10)`

and so on, with the general pattern of

`a_n=a_1+(n-1)(10)=4+10n`

We're given that the sum of the first
`N` consecutive terms is

`S_N=\displaystyle\sum_(n=1)^Na_n=\sum_(n=1)^N(4+10n)=1260`

Recall that

`\displaystyle\sum_(n=1)^N1=N`

`\displaystyle\sum_(n=1)^Nn=\frac{N(N+1)}2`

So we solve for
`N`:

`1260=\displaystyle\sum_(n=1)^N(4+10n)`

`1260=\displaystyle\sum_(n=1)^N4+\sum_(n=1)^N10n`

`1260=\displaystyle4\sum_(n=1)^N1+10\sum_(n=1)^Nn`

`1260=4N+10\frac{N(N+1)}2`

`1260=5N^2+9N`

`5N^2+9N-1260=0\implies N=15`

(there are two solutions, but only one is a positive integer)