Question

Determine the number of terms n in each arithmetic series.

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

$\displaystyle\sum_{n=1}^N1=N$

$\displaystyle\sum_{n=1}^Nn=\dfrac{N(N+1)}2$

So we solve for $N$:

$1260=\displaystyle\sum_{n=1}^N(4+10n)$

$1260=\displaystyle\sum_{n=1}^N4+\sum_{n=1}^N10n$

$1260=\displaystyle4\sum_{n=1}^N1+10\sum_{n=1}^Nn$

$1260=4N+10\dfrac{N(N+1)}2$

$1260=5N^2+9N$

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

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