We're given an arithmetic sequence [tex]\{a_n\}_{n\ge1}[/tex] that starts with [tex]a_1=14[/tex] and a common difference between terms of [tex]d=10[/tex]. Recursively, this sequence is given by
[tex]\begin{cases}a_n=a_{n-1}+10&\text{for }n>1\\a_1=14\end{cases}[/tex]
We can find an explicit formula for the [tex]n[/tex]-th term [tex]a_n[/tex]:
[tex]a_2=a_1+10[/tex]
[tex]a_3=a_2+10=a_1+2(10)[/tex]
[tex]a_4=a_3+10=a_1+3(10)[/tex]
and so on, with the general pattern of
[tex]a_n=a_1+(n-1)(10)=4+10n[/tex]
We're given that the sum of the first [tex]N[/tex] consecutive terms is
[tex]S_N=\displaystyle\sum_{n=1}^Na_n=\sum_{n=1}^N(4+10n)=1260[/tex]
Recall that
[tex]\displaystyle\sum_{n=1}^N1=N[/tex]
[tex]\displaystyle\sum_{n=1}^Nn=\dfrac{N(N+1)}2[/tex]
So we solve for [tex]N[/tex]:
[tex]1260=\displaystyle\sum_{n=1}^N(4+10n)[/tex]
[tex]1260=\displaystyle\sum_{n=1}^N4+\sum_{n=1}^N10n[/tex]
[tex]1260=\displaystyle4\sum_{n=1}^N1+10\sum_{n=1}^Nn[/tex]
[tex]1260=4N+10\dfrac{N(N+1)}2[/tex]
[tex]1260=5N^2+9N[/tex]
[tex]5N^2+9N-1260=0\implies N=15[/tex]
(there are two solutions, but only one is a positive integer)
