Sum of terms of an arithmetical progression
In an arithmetical progression with general term $a_n=5n+2$, how many terms is it necessary to add up so that the result is $6.475$?
We want to find natural $m$ such that the sum of the $m$ first terms of the succession $a_n=5n+2$ is exactly $6.475$, that is $S_m=\sum_{n=1}^m 5n+2 = 6.475$, but we know that:
$$S_m=\dfrac{m\cdot(a_1+a_m)}{2}=\dfrac{m\cdot ((5+2)+(5m+2))}{2}$$
And comparing both expressions, we have:
$$6.475=\dfrac{m(7+5m+2)}{2}$$
So we solve this equation of second grade:
$$5m^2+9m-12.850=0 \Rightarrow m=\left\{ \begin{array}{c} 50 \\ -\dfrac{259}{5} \end{array} \right.$$
Knowing that $m$ must be a positive integer, we can conclude that the solution is $m=50.$
It is necessary to add the first $50$ terms.
Calculate the first term of an arithmetical progression with difference $d=-\dfrac{1}{2}$ if we know that the sum of the $30$ first terms is equal to $13.$
We want to find a real number $a_1$ in such a way that it is the first term of an arithmetical progression of difference $d=-\dfrac{1}{2}$ and that the sum of the $30$ first terms is equal to $13$.
Namely we have the progression $$a_n=a_1+(n-1)\Big(-\dfrac{1}{2}\Big)=a_1+\dfrac{1-n}{2}$$ and the sum of the first $30$ terms is equal to $13$ $$S_30=\sum_{n=1}^30 \Big(a_1+\dfrac{1-n}{2}\Big) = 13$$ and, on the other hand, we have $$S_30=\dfrac{30(a_1+a_30)}{2}=15\Big(a_1+\Big(a_1+\dfrac{1-30}{2}\Big)\Big)$$ putting together both expressions, we obtain: $$15\Big(2a_1-\dfrac{29}{2}\Big)=13$$
And solving this equation: $$a_1=\dfrac{461}{60}$$
$$a_1=\dfrac{461}{60}$$