Problems from Sum of terms of an arithmetical progression

Development:

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 $d=-{\displaystyle \frac{1}{2}}$ and that the sum of the $30$ first terms is equal to $13$.

Namely we have the progression $$a_n=a_1+(n-1)(-{\displaystyle \frac{1}{2}})=a_1+{\displaystyle \frac{1-n}{2}}$$ and the sum of the first $30$ terms is equal to $13$ $$S_{3}0=\sum _{n=1}^{3}0(a_1+{\displaystyle \frac{1-n}{2}})=13$$ and, on the other hand, we have $$S_{3}0={\displaystyle \frac{30(a_1+a_{3}0)}{2}}=15(a_1+(a_1+{\displaystyle \frac{1-30}{2}}))$$ putting together both expressions, we obtain: $$15(2a_1-{\displaystyle \frac{29}{2}})=13$$

And solving this equation: $$a_1={\displaystyle \frac{461}{60}}$$

Solution:

$a_1={\displaystyle \frac{461}{60}}$

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Development:

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:

$$S_m={\displaystyle \frac{m\cdot (a_1+a_m)}{2}}={\displaystyle \frac{m\cdot ((5+2)+(5m+2))}{2}}$$

And comparing both expressions, we have:

$$6.475={\displaystyle \frac{m(7+5m+2)}{2}}$$

So we solve this equation of second grade:

$$5m^2+9m-12.850=0\Rightarrow m=\{\begin{array}{c}50\\ -{\displaystyle \frac{259}{5}}\end{array}$$

Knowing that $m$ must be a positive integer, we can conclude that the solution is $m=50.$

Solution:

It is necessary to add the first $50$ terms.

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