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 $$d=-\dfrac{1}{2}$$ 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: $$$a_1=\dfrac{461}{60}$$$

Solution:

$$a_1=\dfrac{461}{60}$$

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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:

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

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 $$m=50.$$

Solution:

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

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