B1. The sum of the fourth and sixth terms of an arithmetic sequence is the square of the solution to the equation

$$
\sqrt{16 y+9}-\sqrt{y-1}=\sqrt{9 y+10}
$$

The difference between the fifth and third terms is equal to the solution to the equation

$$
4^{y-1}-2^{2 y-3}-16^{4}=4^{y-2}
$$

What is the sum of the first thirty terms of this sequence?

Problems for 4th grade
B1. The equation $\sqrt{16 y+9}-\sqrt{y-1}=\sqrt{9 y+10}$ is squared to give $16 y+9+y-1-2 \sqrt{(16 y+9)(y-1)}=$ $9 y+10$. We rearrange it to the form $-2 \sqrt{16 y^{2}-7 y-9}=-8 y+2$ and divide by -2. The resulting equation $\sqrt{16 y^{2}-7 y-9}=4 y-1$ is squared to give the equation $16 y^{2}-7 y-9=16 y^{2}-8 y+1$. The solution to the equation is $y=10$. The equation $4^{y-1}-2^{2 y-3}-16^{4}=4^{y-2}$ is rearranged to the form $2^{2 y-2}-2^{2 y-3}-2^{2 y-4}=16^{4}$. After factoring out the common factor and rearranging, we get $2^{2 y-4} \cdot(4-2-1)=16^{4}$ and further $2^{2 y-4}=2^{16}$, equating the exponents $2 y-4=16$ and we get the solution $y=10$. Squaring the solution of the first equation and writing the equation $a_{4}+a_{6}=100$, then writing the second equation of the system $a_{5}-a_{3}=10$. Considering the general term of the arithmetic sequence, the equations are written in the form $a_{1}+3 d+a_{1}+5 d=100$ and $a_{1}+4 d-a_{1}-2 d=10$. We get the solutions $d=5$ and $a_{1}=30$. We calculate the sum of the first thirty terms using the formula $S_{n}=\frac{n}{2}\left(2 a_{1}+(n-1) d\right)$ and get $S_{30}=$ $\frac{30}{2}(2 \cdot 30+(30-1) 5)=3075$.

![](https://cdn.mathpix.com/cropped/2024_06_07_37ea02f5a530d200baf4g-24.jpg?height=66&width=1753&top_left_y=1937&top_left_x=163)

![](https://cdn.mathpix.com/cropped/2024_06_07_37ea02f5a530d200baf4g-24.jpg?height=54&width=1739&top_left_y=1995&top_left_x=176)

Solving the equation $4^{y-1}-2^{2 y-3}-16^{4}=4^{y-2}$. . . . . . . . . . . . . . . . . . . . . . . . . . $1^{*}$ point

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Setting up the system of equations $a_{4}+a_{6}=100$ and $a_{5}-a_{3}=10 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots . \ldots . \ldots . \ldots . \ldots$ point

Solving the system of equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 point

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Calculating the sum of terms $S_{30}=\frac{30}{2}(2 \cdot 30+(30-1) 5)=3075 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .1$ point