A high school arithmetic sequences problem from Korea.
Hello,
I’m Jojo Teacher teaching mathematics in Korea.
Today, I have brought some problems related to arithmetic sequences that Korean students study.
Let’s solve them together, starting with easy problems and gradually increasing in difficulty, aiming to solve 5 problems in total.
The first problem is about arithmetic sequences.
For a non-zero common difference arithmetic sequence \{a_n\} , when a_3 \times a_5 = a_7 \times a_9 , find the natural number k for which a_k = 0.
Read the problem carefully and think about it thoroughly before attempting to solve it on your own.
Afterwards, compare your solution with the explanation provided below to further your understanding.
Sure, let’s begin solving the problem.
The general term of an arithmetic sequence is:
\{a_n\} = a + (n-1)d
Given the property of the arithmetic sequence:
a_3 \times a_5 = a_7 \times a_9
Therefore,
(a + 2d)(a + 4d) = (a + 6d)(a + 8d)
Expanding both sides yields:
a^2 + 6ad + 8d^2 = a^2 + 14ad + 48d^2
Collecting like terms results in:
-8ad = 40d^2
-a = 5d
a = -5d
Thus, the general term of the arithmetic sequence is:
a_n = -5d + (n-1)d = (n-6)d
a_k = 0 holds true for the natural number k,
(k-6)d = 0
Since d \neq 0 ,
k – 6 = 0
k = 6
Therefore, the natural number k satisfying a_6 = 0 is \boxed 6.
The second problem is about arithmetic sequences.
For an arithmetic sequence a_n with a positive common difference, where a_1 + a_3 = 0 and |a_1| + |a_3| = 3 , find the value of a_{10}
Read the problem carefully and think about it thoroughly before attempting to solve it on your own.
Afterwards, compare your solution with the explanation provided below to further your understanding.
For an arithmetic sequence \{a_n\} with a positive common difference, given the conditions:
a_1 + a_3 = 0
|a_1| + |a_3| = 3
Find a_{10}
From the first condition a_1 + a_3 = 0 , we have a_3 = -a_1
From the second condition |a_1| + |a_3| = 3
|a_1| + |-a_1| = 3
2|a_1| = 3
|a_1| = \cfrac{3}{2}
So, a_1 = \cfrac{3}{2} \; or \; a_1 = -\cfrac{3}{2}
If a_1 = \cfrac{3}{2} \; then \; a_3 = -\cfrac{3}{2}
Using the formula for the n-th term of an arithmetic sequence, a_n = a_1 + (n-1)d , where d is the common difference:
For a_1 = \cfrac{3}{2} \; and \; a_3 = -\cfrac{3}{2}
d = a_3 – a_1 = -\cfrac{3}{2} – \cfrac{3}{2} = -3
Therefore,
a_n = \cfrac{3}{2} + (n-1)(-3)
a_n = \cfrac{3}{2} – 3n + 3
Now calculate a_{10} :
a_{10} = \cfrac{3}{2} – 3 \cdot 10 + 3
a_{10} = \cfrac{3}{2} – 30 + 3
a_{10} = \cfrac{3}{2} – 27
a_{10} = -\cfrac{51}{2}
Therefore, a_{10} is \boxed{-\cfrac{51}{2}} .
The third problem.
Given two different numbers x and y , consider the two sequences x, a_1, a_2, y and x, b_1, b_2, b_3, y which are both arithmetic sequences. We need to find the value of \cfrac{a_2 – a_1}{b_2 – b_1}.
Analyzing the first sequence x, a_1, a_2, y
Since this sequence is arithmetic, the difference between each consecutive term is constant. Therefore,
a_1 = x + d_1
a_2 = x + 2d_1
y = x + 3d_1
( d_1 is the common difference of the sequence x, a_1, a_2, y )
Analyzing the second sequence x, b_1, b_2, b_3, y
This sequence is also arithmetic, so the difference between each consecutive term is constant. Therefore,
b_1 = x + d_2
b_2 = x + 2d_2
b_3 = x + 3d_2
y = x + 4d_2
( d_2 is the common difference of the sequence x, b_1, b_2, b_3, y )
Since the last terms of both sequences are equal:
x + 3d_1 = x + 4d_2
Eliminating x from both sides, we get:
3d_1 = 4d_2
Thus,
d_1 = \cfrac{4}{3}d_2
Calculating the required values:
a_2 – a_1 = (x + 2d_1) – (x + d_1) = d_1
b_2 – b_1 = (x + 2d_2) – (x + d_2) = d_2
Therefore,
\cfrac{a_2 – a_1}{b_2 – b_1} = \cfrac{d_1}{d_2} = \cfrac{\cfrac{4}{3}d_2}{d_2} = \cfrac{4}{3}
In conclusion,
\cfrac{a_2 – a_1}{b_2 – b_1} = \cfrac{4}{3} .
The Fourth problem .
In an arithmetic sequence where the 10th term is 38 and the 23rd term is 12, find the first term in the sequence that becomes negative.
Let the first term of the arithmetic sequence be denoted as a and the common difference as d .
Using the information given:
a + 9d = 38 \cdots (1)
a + 22d = 12 \cdots (2)
Solve for d
Subtract equation (1) from equation (2)
(a + 22d) – (a + 9d) = 12 – 38
13d = -26
d = -2
Substitute d = -2 back into equation (1) to find a
a + 9(-2) = 38
a – 18 = 38
a = 56
Therefore, the arithmetic sequence is a_n = 56 + (n-1)(-2) .
Find the first term where the sequence becomes negative:
56 + (n-1)(-2) < 0
-2(n-1) < -56
n-1 > 28
n > 29
So, the first term in the sequence that becomes negative is the \boxed{10th} term.
The Fifth problem.
Inserting 25 numbers a_1, a_2, \ldots, a_{25} between -26 and 78 such that they form an arithmetic sequence in ascending order, what is the value of a_1 ?
Given:
b_1 = -26, \quad b_{27} = 78, \quad \text{and} \quad n = 27
Common difference calculation:
b_{27} = b_1 + 26d = 78
26d = 104
d = \cfrac{104}{26} = 4
Finding a_1 :
b_2 = b_1 + d = -26 + 4 = -22
Therefore, a_1 is \boxed{-22} .
Wrapping up the post…
How did you find solving the problem on arithmetic sequences?
Did the problem solve well for you?
I tried to make the problem challenging but not too difficult.
These were problems that Korean students can typically solve.
Later, I’ll try to find more challenging problems to present.
Also, I have a YouTube channel.
Here’s the link to my YouTube channel…
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