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Sagot :
To find the first term and the common difference of the arithmetic sequence, let's define the terms and follow the steps methodically.
1. Let the first term of the arithmetic sequence be [tex]\( a \)[/tex] and the common difference be [tex]\( d \)[/tex].
2. The first six terms of the sequence are:
[tex]\( a, a+d, a+2d, a+3d, a+4d, a+5d \)[/tex].
3. The mean (average) of these first six terms is given as 10.
The mean of these terms can be calculated as follows:
[tex]\[ \text{Mean of first six terms} = \frac{a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d)}{6} = \frac{6a + 15d}{6} = a + \frac{15d}{6} = a + 2.5d \][/tex]
Given that this mean is 10, we can write the equation:
[tex]\[ a + 2.5d = 10 \][/tex]
4. The next ten terms of the sequence are:
[tex]\( a + 6d, a + 7d, a + 8d, a + 9d, a + 10d, a + 11d, a + 12d, a + 13d, a + 14d, a + 15d \)[/tex].
5. The mean of these ten terms is given as 26.
The mean of these terms can be calculated as follows:
[tex]\[ \text{Mean of next ten terms} = \frac{(a + 6d) + (a + 7d) + (a + 8d) + (a + 9d) + (a + 10d) + (a + 11d) + (a + 12d) + (a + 13d) + (a + 14d) + (a + 15d)}{10} = \frac{10a + (6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15)d}{10} = \frac{10a + 105d}{10} = a + 10.5d \][/tex]
Given that this mean is 26, we can write the equation:
[tex]\[ a + 10.5d = 26 \][/tex]
6. Now, we have a system of linear equations:
[tex]\[ \begin{cases} a + 2.5d = 10 \\ a + 10.5d = 26 \end{cases} \][/tex]
7. To solve these equations, we can subtract the first equation from the second equation:
[tex]\[ (a + 10.5d) - (a + 2.5d) = 26 - 10 \][/tex]
Simplifying the above equation:
[tex]\[ 8d = 16 \][/tex]
Solving for [tex]\( d \)[/tex]:
[tex]\[ d = \frac{16}{8} = 2 \][/tex]
8. Substitute [tex]\( d = 2 \)[/tex] back into the first equation:
[tex]\[ a + 2.5 \times 2 = 10 \][/tex]
Simplifying:
[tex]\[ a + 5 = 10 \][/tex]
Solving for [tex]\( a \)[/tex]:
[tex]\[ a = 10 - 5 = 5 \][/tex]
Therefore, the first term ([tex]\( a \)[/tex]) of the arithmetic sequence is 5, and the common difference ([tex]\( d \)[/tex]) is 2.
1. Let the first term of the arithmetic sequence be [tex]\( a \)[/tex] and the common difference be [tex]\( d \)[/tex].
2. The first six terms of the sequence are:
[tex]\( a, a+d, a+2d, a+3d, a+4d, a+5d \)[/tex].
3. The mean (average) of these first six terms is given as 10.
The mean of these terms can be calculated as follows:
[tex]\[ \text{Mean of first six terms} = \frac{a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d)}{6} = \frac{6a + 15d}{6} = a + \frac{15d}{6} = a + 2.5d \][/tex]
Given that this mean is 10, we can write the equation:
[tex]\[ a + 2.5d = 10 \][/tex]
4. The next ten terms of the sequence are:
[tex]\( a + 6d, a + 7d, a + 8d, a + 9d, a + 10d, a + 11d, a + 12d, a + 13d, a + 14d, a + 15d \)[/tex].
5. The mean of these ten terms is given as 26.
The mean of these terms can be calculated as follows:
[tex]\[ \text{Mean of next ten terms} = \frac{(a + 6d) + (a + 7d) + (a + 8d) + (a + 9d) + (a + 10d) + (a + 11d) + (a + 12d) + (a + 13d) + (a + 14d) + (a + 15d)}{10} = \frac{10a + (6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15)d}{10} = \frac{10a + 105d}{10} = a + 10.5d \][/tex]
Given that this mean is 26, we can write the equation:
[tex]\[ a + 10.5d = 26 \][/tex]
6. Now, we have a system of linear equations:
[tex]\[ \begin{cases} a + 2.5d = 10 \\ a + 10.5d = 26 \end{cases} \][/tex]
7. To solve these equations, we can subtract the first equation from the second equation:
[tex]\[ (a + 10.5d) - (a + 2.5d) = 26 - 10 \][/tex]
Simplifying the above equation:
[tex]\[ 8d = 16 \][/tex]
Solving for [tex]\( d \)[/tex]:
[tex]\[ d = \frac{16}{8} = 2 \][/tex]
8. Substitute [tex]\( d = 2 \)[/tex] back into the first equation:
[tex]\[ a + 2.5 \times 2 = 10 \][/tex]
Simplifying:
[tex]\[ a + 5 = 10 \][/tex]
Solving for [tex]\( a \)[/tex]:
[tex]\[ a = 10 - 5 = 5 \][/tex]
Therefore, the first term ([tex]\( a \)[/tex]) of the arithmetic sequence is 5, and the common difference ([tex]\( d \)[/tex]) is 2.
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