If the 4th term of an arithmetic sequence is 24 and the 12th term is 56, whats the first term?
done
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why are you switching names?
To find the first term of an arithmetic sequence, we need to use the formula for the nth term of an arithmetic sequence:
aₙ = a₁ + (n-1)d
where aₙ is the nth term, a₁ is the first term, n is the position of the term, and d is the common difference.
Given that the 4th term is 24 and the 12th term is 56, we can set up two equations:
a₄ = a₁ + (4-1)d
24 = a₁ + 3d ---(1)
a₁₂ = a₁ + (12-1)d
56 = a₁ + 11d ---(2)
Now, we have a system of two equations with two unknowns (a₁ and d). We can solve this system to find the values of a₁ and d. Subtracting equation (1) from equation (2), we get:
56 - 24 = a₁ + 11d - (a₁ + 3d)
32 = 8d
Dividing both sides by 8, we get:
d = 4
Substituting the value of d = 4 into equation (1), we have:
24 = a₁ + 3(4)
24 = a₁ + 12
Subtracting 12 from both sides, we get:
a₁ = 24 - 12
a₁ = 12
Therefore, the first term of the arithmetic sequence is 12.
To find the first term of an arithmetic sequence, we need to use the formula for the nth term of an arithmetic sequence:
\[a_n = a_1 + (n-1)d \]
Where:
- \(a_n\) is the nth term of the sequence,
- \(a_1\) is the first term,
- \(n\) is the position of the term in the sequence, and
- \(d\) is the common difference between consecutive terms.
Given that the 4th term is 24 and the 12th term is 56, we can substitute these values into the formula to create a system of equations.
For the 4th term:
\[ a_4 = a_1 + (4-1)d = 24 \]
For the 12th term:
\[ a_{12} = a_1 + (12-1)d = 56 \]
Now we have a system of two equations with two variables (\(a_1\) and \(d\)). We can solve this system to find the values of \(a_1\) and \(d\).
Subtracting the first equation from the second equation, we eliminate the \(a_1\) term:
\[ (12-1)d - (4-1)d = 56 - 24 \]
\[ 11d - 3d = 32 \]
\[ 8d = 32 \]
\[ d = 4 \]
Now that we have found the value of \(d\), we can substitute it back into either of the original equations to find the value of \(a_1\). Let's use the first equation:
\[ a_1 + (4-1) \cdot 4 = 24 \]
\[ a_1 + 3 \cdot 4 = 24 \]
\[ a_1 + 12 = 24 \]
\[ a_1 = 24 - 12 \]
\[ a_1 = 12 \]
Therefore, the first term of the arithmetic sequence is 12.
a+11d=56---->1
a+3d=24----->2
1-2 we get
8d=32
d=4
put d=4 in 2
a+3(4)=24
a+12=24
a=24-14
a=10