The 5th term of an arithmetic sequence is 11,12th term is 32.What are the first 3 terms?
d = (32-11)/(12-5) = 3
a = 11-4*3 = -1
now you can fill in the rest.
To find the first three terms of an arithmetic sequence, we need to find the common difference (d) first.
Given:
5th term (a₅) = 11
12th term (a₁₂) = 32
Using the formula to calculate the nth term of an arithmetic sequence:
aₙ = a₁ + (n - 1) * d
We can plug in the values we have to solve for d.
For the 5th term:
a₅ = a₁ + (5 - 1) * d
11 = a₁ + 4d
For the 12th term:
a₁₂ = a₁ + (12 - 1) * d
32 = a₁ + 11d
Now we have a system of two equations:
Equation 1: a₁ + 4d = 11
Equation 2: a₁ + 11d = 32
To solve this system, we can use either substitution or elimination. Let's use the elimination method.
Multiply Equation 1 by 11 and Equation 2 by 4 to eliminate the variable a₁:
11(a₁ + 4d) = 11 * 11
4(a₁ + 11d) = 4 * 32
11a₁ + 44d = 121
4a₁ + 44d = 128
Subtract the second equation from the first equation to eliminate the variable d:
11a₁ + 44d - (4a₁ + 44d) = 121 - 128
11a₁ - 4a₁ = -7
7a₁ = -7
a₁ = -1
Now that we have the value of the first term (a₁), we can substitute it back into either Equation 1 or Equation 2 to find the common difference (d).
Using Equation 1:
a₁ + 4d = 11
-1 + 4d = 11
4d = 11 + 1
4d = 12
d = 12/4
d = 3
So, the first term (a₁) is -1 and the common difference (d) is 3.
Now we can find the first three terms of the arithmetic sequence:
1st term: a₁ = -1
2nd term: a₂ = a₁ + d = -1 + 3 = 2
3rd term: a₃ = a₁ + 2d = -1 + 2(3) = -1 + 6 = 5
Therefore, the first three terms of the arithmetic sequence are -1, 2, and 5.