You're on a roll — let's find the first term and common difference for this arithmetic progression.
Let the first term be a and the common difference be d.
The formula for the n-th term is aₙ = a + (n-1)d.
The formula for the sum of the first n terms is Sₙ = (n/2)[2a + (n-1)d].
Step 1: Use the first condition to form an equation.
"The ninth term of an arithmetic progression is three times the third term."
a₉ = a + (9-1)d = a + 8d
a₃ = a + (3-1)d = a + 2d
So, we have:
a + 8d = 3(a + 2d)
a + 8d = 3a + 6d
Rearrange the terms:
8d - 6d = 3a - a
2d = 2a
Divide by 2:
d = a (Equation 1)
Step 2: Use the second condition to form another equation.
"The sum of the first 4 terms is 30."
S₄ = (4/2)[2a + (4-1)d]
S₄ = 2[2a + 3d]
So, we have:
2(2a + 3d) = 30
Divide by 2:
2a + 3d = 15 (Equation 2)
Step 3: Solve the system of equations.
Substitute Equation 1 (d = a) into Equation 2:
2a + 3(a) = 15
2a + 3a = 15
5a = 15
Divide by 5:
a = 3
Now, use Equation 1 to find d:
d = a
d = 3
The first term of the progression is 3 and the common difference is 3.
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