how do i solve the sum of the first two terms of an arithmetic sequence is 16. the sum of the second and third term is 28. determine the first three terms, algebraically
To determine the first three terms of an arithmetic sequence algebraically, we need to make use of the properties of arithmetic sequences.
Let's denote the first term of the sequence as "a" and the common difference between the terms as "d."
We are given two pieces of information:
1. The sum of the first two terms is 16, which can be expressed as:
a + (a + d) = 16
2. The sum of the second and third term is 28, which can be expressed as:
(a + d) + ((a + d) + d) = 28
To solve these equations, we can follow these steps:
Step 1: Simplify the equations:
a + (a + d) = 16 -> 2a + d = 16 (Equation 1)
(a + d) + ((a + d) + d) = 28 -> 3a + 3d = 28 (Equation 2)
Step 2: Solve Equation 1 for "d":
2a + d = 16 -> d = 16 - 2a
Step 3: Substitute the value of "d" in Equation 2:
3a + 3(16 - 2a) = 28
Step 4: Simplify and solve for "a":
3a + 48 - 6a = 28
-3a + 48 = 28
-3a = 28 - 48
-3a = -20
a = -20 / -3
a = 20/3
Step 5: Substitute the value of "a" back into Equation 1 to find "d":
2(20/3) + d = 16
40/3 + d = 16
d = 16 - 40/3
d = 48/3 - 40/3
d = 8/3
Therefore, the first term (a) is 20/3 and the common difference (d) is 8/3.
The first three terms of the arithmetic sequence are:
a = 20/3
a + d = 20/3 + 8/3 = 28/3
(a + d) + d = 28/3 + 8/3 = 36/3 = 12
Hence, the first three terms of the arithmetic sequence are 20/3, 28/3, and 12.