In that case, you must have forgotten your Algebra I. Look at what you know:
a+7d = 2(a+2d)
8/2 (2a+7d) = 39
or, massaged a bit,
a = 3d
8a+28d = 39
a+7d = 2(a+2d)
8/2 (2a+7d) = 39
or, massaged a bit,
a = 3d
8a+28d = 39
\$a_n = a_1 + (n-1)d\$
Where:
- \$a_n\$ is the nth term of the series
- \$a_1\$ is the first term of the series
- \$d\$ is the common difference between consecutive terms
- \$n\$ is the number of terms
Let's break down the information given step by step:
1. The eighth term (\$a_8\$) is twice the third term (\$a_3\$).
=> \$a_8 = 2 \cdot a_3\$
2. The sum of the first eight terms is 39.
=> \$a_1 + a_2 + a_3 + \ldots + a_8 = 39\$
Now, let's solve for the first term (\$a_1\$).
We'll start with the equation for the sum of the eight terms and substitute the values of \$a_1\$, \$a_3\$, and \$a_8\$:
\$a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 = 39\$
Since \$a_2 = a_1 + d\$, \$a_3 = a_1 + 2d\$, and \$a_8 = a_1 + 7d\$, we can rewrite the equation as:
\$a_1 + (a_1 + d) + (a_1 + 2d) + \ldots + (a_1 + 7d) = 39\$
Now, substitute the equation for \$a_8\$:
\$a_1 + (a_1 + d) + (a_1 + 2d) + \ldots + (a_1 + 6d) + 2(a_1 + 2d) = 39\$
Simplify the equation:
\$8a_1 + 28d = 39\$
We also have the equation given in the problem that the eighth term is twice the third term:
\$a_8 = 2 \cdot a_3\$
Substitute the values for \$a_8\$ and \$a_3\$:
\$a_1 + 7d = 2(a_1 + 2d)\$
Simplify the equation:
\$a_1 + 7d = 2a_1 + 4d\$
\$3d = a_1\$
Now we have two equations:
1. \$8a_1 + 28d = 39\$
2. \$3d = a_1\$
Using equation 2, substitute \$a_1\$ with \$3d\$ in equation 1:
\$8(3d) + 28d = 39\$
Simplify and solve for \$d\$:
\$24d + 28d = 39\$
\$52d = 39\$
\$d = \frac{39}{52}\$
Now that we know the value of \$d\$, we can find the value of \$a_1\$ by substituting it into equation 2:
\$3(\frac{39}{52}) = a_1\$
Simplify:
\$a_1 = \frac{117}{52}\$
Therefore, the first term is \$\frac{117}{52}\$ or approximately 2.25.