1. Use the formula for the sum of the first n terms of an arithmetic series:
Sn = n/2[2a + (n-1)d]
where a is the first term and d is the common difference.
2. Calculate the sum of the first 10 terms, S10, using the formula:
S10 = 10/2[2a + (10-1)d]
S10 = 5[2a + 9d]
S10 = 10a + 45d
3. Calculate the sum of the first 5 terms, S5:
S5 = 5/2[2a + (5-1)d]
S5 = 5[2a + 4d]
S5 = 5a + 10d
4. According to the problem, S10 = 4 × S5, so we have:
10a + 45d = 4(5a + 10d)
10a + 45d = 20a + 40d
5. Solve for d by subtracting 20a + 40d from both sides:
5d = 10a
d = 2a
6. Use the 8th term to find a:
a8 = a + (8-1)d
45 = a + 7d
45 = a + 7(2a)
45 = a + 14a
45 = 15a
a = 3
So, the first term of the series is a = 3.
