Additional Mathematics – Series | e-Consult
Series (1 questions)
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Answer
We use the formulas:
\(a_8 = a + 7d = 38\) (1)
\(S_8 = \dfrac{8}{2}\bigl(2a + 7d\bigr) = 200\) (2)
From (2):
\(4(2a + 7d) = 200 \;\Rightarrow\; 2a + 7d = 50\) (3)
Subtract (1) from (3):
\((2a + 7d) - (a + 7d) = 50 - 38\)
\(a = 12\).
Insert \(a = 12\) into (1):
\(12 + 7d = 38 \;\Rightarrow\; 7d = 26 \;\Rightarrow\; d = \dfrac{26}{7}\).
Now the 15th term:
\(a_{15} = a + 14d = 12 + 14\left(\dfrac{26}{7}\right) = 12 + 2 \times 26 = 12 + 52 = 64\).
Summary of calculations:
| Equation | Result |
| \(a_8 = a + 7d = 38\) | \(a + 7d = 38\) |
| \(S_8 = 4(2a + 7d) = 200\) | \(2a + 7d = 50\) |
| Subtract → \(a = 12\) | \(d = \dfrac{26}{7}\) |
| \(a_{15} = a + 14d\) | \(a_{15} = 64\) |