If \(a_3+a_8=7\) and \(a_7+a_{14}=-3\), find \(a_{10}\).
\(a_{10}=-1\)
Step 1: Recall the formula for the \(n^{th}\) term of an arithmetic progression (AP):
\(a_n = a + (n-1)d\), where:
Step 2: Write expressions for \(a_3\) and \(a_8\):
\(a_3 = a + (3-1)d = a + 2d\)
\(a_8 = a + (8-1)d = a + 7d\)
Step 3: Add them because the question says \(a_3 + a_8 = 7\):
\((a + 2d) + (a + 7d) = 7\)
\(2a + 9d = 7\) …(Equation 1)
Step 4: Write expressions for \(a_7\) and \(a_{14}\):
\(a_7 = a + (7-1)d = a + 6d\)
\(a_{14} = a + (14-1)d = a + 13d\)
Step 5: Add them because the question says \(a_7 + a_{14} = -3\):
\((a + 6d) + (a + 13d) = -3\)
\(2a + 19d = -3\) …(Equation 2)
Step 6: Solve the two equations:
Equation (1): \(2a + 9d = 7\)
Equation (2): \(2a + 19d = -3\)
Subtract Equation (1) from Equation (2):
\((2a + 19d) - (2a + 9d) = -3 - 7\)
\(10d = -10\)
\(d = -1\)
Step 7: Put \(d = -1\) into Equation (1):
\(2a + 9(-1) = 7\)
\(2a - 9 = 7\)
\(2a = 16\)
\(a = 8\)
Step 8: Find \(a_{10}\):
\(a_{10} = a + (10-1)d = a + 9d\)
\(a_{10} = 8 + 9(-1) = 8 - 9 = -1\)
Final Answer: \(a_{10} = -1\)