The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
-13, -8, -3
Step 1: Recall the nth term of an AP.
For an arithmetic progression (AP), the nth term is
\[a_n = a + (n - 1)d\]
where \(a\) is the first term and \(d\) is the common difference.
Step 2: Express the 4th, 8th, 6th and 10th terms.
4th term: \(a_4 = a + 3d\)
8th term: \(a_8 = a + 7d\)
6th term: \(a_6 = a + 5d\)
10th term: \(a_{10} = a + 9d\)
Step 3: Use the given sums.
Sum of 4th and 8th terms is 24:
\[a_4 + a_8 = 24\]
Substitute the expressions:
\[(a + 3d) + (a + 7d) = 24\]
\[2a + 10d = 24 \quad ...(1)\]
Sum of 6th and 10th terms is 44:
\[a_6 + a_{10} = 44\]
Substitute:
\[(a + 5d) + (a + 9d) = 44\]
\[2a + 14d = 44 \quad ...(2)\]
Step 4: Solve the two equations for \(d\).
Subtract equation (1) from equation (2):
\[(2a + 14d) - (2a + 10d) = 44 - 24\]
\[2a + 14d - 2a - 10d = 20\]
\[4d = 20\]
So,
\[d = \dfrac{20}{4} = 5\]
Step 5: Find the first term \(a\).
Use equation (1): \(2a + 10d = 24\).
Substitute \(d = 5\):
\[2a + 10 \times 5 = 24\]
\[2a + 50 = 24\]
\[2a = 24 - 50 = -26\]
\[a = -13\]
Step 6: Write the first three terms.
First term: \(a_1 = a = -13\)
Second term: \(a_2 = a + d = -13 + 5 = -8\)
Third term: \(a_3 = a + 2d = -13 + 10 = -3\)
Conclusion: The first three terms of the AP are -13, -8, -3.