1. **State the problem:** We are given an arithmetic progression (A.P.) where the sum of the 5th and 12th terms is 93, and the 8th term is 43. We need to find the first four terms of this A.P. 2. **Recall the formula for the nth term of an A.P.:** $$a_n = a + (n-1)d$$ where $a$ is the first term and $d$ is the common difference. 3. **Write the given conditions using the formula:** - 5th term: $a_5 = a + 4d$ - 12th term: $a_{12} = a + 11d$ - 8th term: $a_8 = a + 7d$ Given: $$a_5 + a_{12} = 93$$ $$a_8 = 43$$ 4. **Express the sum of the 5th and 12th terms:** $$a_5 + a_{12} = (a + 4d) + (a + 11d) = 2a + 15d = 93$$ 5. **Express the 8th term:** $$a_8 = a + 7d = 43$$ 6. **Solve the system of equations:** From the 8th term: $$a = 43 - 7d$$ Substitute into the sum equation: $$2(43 - 7d) + 15d = 93$$ $$86 - 14d + 15d = 93$$ $$86 + d = 93$$ $$d = 93 - 86 = 7$$ 7. **Find the first term $a$:** $$a = 43 - 7 \times 7 = 43 - 49 = -6$$ 8. **Find the first four terms:** - $a_1 = a = -6$ - $a_2 = a + d = -6 + 7 = 1$ - $a_3 = a + 2d = -6 + 14 = 8$ - $a_4 = a + 3d = -6 + 21 = 15$ **Final answer:** The first four terms of the A.P. are $$-6, 1, 8, 15$$.