1. **Problem:** If the 4th term of an Arithmetic Progression (AP) is 14 and its 12th term is 70, find the 1st and 5th terms. 2. **Formula for the nth term of an AP:** $$a_n = a + (n-1)d$$ where $a$ is the first term and $d$ is the common difference. 3. **Given:** - $a_4 = 14$ - $a_{12} = 70$ Using the formula: $$a_4 = a + 3d = 14$$ $$a_{12} = a + 11d = 70$$ 4. **Subtract the first equation from the second to eliminate $a$:** $$a + 11d - (a + 3d) = 70 - 14$$ $$\cancel{a} + 11d - \cancel{a} - 3d = 56$$ $$8d = 56$$ 5. **Solve for $d$:** $$d = \frac{56}{8} = 7$$ 6. **Substitute $d=7$ back into the first equation to find $a$:** $$a + 3(7) = 14$$ $$a + 21 = 14$$ $$a = 14 - 21 = -7$$ 7. **Find the 5th term:** $$a_5 = a + 4d = -7 + 4(7) = -7 + 28 = 21$$ **Final answer:** - First term $a = -7$ - Fifth term $a_5 = 21$