**1. Find four consecutive terms in an A.P. whose sum is 12 and the sum of 3rd and 4th term is 14.**

**Sol. Let the four consecutive terms in an A.P. be a – 3d, a – d, a + d, a + 3d**

**As per the first condition,**

**a – 3d + a – d + a + d + a + 3d = 12**

**∴**

**4a = 12**

**∴**

**a =**

^{12}/_{4}

**∴**

**a = 3 ...........eq. (i)**

**As per the second condition,**

**a + d + a + 3d = 14**

**∴**

**2a + 4d = 14**

**∴**

**2 (3) + 4d = 14 [From eq. (i)]**

**∴**

**6 + 4d = 14**

**∴**

**4d = 14 – 6**

**∴**

**4d = 8**

**∴**

**d =**

^{8}/_{4}

**∴**

**d = 2**

**∴**

**a – 3d = 3 – 3 (2) = 3 – 6 = – 3**

**a – d = 3 – 2 = 1**

**a + d = 3 + 2 = 5**

**a + 3d = 3 + 3 (2) = 9**

**∴ The four consecutive terms of A.P. are – 3, 1, 5 and 9.**