Question

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

(Assume the four consecutive terms in A.P. are a – d, a, a + d, a +2d) 

Answer 1

Assume that the four consecutive terms in A.P. are a – d, a, a + d, a +2d . 

It is given that,
Sum of four consecutive terms = 12
Sum of 3^rd and 4^th term = 14

\[\left( a - d \right) + a + \left( a + d \right) + \left( a + 2d \right) = 12\]

\[ \Rightarrow 4a + 2d = 12\]

\[ \Rightarrow 2a + d = 6\]

\[ \Rightarrow 2a = 6 - d     . . . \left( 1 \right)\]

\[\left( a + d \right) + \left( a + 2d \right) = 14\]

\[ \Rightarrow 2a + 3d = 14\]

\[ \Rightarrow 6 - d + 3d = 14\]

\[ \Rightarrow 2d = 14 - 6\]

\[ \Rightarrow 2d = 8\]

\[ \Rightarrow d = 4 \left( \text { from} \left( 1 \right) \right)\]

\[ \Rightarrow 2a = 6 - d\]

\[ \Rightarrow 2a = 6 - 4\]

\[ \Rightarrow 2a = 2\]

\[ \Rightarrow a = 1\]

Hence, the terms are –3, 1, 5 and 9.