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Question
Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.
(Assume the four consecutive terms in A.P. are a – d, a, a + d, a +2d)
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Solution
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 3rd and 4th 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.
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