Question

2. Find four consecutive terms in an A.P whose sum is 12 and sum

of 3rd and 4th term is 14



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Answer 1

Answer:

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Answer 2

Answer:

Let the four consecutive terms in 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.1)

As per the second condition,

a+d+a+3d=14

∴ 2a+4d= 14

∴2(3)+4d=14 (from eq.1)

∴ 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+2(3)= 9

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

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