Question

Find four consecutive terms in an AP whose sum is 12 and some of 3rd and 4th term is 14 (assume that the four consecutive terms are:
a-d, a, a+d, a+2d)

Answer 1

4 consecutive terms are a-d, a, a+d, a+2d
Sum = 12 => a-d+ a+a+d+ a+2d=12=>4a+2d = 12 (1)
Sum of 3rd and 4th term = 14
nth term = a+(n-1)d => 3rd term = a+2d and 4th term = a+3d
So a+2d + a+3d = 2a+5d = 14 (2)
Multiply (2) by -2 and add to (1)
-4a-10d=-28
4a+2d = 12
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-8d = -16 => d = 2
Substiyute in (1) to get 4a+2(2) =12=> 4a +4 = 12=> 4a=8=>a=2
So the 4 consecutive terms are: 0, 2, 4, 6 Answer