If a + 1, 3a, 4a + 2 are in AP, then the value of a is
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If a + 1, 3a, 4a + 2 are in AP, then the value of a is
3
2
1
0
a + 1, 3a, 4a + 2 are in AP. If a, b, c are in AP, then 2b = a + c. Therefore, 2(3a) = a + 1 + 4a + 2 6a = 5a + 3 6a – 5a = 3 a = 3
In an arithmetic progression (AP), the difference between consecutive terms is constant, so 3a − (a+1) = (4a+2) − 3a. Simplifying gives 2a − 1 = a + 2, so a = 3. Substituting back confirms the three terms 4, 9, 14 do indeed have a common difference of 5.