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Given that 4, p, 9, 13 are consecutive terms of an arithmetic progression, fine the values of p and 9 ?

9 is always 9 ...

d = 13-9 = 4
but the sequence 4,8,9,13 is not an AP
whassup?

if by 9 you mean q, then
3d = 13-4 = 9
so d = 3
you can use that to get p and q

Did you know?

Did you know that in an arithmetic progression, consecutive terms have a constant difference between them? In the given sequence, the common difference can be found by subtracting the second term (p) from the first term (4), resulting in a difference of (4 - p). To find the missing value of p, we can subtract this difference from the second term (p) to obtain (p - (4 - p)) = (2p - 4). Therefore, the value of p is (2p - 4), and we can replace p with this expression to get the missing number as (2(2p-4) - p) = (4p - 8 - p) = (3p - 8). Similarly, to find the missing value of 9, we can add the common difference (4 - p) to the previous term (13) to obtain (13 + (4 - p)) = (17 - p). So, the missing values in the sequence are p = (2p - 4) and 9 = (17 - p).