koala wrote:
KarunMendiratta wrote:
Explanation:
Let the terms be (x - 3), (x - 1), (x + 1), and (x + 3)
So, (x - 3) + (x- 1) + (x + 1) + (x + 3) = 28
x = 7
Therfore, terms are 4, 6, 8, and 10
Let us check if x = 7 satisfies the second scenario;
\(\frac{(4)(10)}{(6)(8)} = \frac{40}{48} = \frac{5}{6}\)
We have a match
Col. A: 10
Col. B: 11
Hence, option B
How did you arrived at the decision of being the terms equal to (x - 3), (x - 1), (x + 1), and (x + 3)?
While solving the Qs from AP and/or GP, assume the consecutive numbers as ... (a - 3d), (a - 2d), (a- d), a, (a + d), (a + 2d), (a + 3d), ....
This will help you save a lot of time.