If we want to choose 4 consecutive number then why do we choose (a - 3d), (a - d), (a + d), (a + 3d), why we have not chosen (a-2d), (a-d) ,(a+d), (a+2d).

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Solution

When you want to choose 4 consecutive numbers, you must ensure that the difference between any two numbers is always 1.
Hence, a - 3d, a - d, a + d, and a + 2d are not consecutive numbers. This has common difference of 2d.
Even the pattern suggested by you is also not right, since the common difference is d. (unless d = 1, ths is not right).
So, 4 consecutive numbers can be represented as :
x, x + 1, x + 2 and x + 3 OR
x - 2, x - 1, x and x + 1 etc.,
Please let us know how did you assume that the 4 consecutive numbers are written as (a - 3d), (a - d), (a + d), (a + 3d)

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Sir for when sum of 4 unknown consencutive terms of an ap is given then we take it as a-3d,a-d,a+d,a+3d.why dont we take it as a-2d,a-d,a+d,a+2d.

Does this have to do any thing with odd and even.

Please explain.

Thak you.

Which of the following series forms an AP?

$(i) a - 2 d, a - d, a, a + d, a + 2 d$

$(i i) a, a + d, a + 2 d, a + 3 d, a + 4 d$

$(i i i) a - 3 d, a - d, a + d, a + 3 d$

∣ ∣ ∣ ∣ ∣ \begin{matrix} (a + d) (a + 2 d) & a + 2 d & a 2 d (a + 2 d) & d & d 2 d (a + 3 d) & d & d \end{matrix} ∣ ∣ ∣ ∣ ∣

a > 0, d > 0

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{a} & \frac{1}{a (a + d)} & \frac{1}{(a + d) (a + 2 d)} \frac{1}{a (a + d)} & \frac{1}{(a + d) (a + 2 d)} & \frac{1}{(a + 2 d) (a + 3 d)} \frac{1}{(a + 2 d)} & \frac{1}{(a + 2 d) (a + 3 d)} & \frac{1}{(a + 3 d) (a + 4 d)} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{a} & \frac{1}{a (a + d)} & \frac{1}{(a + d) (a + 2 d)} \frac{1}{(a + d)} & \frac{1}{(a + d) (a + 2 d)} & \frac{1}{(a + 2 d) (a + 3 d)} \frac{1}{(a + 2 d)} & \frac{1}{(a + 2 d) (a + 3 d)} & \frac{1}{(a + 3 d) (a + 4 d)} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣

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