Question

7. Prove that no matter what the real numbersa andet ene the sequence with nth term
a + nb is always an A.P. What is the common difference?​

Answer 1

Step-by-step explanation:

Let a + nb represent an A.P

Thus, when n = 1

a1 = a + (1)b

a1 = a + b

when n = 2

a2 = a + (2)b

a2 = a + 2b

Similarly,

a3 = a + 3b

a4 = a + 4b

Now, if they form an A.P, then their common difference must be equal.

∴ a2 - a1 = a3 - a2 = a4 - a3

Now,

a2 - a1 = (a + 2b) - (a + b)

a2 - a1 = a + 2b - a - b

a2 - a1 = b

And,

a3 - a2 = (a + 3b) - (a + 2b)

a3 - a2 = a + 3b - a - 2b

a3 - a2 = b

Also,

a4 - a3 = (a + 4b) - (a + 3b)

a4 - a3 = a + 4b - a - 3b

a4 - a3 = b

∴ a2 - a1 = a3 - a2 = a4 - a3 = b

Thus,

they will always follow an A.P for any real numbers of a and b because they will form the same common difference among all the consecutive terms.

Hope it helped and you understood it........All the best