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?
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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?
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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