Question
The $n^{th}$ term of a sequence is $8 - 5$n. Show that the sequence is an $A.P.$
✓
Answer
$t_n = 8 - 5n$
Replacing n by $(n + 1)$, we get
$t_{n+1} = 8 - 5(n + 1) = 8 - 5n - 5 = 3 - 5n$
Now,
$t_{n+1} - t_n = (3 - 5n) - (8 - 5n) = -5$
Since, $(t_{n+1} - t_n)$ is independent of n and is therefore a constant.
Hence, the given sequence is an $A.P.$
Replacing n by $(n + 1)$, we get
$t_{n+1} = 8 - 5(n + 1) = 8 - 5n - 5 = 3 - 5n$
Now,
$t_{n+1} - t_n = (3 - 5n) - (8 - 5n) = -5$
Since, $(t_{n+1} - t_n)$ is independent of n and is therefore a constant.
Hence, the given sequence is an $A.P.$
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