If $n^{t h}$ term of a sequence is $8 - 5 n$ . Show that the sequence is an $A . P .$

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Solution

The $n^{t h}$ term is given as,

$a_{n} = 8 - 5 n$

$a_{1} = 8 - 5 (1)$

$= 8 - 5$

$= 3$

$a_{2} = 8 - 5 (2)$

$= 8 - 10$

$= - 2$

$a_{3} = 8 - 5 (3)$

$= 8 - 15$

$= - 7$

$a_{2} - a_{1} = - 2 - 3$

$= - 5$

$a_{3} - a_{2} = - 7 - (- 2)$

$= - 7 + 2$

$= - 5$

Since the common difference is same, therefore, the sequence is in A.P.

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