Show that the sequence defined by $a_n = 5n – 7$ is an A.P., find its common difference.
Given:
$a_n = 5n – 7$
To do:
We have to show that the sequence defined by $a_n = 5n – 7$ is an A.P. and find its common difference.
Solution:
To  show that the sequence defined by $a_n = 5n – 7$ is an A.P., we have to show that the difference between any two consecutive terms is equal.
Let us find the first few terms of the sequence by substituting $n=1, 2, 3....$
When $n=1$,
$a_1=5(1)-7$
$=5-7$
$=-2$
$a_2=5(2)-7$
$=10-7$
$=3$
$a_3=5(3)-7$
$=15-7$
$=8$
$a_4=5(4)-7$
$=20-7$
$=13$
Here,
$d=a_2-a_1=3-(-2)=3+2=5$
$d=a_3-a_2=8-3=5$
$d=a_4-a_3=13-8=5$
$d=a_2-a_1=a_3-a_2=a_4-a_3$
Hence, the given sequence is an A.P. and the common difference is $5$.
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