Show that a1 , a2 , . . ., an, . . . form an AP where an is defined as below :
an = 9 – 5n
It is given that :
an = 9 – 5n
a1 = 9 – 5(1) = 9 – 5 = 4
a2 = 9 – 5 (2) = 9 – 10 = –1
and a3 = 9 – 5(3) = 9 – 15 = –6
Now, we have following numbers :
4, –1,–6, –11,.........
Since, the difference between each pair of consecutive terms are constant.
So, the given numbers form an A.P.
Here, a = 4, d = –5, n = 15