Formula Used.
dn = an + 1 – an
an = a + (n–1)d
In the above sequence,
a =
;
d1 = a2–a1 =
= 1
d2 = a3–a2 =
= 1
d3 = a4–a3 =
= 1
The difference in sequence is same and comes to be (1).
∴ The above sequence is A.P
The nth term of A.P is an = a + (n–1)d
an = a + (n–1)d =
+ (n–1)(1)
=
+ n–1
= n–![]()
Couldn't generate an explanation.
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