According to your given, the sequence would be
a, 5a, 9a, 13a, 17a,... , 37a
a + (n-1)(4a) = 37a
1 + (n-1)(4) = 37
4n - 4 = 36
4n = 40
n = 10
find the number of term in an A.P given that its first and last term are a and 37a respectively and that its common difference is 4a
2 answers
In an A.P.,Un=a+(n+1)d
First term=a
Common difference=d=4a
Last term In=37a
Un=37a
a+(n-1)(4a)=37a
(n-1)(4a)=36a
n-1=9
n=10
Therefore there are 10 terms in that A.P.
First term=a
Common difference=d=4a
Last term In=37a
Un=37a
a+(n-1)(4a)=37a
(n-1)(4a)=36a
n-1=9
n=10
Therefore there are 10 terms in that A.P.
2 answers