37a = a + 9*4a
so, ...
find the no of terms in an AP given that its first and last terms are a and 37a respectively and that its common difference is 4a
2 answers
term(1) = a
term(n)
= a + (n-1)d
= a + (n-1)(4a) = 37a
(n-1)(4a) = 36a
n-1 = 9
n = 10
term(n)
= a + (n-1)d
= a + (n-1)(4a) = 37a
(n-1)(4a) = 36a
n-1 = 9
n = 10
1 answer