term(n) = a + (n-1)d
for yours:
term(n) = 2an+b
term(n-1) = 2a(n-1) + b = 2an - 2a + b
term(n+1) = 2a(n+1) + b = 2an + 2a + b
common difference = term(n) - term(n-1)
= 2an + b - (2an - 2a + b)
= 2a
or common difference = term(n+1) - term(n)
= 2an+2a + b - (2an + b)
= 2a
yes, it is
If n th term of a sequence is 2an+b, where a, b are constants, is this sequence an a.p.?
1 answer