Question

Justify whether it is true to say that the sequence having following n^th term is an A.P.

a_n = 3n² + 5

Answer 1

Given: a_n = 3n² + 5

Step-wise calculation:

1. Compute a_{n + 1} by replacing n with (n + 1): 

a_{n + 1} = 3(n + 1)² + 5 

= 3(n² + 2n + 1) + 5

= 3n² + 6n + 8

2. Compute the difference a_{n + 1} – a_n: 

a_{n + 1} – a_n = (3n² + 6n + 8) – (3n² + 5) 

= 6n + 3

3. For an A.P. the difference a_{n + 1} – a_n must be a constant independent of n; the standard test is to check whether a_{n + 1} – a_n is independent of n.

a_{n + 1} – a_n = 6n + 3 depends on n (not constant), so the sequence is not an arithmetic progression. This agrees with the observation that an A.P. has a general term linear in n, whereas 3n² + 5 is quadratic.