Question

Show that the sequence defined by $a_n=3n^2-5$ is not an A.P

Answer 1

Ans.

given sequence is

$a_n=3n^2-5....eq\left(1\right)$

first term of this A.P on puting $n=1$ in eq(1)

$a_1=3\times 1^2-5$

$a_1=3-5$

$a_1=-2......eq\left(2\right)$

second term of this A.P on putting $n=2$ in eq(1)

$a_2=3\times 2^2-5$

$a_2=12-5$

$a_2=7......eq\left(3\right)$

third term of this A.P on putting $n=3$ in eq(1)

$a_3=3\times 3^2-5$

$a_3=27-5$

$a_3=22......eq\left(4\right)$

hence the common difference of first and second term is

$d_1=a_2-a_1=7-(-2)$

$d_1=9$

and

the common difference of second and third term is

$d_2=a_3-a_2=22-7$

$d_2=15$

$since,d_1\ne d_2$

hence the common difference is not constant, hence this sequence dose't form an A.P