Question

Check that the list of number defined by the

n-th term 2n² + 1 is an AP or not.
​

Answer 1

Question :

Check the number defined by the nth term 2n² + 1 is an A.P or not.

Answer :

$\boxed{\sf{a_n \: = \: 2n^2 \: + \: 1}}$

Put n = 1

⇒a1 = 2(1)² + 1

⇒a1 = 2 + 1

⇒a1 = 3

$\rule{150}{0.5}$

Put n = 2

⇒a2 = 2(2)² + 1

⇒a2 = 2(4) + 1

⇒a2 = 8 + 1

⇒a2 = 9

$\rule{150}{0.5}$

Put n = 3

⇒a3 = 2(3)² + 1

⇒a3 = 2(9) + 1

⇒a3 = 18 + 1

⇒a3 = 19

_____________________________

If the common difference between the terms will be same then it will be an A.P

⇒d1 = a2 - a1

⇒d1 = 9 - 3

⇒d1 = 6

_________

⇒d2 = a3 - a2

⇒d2 = 19 - 9

⇒d2 = 10

$\rule{200}{2}$

As, d1 ≠ d2

So, this is not defined for A.P

Answer 2

★$\bold{\Large{\underline{\underline{\rm{\red{Question:-}}}}}}$

Check the number defined by the nth term 2n² + 1 is an A.P or not.

★$\bold{\Large{\underline{\underline{\rm{\pink{AnSwer:-}}}}}}$

According to the question,

$\sf \longmapsto \: an \: = 2 {n}^{2} + 1 ...(1)$

━━━━━━━━━━━━━━━━━━━━━━━━━━

Substitute n = 1 in equation (1),

$\sf \longmapsto \: a1 = 2 ({1})^{2} + 1 \\ \\ \sf \longmapsto \:a1 = 2 + 1 \\ \\ \sf \longmapsto \:a1 = 3$

━━━━━━━━━━━━━━━━━━━━━━━━━━

Similarly,

Substitute n=2 in equation (1),

$\sf \longmapsto \:a2 = 2 ({2})^{2} + 1 \\ \\ \sf \longmapsto \:a2 = 2 \times 4 + 1 \\ \\ \sf \longmapsto \:a2 = 8 + 1 \\ \\ \sf \longmapsto \:a2 = 9$

━━━━━━━━━━━━━━━━━━━━━━━━━━

Also,

Substitute n=3 in equation (1),

$\sf \longmapsto \:a3 = 2 ({3})^{2} + 1 \\ \\ \sf \longmapsto \:a3 = 2 \times 9 + 1 \\ \\ \sf \longmapsto \:a3 = 18 + 1 \\ \\ \sf \longmapsto \:a3 = 19$

━━━━━━━━━━━━━━━━━━━━━━━━━━

To be an AP the common difference of the terms should be same. So,

$\sf \longmapsto \:d1 = a2 - a1 \\ \\ \\ \sf \longmapsto \:d1 = 9 - 3 \\ \\ \\ \sf \longmapsto \:d1 = 6$

━━━━━━━━━━━━━━━━━━━━━━━━━━

Now,

$\sf \longmapsto \:d2 = a3 - a2 \\ \\ \\ \sf \longmapsto \:d2 = 19 - 9 \\ \\ \\ \sf \longmapsto \:d2 = 10$

Here, d1 ≠ d2

So ,we cannot define it as AP .