Question

Prove that nth term of an AP can't be n^2+n+1.​

Answer 1

Solution :-

Now we know that what is A.P

A.P is series in which the difference between the two consecutive terms is constant.

$a_{n} - a_{n-1} = d$

Now as given expression for nth term

= n² + n + 1

So

(n-1) th term

= (n-1)² +(n-1) + 1

= n² - 2n + 1 + n - 1 + 1

= n² -n + 1

Now we will find out the common difference :-

nth term - (n-1)th term

→ n² + n + 1 - ( n² -n + 1 )

→ n² + n + 1 - n² + n - 1

→n² - n² + n + n + 1 - 1

→ 2n

Now as d is not a constant value i.e it is dependent upon a variable quantity "n"

So n² + n + 1 cannot be nth term of an AP

Answer 2

Answer:

Solution :-

Now we know that what is A.P

A.P is series in which the difference between the two consecutive terms is constant.

Now as given expression for nth term

= n² + n + 1

So

(n-1) th term

= (n-1)² +(n-1) + 1

= n² - 2n + 1 + n - 1 + 1

= n² -n + 1

Now we will find out the common difference :-

nth term - (n-1)th term

→ n² + n + 1 - ( n² -n + 1 )

→ n² + n + 1 - n² + n - 1

→n² - n² + n + n + 1 - 1

→ 2n

Now as d is not a constant value i.e it is dependent upon a variable quantity "n"

So n² + n + 1 cannot be nth term of an AP

Step-by-step explanation: