Question

Prove that the 11th term of ap cannot be n^2+1 .justify your answer

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Answer 1

Answer:

If n th term is n²+1 then, the difference between nth term and n+1 th term is:

(n+1)² + 1 - n² - 1 = 2n +1

So the common difference changes with n. and is not a constant. Hence, the nth term cannot be n²+1 in an AP.

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Let the arithmetic series be:

a , a + d, a + 2 d, a + 3 d, ..... a + (n - 1) d

Let assume that the nth term is equal to n²+1,

So a + (n - 1) d = n² + 1

n² - n d + (1+d -a) = 0

n is an integer if d² - 4 d + 4a -4 is real, a perfect square and then its square root added to +d or -d is an even integer.

d² - 4 d + 4 (a-1) is real and perfect square only if the discriminant is 0.

4² - 16 (a - 1) = 0

=> a = 2

Now, d + (d-2) = 2 (d - 1) must be an even integer. Which is true if d is an integer.

OR, d - (d - 2) must be an even integer, which is true,

now n = d - 1 or 1

case 1) n = d - 1 => d = n+1 and a = 2

2, 2+n+1, 2+2(n+1), 2 + 3 (n+1) , ....

2, 5, 10, 17, ... This is not an Arithmetic Progression.

case 2) 2, 2+d, 2+2 d, 2+3d , 2+4 d

It seems only one term or two terms can satisfy the condition that n the term is n² + 1. Like in

2, 5, 8, 11 .... : only the first two terms satisfy the condition.

or 2, 6, 10, 14, ... : the first and 3 rd terms satisfy the condition.

So it is clear and proved that the nth term of an AP cannot be n²+1.