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2015-06-10T15:22:31+05:30

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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=\frac{d+-\sqrt{d^2-4d+4a-4}}{2}

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.

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