Question

Show that 105 is not a term of the AP. 4, 6, 8...​

Answer 1

Answer:

An = 105 , a = 4 and d= 6-4 = 2

an = a + (n-1) d

105 = 4+(n-1) 2

105-4=2n-2

101+2=2n

103/2= n

n=51.5

So, n has not a proper value

therefore 105 is not a term of this AP

Answer 2

First term

a = 4

common difference(d) = 2

let 105 be the Nth term

an = a+(n-1)d

105 = 4+(n-1)2

105 = 4+2n-2

105-+2-4 = 2n

105-2 = 2n

103

$\frac{103}{2} = n$

Now, n, or term number, has to be an integer for the number to be a term of the A.P. Here, it's 51.5 which isn't an integer. So 105 isn't a term of this A.P.