Question

What is the nth term of the quadratic sequence: 6,20,40,66,98,136?

Answer 1

Answer:

$3n^2+5n-2$

Step-by-step explanation:

Consider the quadratic sequence :  6, 20, 40, 66, 98, 136

A quadratic sequence is of form $an^2+bn+c$.

For n = 1 :

$6=a+b+c$    ...(i)

For n = 2 :

$20=4a+2b+c$    ...(ii)

For  n = 3 :

$40=9a+3b+c$     ...(iii)

On subtracting equation (ii) from (iii), we get

$5a+b=20$      ...(iv)

On subtracting equation (i) from (ii), we get

3a+b=14                    ...(v)

On subtracting equation (iv) and (v) , we get

2a=6 which implies a=3

So, from equation (v), we get $b=14-3a=14-9=5$

From equation (i), we get

c=6-a-b=6-3-5=-2

On putting a = 3 , b = 5, c = -2 in $an^2+bn+c$, we get

$3n^2+5n-2$