Question

Find the first three terms of the sequence below.
Tn=n^2−3n−6

Answer 1

Answer:

The first three terms of the given sequence are  -8, -8, -6.

Step-by-step explanation:

Given,

         the general term of the sequence, $T_{n}$ = $n^{2}$ - 3n - 6

To find the terms, putting the value of n=1, 2, 3 we get,

                  $T_{1}$ = $1^{2}$ - 3(1) - 6 = 1 - 3 - 6 = - 8

                  $T_{2}$ = $2^{2}$ - 3(2) - 6 = 4 - 6 - 6 = - 8

                  $T_{3}$ = $3^{2}$ - 3(3) - 6 = 9 - 9 - 6 = - 6

In this way, we can find all the other terms of the sequence too.

Hence, the required first three terms of the given sequence are -8, -8, -6.