The first four terms of an artithmetic sequence are 2, a-b, 2a+b+7, and a-3b, where a and b are constants

Question

The first four terms of an artithmetic sequence are 2, a-b, 2a+b+7, and a-3b, where a and b are constants.

Find a and b. 

Answer 1

Since the sequence is arithmetic, consecutive terms always differ by the same constant.

 

So, (a-b) - 2 = (2a+b+7) - (a-b)   and (2a+b+7) - (a-b) = (a-3b) - (2a+b+7)

 

Simplify to get a-b-2 = a+2b+7  and  a+2b+7 = -a-4b-7

 

From the first equation, we get -3b = 9.  So, b = -3.

 

Substituting for b into the equation a+2b+7 = -a-4b-7, we have a+1 = -a+5

 

                                                                                                2a = 4

 

                                                                                                 a = 2

 

The first 4 terms of the sequence are 2, 5, 8, 11.