Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2006 | June | Q#4

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Question

A sequence a1, a2, a3, . . . is defined by

a1 = 3

an+1 = 3an– 5, n1.

a.   Find the value of a2 and the value of a3.

b.   Calculate the value of

Solution

a.
 

We are given that;

an+1 = 3an – 5

a1 = 3

Therefore, we can find the required terms as follows.

First we find a2.

an+1 = 3an – 5

a1+1 = 3a1 – 5

a2 = 3a1 – 5

We are given that a1 =3, therefore;

a2 = 3(3) – 5

a2 = 9 – 5

a2 = 4

Next we find a3.

an+1 = 3an – 5

a2+1 = 3a2 – 5

a3 = 3a2 – 5

We have found that a2 = 4, therefore;

a3 = 3(4) – 5

a3 = 12 – 5

a3 = 7

b.
 

We are required to find .

It is evident that we are required to find the sum of first 5 terms of arithmetic series given and found  in (a).

We can also find a4 and a5 as in (a).

We find a4.

an+1 = 3an – 5

a3+1 = 3a3 – 5

a4 = 3a3 – 5

We have found in (a) that a3 = 7, therefore;

a4 = 3(7) – 5

a4 = 21 – 5

a4 = 16

Next, we find a5.

an+1 = 3an – 5

a4+1 = 3a4 – 5

a5 = 3a4 – 5

We have found in (a) that a4 = 16, therefore;

a5 = 3(16) – 5

a5 = 48 – 5

a5 = 43

Now we can find;

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