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

Hits: 37

**Question**

A sequence a_{1}, a_{2}, a_{3}, . . . is defined by

a_{1} = 3

a_{n+1} = 3a_{n}– 5, n1.

**a. **Find the value of a_{2} and the value of a_{3}.

**b. **Calculate the value of

**Solution**

**a.
**

We are given that;

a_{n+1} = 3a_{n} – 5

a_{1} = 3

Therefore, we can find the required terms as follows.

First we find a_{2}.

a_{n+1} = 3a_{n} – 5

a_{1+1} = 3a_{1} – 5

a_{2} = 3a_{1} – 5

We are given that a_{1} =3, therefore;

a_{2} = 3(3) – 5

a_{2} = 9 – 5

a_{2} = 4

Next we find a_{3}.

a_{n+1} = 3a_{n} – 5

a_{2+1} = 3a_{2} – 5

a_{3} = 3a_{2} – 5

We have found that a_{2} = 4, therefore;

a_{3} = 3(4) – 5

a_{3} = 12 – 5

a_{3} = 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 a_{4} and a_{5 }as in (a).

We find a_{4}.

a_{n+1} = 3a_{n} – 5

a_{3+1} = 3a_{3} – 5

a_{4} = 3a_{3} – 5

We have found in (a) that a_{3} = 7, therefore;

a_{4} = 3(7) – 5

a_{4} = 21 – 5

a_{4} = 16

Next, we find a_{5}.

a_{n+1} = 3a_{n} – 5

a_{4+1} = 3a_{4} – 5

a_{5} = 3a_{4} – 5

We have found in (a) that a_{4} = 16, therefore;

a_{5} = 3(16) – 5

a_{5} = 48 – 5

a_{5} = 43

Now we can find;

## Comments