FPM(2023) January Paper 01 que 1 and solution

Question 1

The $n^{th}$ term of an arithmetic series is $a_n,$ where $a_{10} + a_{11} + a_{12} = 129$ and $a_{19} + a_{20} + a_{21} = 237.$

Find $a_1$

Step 1: Expressing the terms in the arithmetic series

In an arithmetic series, the $n$th term is given by the formula:

$$ a_n = a_1 + (n - 1)d $$

where $a_1$ is the first term and $d$ is the common difference.

We are given two equations:

$$ a_{10} + a_{11} + a_{12} = 129 $$

and

$$ a_{19} + a_{20} + a_{21} = 237 $$

Let's express each of these terms using the general formula for $a_n$.

Step 2: Substituting the expressions for the terms

For the first equation $a_{10} + a_{11} + a_{12} = 129$, we substitute the general formula for each term:

$$ a_{10} = a_1 + 9d, \quad a_{11} = a_1 + 10d, \quad a_{12} = a_1 + 11d $$

Thus, the equation becomes:

$$ (a_1 + 9d) + (a_1 + 10d) + (a_1 + 11d) = 129 $$

Simplifying:

$$ 3a_1 + 30d = 129 $$

Divide through by 3:

$$ a_1 + 10d = 43 \quad \text{(Equation 1)} $$

For the second equation $a_{19} + a_{20} + a_{21} = 237$, we substitute the general formula for each term:

$$ a_{19} = a_1 + 18d, \quad a_{20} = a_1 + 19d, \quad a_{21} = a_1 + 20d $$

Thus, the equation becomes:

$$ (a_1 + 18d) + (a_1 + 19d) + (a_1 + 20d) = 237 $$

Simplifying:

$$ 3a_1 + 57d = 237 $$

Divide through by 3:

$$ a_1 + 19d = 79 \quad \text{(Equation 2)} $$

Step 3: Solving the system of equations

Now, we solve the system of equations:

$$ a_1 + 10d = 43 \quad \text{(Equation 1)} $$ $$ a_1 + 19d = 79 \quad \text{(Equation 2)} $$

Subtract Equation 1 from Equation 2:

$$ (a_1 + 19d) - (a_1 + 10d) = 79 - 43 $$

Simplifying:

$$ 9d = 36 $$

Solve for $d$:

$$ d = 4 $$

Step 4: Finding $a_1$

Substitute $d = 4$ into Equation 1:

$$ a_1 + 10(4) = 43 $$

Simplifying:

$$ a_1 + 40 = 43 $$

Solving for $a_1$:

$$ a_1 = 43 - 40 = 3 $$

Thus, the first term of the arithmetic series is $\boxed{3}$.