6 (a) The ninth term of an arithmetic series is 3 The sum of the first n terms of the series is $S_n$, and $S_{21} = 42$ Find the first term and common difference of the series. 6 (b) A second arithmetic series has first term −18 and common difference $rac{3}{4}$ The sum of the first n terms of this series is $T_n$ Find the value of n such that $T_n = S_n$

A-Level AQA Maths Mechanics Quantities, Units & Modelling: 6 (a) The ninth term of an arith

6 (a) The ninth term of an arithmetic series is 3 The sum of the first n terms of the series is $S_n$, and $S_{21} = 42$ Find the first term and common difference of the series - AQA - A-Level Maths Mechanics - Question 6 - 2021 - Paper 1

Question 6

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6 (a) The ninth term of an arithmetic series is 3 The sum of the first n terms of the series is $S_n$, and $S_{21} = 42$ Find the first term and common difference of... show full transcript

Worked Solution & Example Answer:6 (a) The ninth term of an arithmetic series is 3 The sum of the first n terms of the series is $S_n$, and $S_{21} = 42$ Find the first term and common difference of the series - AQA - A-Level Maths Mechanics - Question 6 - 2021 - Paper 1

Step 1

Find the first term and common difference of the series.

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Answer

Given that the ninth term is 3, we can express this in terms of the first term $a$ and the common difference $d$ as follows:

$a + 8d = 3$

Additionally, we know from the sum of the first $n$ terms formula:

$S_n = \frac{n}{2}(2a + (n-1)d)$

For $n = 21$ , we have:

$S_{21} = \frac{21}{2}(2a + 20d) = 42$

Multiplying both sides by 2 gives:

$21(2a + 20d) = 84$

Dividing the equation by 21 simplifies to:

$2a + 20d = 4$

Now, we have a system of equations:

$a + 8d = 3$
$2a + 20d = 4$

We can solve these equations. First, let's express $a$ from the first equation:

$a = 3 - 8d$

Substituting this into the second equation:

$2(3 - 8d) + 20d = 4$

Expanding this gives:

$6 - 16d + 20d = 4$

Combining like terms results in:

$4d = -2$

Thus, we find:

$d = -0.5$

Now substitute $d$ back into $a$ :

$a = 3 - 8(-0.5) = 3 + 4 = 7$

Therefore, the first term is $a = 7$ and the common difference is $d = -0.5$ .

Step 2

Find the value of n such that T_n = S_n.

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Answer

The sum of the first $n$ terms of the second arithmetic series is given by:

$T_n = \frac{n}{2}(2a + (n-1)d)$

Here, the first term $a = -18$ and the common difference $d = \frac{3}{4}$ :

So, substituting these values:

$T_n = \frac{n}{2}(2(-18) + (n-1) \cdot \frac{3}{4})$

This simplifies to:

$T_n = \frac{n}{2}(-36 + \frac{3}{4}(n-1))$

The sum $S_n$ of the first $n$ terms of the first series is:

$S_n = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(2(7) + (n-1)(-0.5))$

This simplifies to:

$S_n = \frac{n}{2}(14 - 0.5(n-1))$

Setting $T_n = S_n$ :

$\frac{n}{2}(-36 + \frac{3}{4}(n-1)) = \frac{n}{2}(14 - 0.5(n-1))$

Canceling out the $rac{n}{2}$ (assuming $n \neq 0$ ) gives:

$-36 + \frac{3}{4}(n-1) = 14 - 0.5(n-1)$

Expanding both sides leads to:

$-36 + \frac{3}{4}n - \frac{3}{4} = 14 - 0.5n + 0.5$

Combining like terms results in:

$\frac{3}{4}n + 0.5n = 14 + 36 + 0.5 + \frac{3}{4}$

This can be simplified and solved for $n$ . Ultimately, after performing the calculations, we find that:

$n = 41.$

6 (a) The ninth term of an arithmetic series is 3 The sum of the first n terms of the series is $S_n$, and $S_{21} = 42$ Find the first term and common difference of the series - AQA - A-Level Maths Mechanics - Question 6 - 2021 - Paper 1

Worked Solution & Example Answer:6 (a) The ninth term of an arithmetic series is 3 The sum of the first n terms of the series is $S_n$, and $S_{21} = 42$ Find the first term and common difference of the series - AQA - A-Level Maths Mechanics - Question 6 - 2021 - Paper 1

Find the first term and common difference of the series.

Find the value of n such that T_n = S_n.

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