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. --- 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 Pure Arithmetic Sequences & Series: The ninth term of an arithmetic

The ninth term of an arithmetic series is 3 - AQA - A-Level Maths Pure - Question 6 - 2021 - Paper 1

Question 6

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:The ninth term of an arithmetic series is 3 - AQA - A-Level Maths Pure - Question 6 - 2021 - Paper 1

Step 1

Find the first term and common difference of the first arithmetic series

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Answer

Let the first term be $a$ and the common difference be $d$ .

From the information given, we can express the following:

The ninth term of the series is: $T_9 = a + 8d = 3 ag{1}$
The sum of the first $21$ terms is: $S_{21} = \frac{21}{2} (2a + 20d) = 42 ag{2}$

From (2), simplifying we have: $21 (2a + 20d) = 84$ Therefore, $2a + 20d = 4 ag{3}$

Now, we have two equations (1) and (3) to solve:

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

From (1): $a = 3 - 8d$

Substituting into (3): $2(3 - 8d) + 20d = 4$ $6 - 16d + 20d = 4$ $4d = -2$ $d = -0.5$

Substituting back to find $a$ : $a = 3 - 8(-0.5) = 3 + 4 = 7$

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

Step 2

Find the value of n such that T_n = S_n

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Answer

For the second arithmetic series, first we find $T_n$ :

The first term is $-18$ and the common difference is $rac{3}{4}$ .

The sum of the first $n$ terms is given by: $T_n = \frac{n}{2}(2a + (n-1)d)$

Substituting values: $T_n = \frac{n}{2}(2(-18) + (n-1)(\frac{3}{4}))$ $T_n = \frac{n}{2}(-36 + \frac{3}{4}(n-1))$

We need to equate $T_n$ with $S_n$ . For $S_n$ , we previously found: $S_n = \frac{n}{2}(2a + (n-1)d)$

Setting equal: $T_n = S_n$

Solving the resulting equation gives: $n = 41$

The ninth term of an arithmetic series is 3 - AQA - A-Level Maths Pure - Question 6 - 2021 - Paper 1

Worked Solution & Example Answer:The ninth term of an arithmetic series is 3 - AQA - A-Level Maths Pure - Question 6 - 2021 - Paper 1

Find the first term and common difference of the first arithmetic series

Find the value of n such that T_n = S_n

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