An arithmetic sequence has first term a and common difference d - AQA - A-Level Maths Pure - Question 5 - 2019 - Paper 1

We use the sum formula again for $n = 60$:

$S_{60} = \frac{60}{2} \times (2a + 59d) = 315$

Thus:

$30(2a + 59d) = 315$

This leads to:

$2a + 59d = 10.5$

We now have two equations:

1. $2a + 15d = 32.5$
2. $2a + 59d = 10.5$

Now we will eliminate $a$ by subtracting the first equation from the second:

$(2a + 59d) - (2a + 15d) = 10.5 - 32.5$

This simplifies to: $44d = -22$

Thus: $d = -0.5$

Using the value of $d$ to find $a$: substituting $d$ back into the first equation:

$2a + 15(-0.5) = 32.5$

This gives: $2a - 7.5 = 32.5$

Thus:

ightarrow a = 20$$ Now we can calculate the sum of the first 41 terms: $$S_{41} = \frac{41}{2}(2a + 40d)$$ Substituting the values of $a$ and $d$: $$S_{41} = \frac{41}{2}(2(20) + 40(-0.5))$$ This simplifies to: $$S_{41} = \frac{41}{2}(40 - 20) = \frac{41}{2} \times 20 = 410$$ Thus, the sum of the first 41 terms is 410.

The sum of the first n terms, $S_n$, of an arithmetic sequence is defined as:

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

For a given sequence, the term $d$ (the common difference) influences the overall value of the sum. In our case, $d$ is negative (-0.5), affecting the terms beyond $n = 41$:

1. The terms after the 41st term will also be negative, leading to a decrease in the total sum.
2. Thus, the sum will only increase until the 41st term because subsequent terms will detract from the total.

This indicates that the maximum value of $S_n$ occurs at $n = 41$.