The first term of an arithmetic sequence is 30 and the common difference is −1.5 (a) Find the value of the 25th term - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 2

To find the value of $r$ such that the $r$th term is 0, we use the nth term formula:

$a_r = a + (r-1)d$

Setting this to zero, we have:

$0 = 30 + (r-1)(-1.5)$

Rearranging the equation yields:

$(r-1)(-1.5) = -30$ $(r-1) = \frac{-30}{-1.5}$ $(r-1) = 20$

Adding 1 to both sides results in:

$r = 21$

Therefore, the value of $r$ is 21.

The sum of the first $n$ terms of an arithmetic sequence is given by:

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

Substituting the values we have:

• $a = 30$,
• $d = -1.5$.

Thus, we can express this as:

$S_n = \frac{n}{2} (2 \cdot 30 + (n-1)(-1.5))$
$S_n = \frac{n}{2} (60 - 1.5(n-1))$
$S_n = \frac{n}{2} (60 - 1.5n + 1.5)$
$S_n = \frac{n}{2} (61.5 - 1.5n)$

To find the largest positive value of $S_n$, we need to consider the function:

$S_n = \frac{61.5n - 1.5n^2}{2}$

This is a quadratic equation in standard form, which opens downwards. The maximum value occurs at the vertex, given by:

$n = -\frac{b}{2a} = -\frac{61.5}{-3} = 20.5$

As $n$ must be an integer, we analyze $n = 20$ and $n = 21$:

For $n = 20$: $S_{20} = \frac{20}{2}(60 - 1.5(19)) = 10(60 - 28.5) = 10 imes 31.5 = 315$

For $n = 21$: $S_{21} = \frac{21}{2}(60 - 1.5(20)) = \frac{21}{2}(60 - 30) = \frac{21}{2} imes 30 = 315$

Since both provide the same sum, the largest positive value of $S_n$ is 315.