Home / IB DP Math AA-Topic: SL 5.3 Derivative of f(x): IB Style Question SL Paper 1

IB DP Math AA-Topic: SL 5.3 Derivative of f(x): IB Style Question SL Paper 1

Question

Consider the arithmetic sequence $a, p, q, …$ where $a, p, q \neq 0$.

(a) Show that $2p – q = a$.

Consider the geometric sequence $a, s, t, …$ where $a, s, t \neq 0$.

(b) Show that $s^2 = at$.

The first term of both sequences is $a$.

It is given that $q = t = 1$.

(c) Show that $p > \frac{1}{2}$.

Consider the case where $a = 9, s > 0$ and $q = t = 1$.

(d) Write down the first four terms of the

(i) arithmetic sequence;

(ii) geometric sequence.

The arithmetic and the geometric sequence are used to form a new arithmetic sequence $u_n$.

The first three terms of $u_n$ are $u_1 = 9 + \ln 9$, $u_2 = 5 + \ln 3$, and $u_3 = 1 + \ln 1$.

(e) (i) Find the common difference of the new sequence in terms of $\ln 3$.

(ii) Show that $\sum_{i=1}^{10} u_i = -90 – 25 \ln 3$.

▶️Answer/Explanation

Detailed solution

(a) Consider the following in the arithmetic sequence

{First term:} \( a \)

{Second term:} \( p = a + d \)

{Third term:} \( q = p + d = (a + d) + d = a + 2d \)

So, we have:
\[
p = a + d
\]
\[
q = a + 2d
\]

We need to evaluate \( 2p – q \):

\[
2p – q = 2(a + d) – (a + 2d)
\]

\[
= (2a + 2d) – (a + 2d)
\]

\[
= 2a + 2d – a – 2d
\]

\[
= a
\]

\[
2p – q = a
\]

Hence proved.

(b) Consider the following in the geometric sequence

{First term:} \( a \)

{Second term:} \( s = a \cdot r \)

{Third term:} \( t = s \cdot r = (a \cdot r) \cdot r = ar^2 \)

\[
s^2 = (ar)^2 = a^2r^2
\]

\[
at = a \cdot t = a \cdot (ar^2) = a^2r^2
\]

\[
s^2 = a^2r^2
\]

\[
at = a^2r^2
\]

\[
s^2 = at
\]

Hence proved.

(c) Substitute \( q = t = 1 \) in

\[
2p – q = a \quad \text{and} \quad s^2 = at
\]

we get

\[
2p – 1 = a \quad \dots (1)
\]

and

\[
s^2 = a \quad \dots (2)
\]

therefore

\[
2p – 1 = s^2 \quad \dots (3)
\]

From \( q = 2p – 1 \),

From (2), \( s^2 = a \)

Substituting in from (3):

\[
2p – 1 = a
\]

\[
s^2 – 1 \geq 0
\]

Since \( s^2 \) is the square of a real number, it must be non-negative:

\[
s^2 – 1 \geq 0
\]

\[
s^2 \geq 1
\]

\[
p \geq 1
\]

\[
2p – 1 \geq 1
\]

If \( p = \frac{1}{2} \),

\[
a = 2 \cdot \frac{1}{2} – 1 = 1 – 1 = 0
\]

But \( a \neq 0 \), so \( p = \frac{1}{2} \) is not allowed. Since \( p \geq \frac{1}{2} \) and \( p \neq \frac{1}{2} \),

\[
p > \frac{1}{2}
\]

Hence proved.

(d) In the arithmetic sequence

{First term:} \( a = 9 \)

{Third term:} \( q = 1 \)

{Second term:} \( p \) (find using part (a))

{Fourth term:} Use common difference

From \( 2p – q = a \):

\[
2p – 1 = 9
\]

\[
2p = 10
\]

\[
p = 5
\]

Common difference:

\[
d = p – a = 5 – 9 = -4
\]

{First:} \( a = 9 \)

{Second:} \( p = 5 \)

{Third:} \( q = 1 \)

{Fourth:} \( 1 + d = 1 – 4 = -3 \)

Hence, the first four terms of the arithmetic sequence are \( 9, 5, 1, -3 \).

(ii) In the geometric sequence

{First term:} \( a = 9 \)

{Third term:} \( t = 1 \)

{Second term:} \( s \) (find using part (b))

{Fourth term:} Use common ratio

From \( p^2 = at \):

\[
s^2 = 9 \times 1 = 9
\]

\[
s = \pm 3
\]

Since \( a > 0 \), \( s > 0 \), so \( s = 3 \).

Common ratio:

\[
r = \frac{s}{a} = \frac{3}{9} = \frac{1}{3}
\]

{First:} \( a = 9 \)

{Second:} \( s = 3 \)

{Third:} \( t = 1 \)

{Fourth:} \( t \cdot r = 1 \times \frac{1}{3} = \frac{1}{3} \)

Hence, the first four terms of the geometric sequence are \( 9, 3, 1, \frac{1}{3} \).

(e) (i) Finding the common difference

\[
d = u_2 – u_1 = 5 + \ln 3 – 9 – \ln 9
\]

OR

\[
d = u_4 – u_3 = 1 + \ln 1 – 5 – \ln 3
\]

\[
\ln 9 = 2 \ln 3 \quad \text{or} \quad \ln 1 = 0
\]

\[
\ln 9 – \ln 3 = \ln \left( \frac{9}{3} \right) = \ln 3
\]

\[
d = -4 – \ln 3
\]

(ii) The required sum of terms of the arithmetic sequence

Sum of first \( n \) terms:

\[
S_n = \frac{n}{2} (u_1 + u_n)
\]

For \( n = 10 \):

\[
u_1 = 9 + \ln 3
\]

\[
u_{10} = 9 + 2 \ln 3 + (10 – 1)(-4 – \ln 3)
\]

\[
= 9 + 2 \ln 3 – 36 – 9 \ln 3
\]

\[
= -27 – 7 \ln 3
\]

\[
S_{10} = \frac{10}{2} [(9 + 2 \ln 3) + (-27 – 7 \ln 3)]
\]

\[
= 5(9 – 27 + 2 \ln 3 – 7 \ln 3)
\]

\[
= 5(-18 – 5 \ln 3)
\]

\[
= -90 – 25 \ln 3
\]

Hence proved.

………………….Markscheme…………………

Solution: –

(a) attempt to find a difference

$d=p-a, 2d=q-a, d=q-p$ OR $p=a+d, q=a+2d, p+d=q$

correct equation

$p-a=q-p$ OR $q-a=2(p-a)$ OR $p=\frac{a+q}{2}$ (or equivalent)

$2p-q=a$

(b) attempt to find a ratio

$r=\frac{s}{a}, r=\frac{t}{s}, r=\frac{1}{t}$ OR $s=ar, t=ar^2, t=sr$

correct equation

$(\frac{s}{a})^3=\frac{t}{a}$ OR $\frac{s}{a}=\frac{t}{s}$ (or equivalent)

$s^2=at$

(c) EITHER

$2p-1=s^2$ (or equivalent)

$(s^2>0)\Rightarrow2p-1>0$ OR $\sqrt{2p-1}=s\Rightarrow2p-1>0$ OR $p=\frac{s^2+1}{2}$ (and $s^2>0$)

OR

$2p-1=a$ and $s^2=a$

$(s^2>0, so) a>0\Rightarrow2p-1>0$ OR $p=\frac{a+1}{2}$ and $a>0$

$\Rightarrow p>\frac{1}{2}$

(d) (i) 9, 5, 1, -3

(ii) 9, 3, 1, \frac{1}{3}

(e) (i) attempt to find the difference between two consecutive terms

$d=u_2-u_1=5+\ln3-9-\ln9$ OR $d=u_3-u_2=1+\ln1-5-\ln3$

$\ln9=2\ln3$ OR $\ln1=0$ OR $\ln3-\ln9=\ln\frac{1}{3}(=\ln3^{-1}=-\ln3)$ (seen anywhere)

$d=-4-\ln3$

(ii) METHOD 1

attempt to substitute first term and their common difference into $S_{10}$

$\frac{10}{2}(2(9+\ln9)+9(-4-\ln3))$ OR $\frac{10}{2}(2(9+2\ln3)+9(-4-\ln3))$ (or equivalent)

$=5(-18-5\ln3)$ (or equivalent in terms of $\ln3$)

$\sum_{i=1}^{10}u_i=-90-25\ln3$

METHOD 2

$u_{10}=9+\ln9+9(-4-\ln3)=(-27+\ln9-9\ln3)$

attempt to substitute first term and their $u_{10}$ into $S_{10}$

$\frac{10}{2}(2(9+\ln9)+9(-4-\ln3))$ OR $\frac{10}{2}(9+\ln9-27+\ln9-9\ln3)$ OR

$\frac{10}{2}(2(9+2\ln3)+9(-4-\ln3))$ OR $\frac{10}{2}(9+\ln9-27-7\ln3)$ (or equivalent)

$=5(-18-5\ln3)$ (or equivalent in terms of $\ln3$)

$\sum_{i=1}^{10}u_i=-90-25\ln3$

Scroll to Top