**Question**

The equation of a curve is \(y=x^{3}+px^{2}\) ,where p is a positive constant .

(i)Show that the origin is a stationary point on the curve and find the coordinates of the other stationary point in terms of p.

(ii)Find the nature of each of the stationary points.

Another curve has equation \(y=x^{3}+px^{2}+px\)

(iii)Find the set of values of p for which this curve has n o stationary points.

**▶️Answer/Explanation**

(i)Differentiate the curve to find the stationary points. For \(y = x^3 + px^2\), the derivative is:

\( \frac{dy}{dx} = 3x^2 + 2px \)

To find stationary points, set the derivative to zero:

\( 3x^2 + 2px = 0 \)

\( x(3x + 2p) = 0 \)

This gives two solutions: \(x = 0\) and \(x = -\frac{2p}{3}\).

At \(x = 0\), \(y = 0^3 + p \cdot 0^2 = 0\). So, the origin \((0, 0)\) is a stationary point.

At \(x = -\frac{2p}{3}\), \(y = \left(-\frac{2p}{3}\right)^3 + p \left(-\frac{2p}{3}\right)^2\). Simplifying, the coordinates of the other stationary point are:

\( \left(-\frac{2p}{3}, \frac{4p^3}{27} – \frac{4p^3}{9}\right) = \left(-\frac{2p}{3}, -\frac{4p^3}{27}\right) \)

(ii) The second derivative of \(y\) is:

\( \frac{d^2y}{dx^2} = 6x + 2p \)

At the origin \((0, 0)\), the second derivative is \(2p\). Since \(p\) is a positive constant, \(2p > 0\), indicating a local minimum at the origin.

At the point \(\left(-\frac{2p}{3}, -\frac{4p^3}{27}\right)\), the second derivative is \(6(-\frac{2p}{3}) + 2p = -4p\). Since \(p\) is positive, \(-4p < 0\), indicating a local maximum at this point.

(iii) For the curve \(y = x^3 + px^2 + px\), differentiating and set the derivative equal to zero:

\( \frac{dy}{dx} = 3x^2 + 2px + p \)

For no stationary points, the quadratic equation \(3x^2 + 2px + p = 0\) should have no real roots. This occurs when the discriminant is negative:

\( \Delta = (2p)^2 – 4 \cdot 3 \cdot p = 4p^2 – 12p < 0 \)

\( 4p(p – 3) < 0 \)

This inequality is satisfied when \(0 < p < 3\).

Therefore, the curve \(y = x^3 + px^2 + px\) has no stationary points when \(p\) is in the range \(0 < p < 3\).

**Question**

An arithmetic progression has first term a and common difference d.It is given that the sum of the first 200 terms is 4 times the sum of first 100 terms.

(i)Find d in terms of a.

(ii) Find the 100th terms of a.

**▶️Answer/Explanation**

(i)\( S_n = \frac{n}{2} [2a + (n – 1)d] \)

Where:

\( S_n \) is the sum of the first \( n \) terms,

\( a \) is the first term,

\( d \) is the common difference, and

\( n \) is the number of terms.

\( S_{200} = 4 \times S_{100} \)

\( \frac{200}{2} [2a + (200 – 1)d] = 4 \times \frac{100}{2} [2a + (100 – 1)d] \)

\( 100 [2a + 199d] = 4 \times 50 [2a + 99d] \)

\( 100 [2a + 199d] = 200 [2a + 99d] \)

\( 200a + 19900d = 400a + 19800d \)

\( 19900d – 19800d = 400a – 200a \)

\( 100d = 200a \)

\( d = 2a \)

(ii)\( T_n = a + (n – 1)d \)

\( T_{100} = a + (100 – 1)(2a) \)

\( T_{100} = a + 99 \times 2a \)

\( T_{100} = a + 198a \)

\( T_{100} = 199a \)

**Question**

The third term of a geometric progression is -108 and the sixth term is 32.Find

**(i)the common ratio.**

**(ii)the first term.**

(iii)the sum to infinity.

**▶️Answer/Explanation**

(i)The \( n \)th term of a geometric progression (GP),

\( T_n = ar^{n-1} \)

\( T_n \) is the \( n \)th term,

\( a \) is the first term, and

\( r \) is the common ratio.

The third term (\( T_3 \)) is -108,

The sixth term (\( T_6 \)) is 32.

\( T_3 = ar^2 = -108 \)

\( T_6 = ar^5 = 32 \)

\( \frac{T_6}{T_3} = \frac{ar^5}{ar^2} \)

\( \frac{32}{-108} = r^3 \)

\( -\frac{32}{108} = r^3 \)

\( -\frac{4}{13.5} = r^3 \)

\( r = \sqrt[3]{-\frac{4}{13.5}} \)

\( r = -\frac{2}{3} \)

(ii) \( -108 = ar^2 \)

\( -108 = a\left(-\frac{2}{3}\right)^2 \)

\( -108 = a\left(\frac{4}{9}\right) \)

\( a = \frac{-108 \times 9}{4} \)

\( a = -243 \)

(iii)\( S_{\infty} = \frac{a}{1 – r} \)

But this formula is only valid if \( |r| < 1 \), which is true in our case since \( r = -\frac{2}{3} \).

\( S_{\infty} = \frac{-243}{1 – (-\frac{2}{3})} \)

\( S_{\infty} = \frac{-243}{1 + \frac{2}{3}} \)

\( S_{\infty} = \frac{-243}{\frac{5}{3}} \)

\( S_{\infty} = -243 \times \frac{3}{5} \)

\( S_{\infty} = -145.8 \)

**Question**

(i)In the expression\(\left ( 1-px \right )^{6}\) ,p is a non-zero constant.Find the first three terms when\(\left ( 1-px \right )^{6}\) is expanded in ascending powers of x.

(ii)It is given that the coefficient of \(x^{2}\) in the expansion of \(\left ( 1-x \right )\left ( 1-px \right )^{6}\)

is zero .Find the value of p.

**▶️Answer/Explanation**

(i)The binomial theorem states that:

\( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)

In our case, \( a = 1 \) and \( b = -px \).

For \( k = 0 \):

\( \binom{6}{0} \cdot 1^6 \cdot (-px)^0 = 1 \)

For \( k = 1 \):

\( \binom{6}{1} \cdot 1^5 \cdot (-px)^1 = -6px \)

For \( k = 2 \):

\( \binom{6}{2} \cdot 1^4 \cdot (-px)^2 = 15p^2x^2 \)

So, the first three terms of \((1 – px)^6\) when expanded are \( 1 – 6px + 15p^2x^2 \).

(ii) The given expression is \((1 – x)(1 – px)^6\). We already have the first three terms of \((1 – px)^6\) from part (i). Let’s multiply \((1 – x)\) with these terms:

\( (1 – x)(1 – 6px + 15p^2x^2) \)

\( -x \times 1 = -x \)

\( -x \times (-6px) = 6p x^2 \)

\( 1 \times 15p^2x^2 = 15p^2x^2 \)

The \( x^2 \) terms are \( 6p x^2 + 15p^2x^2 \). The coefficient of \( x^2 \) is \( 6p + 15p^2 \).

\( 6p + 15p^2 = 0 \)

\( p(6 + 15p) = 0 \)

\( p = 0 \) (but \( p \) is a non-zero constant, so we discard this solution

\( 6 + 15p = 0 \)

\( 15p = -6 \)

\( p = -\frac{6}{15} = -\frac{2}{5} \)

So, the value of \( p \) is \( -\frac{2}{5} \).

**Question**

**(a)The first two terms of an arithmetic progression are 1 and \(\cos ^{2}x\) respectively. Show that the sum of the first ten terms can be expressed in the form \(a-b\sin ^{2}x\),where a and b are constants to be found.**

(b)The first two terms of a geometric progression are 1 and \(\frac{1}{3}\tan ^{2}\Theta \) respectively, where \(0< \Theta < \frac{1}{2}\pi \)

(i) Find the set of values of \(\Theta\) for which the progression is convergent.

(ii)Find the exact value of the sum to infinity when \(\Theta =\frac{1}{6}\pi \)

**▶️Answer/Explanation**

(a) Given the first two terms of an arithmetic progression (AP) are 1 and \(\cos^2 x\) respectively, we can find the common difference \( d \) and then the sum of the first ten terms.

The common difference \( d \) in an AP is given by the difference between the second and the first term:

\( d = \cos^2 x – 1 \)

The sum of the first \( n \) terms of an AP is given by:

\( S_n = \frac{n}{2} [2a + (n – 1)d] \)

Here, \( a = 1 \) (the first term) and \( n = 10 \). Plugging these values in, we get:

\( S_{10} = \frac{10}{2} [2 \times 1 + (10 – 1)(\cos^2 x – 1)] \)

\( S_{10} = 5 [2 + 9(\cos^2 x – 1)] \)

\( S_{10} = 5 [2 + 9\cos^2 x – 9] \)

\( S_{10} = 5 [9\cos^2 x – 7] \)

\( S_{10} = 45\cos^2 x – 35 \)

\( S_{10} = 45(1 – \sin^2 x) – 35 \)

\( S_{10} = 45 – 45\sin^2 x – 35 \)

\( S_{10} = 10 – 45\sin^2 x \)

Therefore, the sum of the first ten terms can be expressed in the form \( a – b\sin^2 x \) where \( a = 10 \) and \( b = 45 \).

(b)(i) The first two terms of the geometric progression (GP) are 1 and \( \frac{1}{3}\tan^2 \Theta \) respectively.

The common ratio \( r \) of the GP is the second term divided by the first term:

\( r = \frac{1}{3}\tan^2 \Theta \)

\( |r| < 1 \)

\( \left|\frac{1}{3}\tan^2 \Theta\right| < 1 \)

\( \frac{1}{3}\tan^2 \Theta < 1 \)

\( \tan^2 \Theta < 3 \)

\(\tan \Theta < \sqrt{3} \)

\( \Theta < \arctan(\sqrt{3}) \)

\( \Theta < \frac{\pi}{3} \)

Given that \( 0 < \Theta < \frac{\pi}{2} \), the set of values for \( \Theta \) for which the progression is convergent is \( 0 < \Theta < \frac{\pi}{3} \).

(ii) Sum to Infinity When \( \Theta = \frac{\pi}{6} \)

For a convergent GP, the sum to infinity is given by:

\( S_{\infty} = \frac{a}{1 – r} \)

Where \( a = 1 \) (the first term) and \( r \) is the common ratio.

When \( \Theta = \frac{\pi}{6} \), we find \( r \):

\( r = \frac{1}{3}\tan^2 \left(\frac{\pi}{6}\right) \)

\( r = \frac{1}{3}\left(\sqrt{3}/3\right)^2 \)

\( r = \frac{1}{3} \times \frac{1}{3} \)

\( r = \frac{1}{9} \)

\( S_{\infty} = \frac{1}{1 – \frac{1}{9}} \)

\( S_{\infty} = \frac{1}{\frac{8}{9}} \)

\( S_{\infty} = \frac{9}{8} \)

Thus, the exact value of the sum to infinity when \( \Theta = \frac{\pi}{6} \) is \( \frac{9}{8} \).

**Question**

(a)A geometric has first 3a and common ratio r. A second geometric progression has first term a and common ratio -2r.The two progressions have the same sum to infinity. Find the value of r.

**(b) The first two terms of an arithmetic progression are 15 and 19 respectively. The first two terms of a second arithmetic progression are 420 and 415 respectively.The two progression have sum of first n terms. Find the value of n.**

**▶️Answer/Explanation**

(a)First progression: First term \(3a\), common ratio \(r\).

Second progression: First term \(a\), common ratio \(-2r\).

\(S\), is given by \(S = \frac{a}{1 – r}\), where \(a\) is the first term and \(r\) is the common ratio.

For the first progression, the sum to infinity is:

\( S_1 = \frac{3a}{1 – r} \)

For the second progression, the sum to infinity is:

\( S_2 = \frac{a}{1 – (-2r)} = \frac{a}{1 + 2r} \)

\( \frac{3a}{1 – r} = \frac{a}{1 + 2r} \)

\( \frac{3a}{1 – r} = \frac{a}{1 + 2r} \)

\( 3(1 + 2r) = 1 – r \)

\( 3 + 6r = 1 – r \)

\( 6r + r = 1 – 3 \)

\( 7r = -2 \)

\( r = \frac{-2}{7} \)

So, \(r = -\frac{2}{7}\).

(b) First progression: First term \(15\), second term \(19\).

Second progression: First term \(420\), second term \(415\).

\(S_n = \frac{n}{2}(2a + (n – 1)d)\), where \(a\) is the first term and \(d\) is the common difference.

For the first progression, the common difference \(d_1\) is \(19 – 15 = 4\). Thus, the sum of the first \(n\) terms is:

\( S_{n1} = \frac{n}{2}(2 \cdot 15 + (n – 1) \cdot 4) \)

For the second progression, the common difference \(d_2\) is \(415 – 420 = -5\). Thus, the sum of the first \(n\) terms is:

\( S_{n2} = \frac{n}{2}(2 \cdot 420 + (n – 1) \cdot (-5)) \)

Since the sums are equal, set \(S_{n1} = S_{n2}\)

Equating the sums of the first \(n\) terms for both progressions:

To solve the equation

\( \frac{1}{2}[2 \times 15 + (n – 1)4] = \frac{1}{2} n[2 \times 420 + (n – 1)(-5)] \)

for \( n \),

\( \frac{1}{2}[30 + 4n – 4] = \frac{1}{2}[26 + 4n] = 13 + 2n \)

\( \frac{1}{2} n[840 – 5n + 5] = \frac{1}{2} n[845 – 5n] = 422.5n – \frac{5}{2}n^2 \)

\( 13 + 2n = 422.5n – \frac{5}{2}n^2 \)

\( \frac{5}{2}n^2 – 420.5n + 13 = 0 \)

\( 5n^2 – 841n + 26 = 0 \)

Using the quadratic formula:

\( n = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)

where \( a = 5 \), \( b = -841 \), and \( c = 26 \).

\( \text{Discriminant} = (-841)^2 – 4 \times 5 \times 26 \)

**Question**

(i)Find the first 3 terms in the expansion of \(\left ( 2x-x^{2} \right )^{6}\) in ascending powers of x.

(ii)Hence find the coefficient of \(x^{8}\) in the expansion of \(\left ( 2+x \right )\left ( 2x-x^{2} \right )^{6}\).

**▶️Answer/Explanation**

(i) The binomial theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\), where \(\binom{n}{k}\) is the binomial coefficient.

For \((2x – x^2)^6\), \(a = 2x\) and \(b = -x^2\). We are interested in the first three terms (ascending powers of \(x\)), so we’ll calculate for \(k = 0, 1, 2\):

\(k = 0\):

\( \binom{6}{0} (2x)^6 (-x^2)^0 = 1 \times 64x^6 \times 1 = 64x^6 \)

\(k = 1\):

\( \binom{6}{1} (2x)^5 (-x^2)^1 = 6 \times 32x^5 \times (-x^2) = -192x^7 \)

\(k = 2\):

\( \binom{6}{2} (2x)^4 (-x^2)^2 = 15 \times 16x^4 \times x^4 = 240x^8 \)

So, the first three terms in ascending powers of \(x\) are \(64x^6\), \(-192x^7\), and \(240x^8\).

(ii)Coefficient of \(x^8\) in \((2 + x)(2x – x^2)^6\)

Now, we’ll multiply these terms by each term in \((2 + x)\) and look for terms that result in \(x^8\).

From \((2 + x)\), we have two terms: \(2\) and \(x\).

Multiplying \(2\) with \(240x^8\) from \((2x – x^2)^6\) gives \(480x^8\).

Multiplying \(x\) with \(-192x^7\) from \((2x – x^2)^6\) gives \(-192x^8\).

Thus, the total contribution to the coefficient of \(x^8\) is:

\( 480x^8 – 192x^8 = 288x^8 \)

Therefore, the coefficient of \(x^8\) in the expansion of \((2 + x)(2x – x^2)^6\) is \(288\).

**Question**

The first term of an arithmetic progression is 61 and the second term is 57.The sum of the first n terms is n.Find the value of the positive integers n.

**▶️Answer/Explanation**

The sum of the first \( n \) terms of an arithmetic progression (AP):

\( S_n = \frac{n}{2} \times (2a + (n – 1)d) \)

where \( S_n \) is the sum of the first \( n \) terms, \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms.

– The first term \( a = 61 \)

– The second term is 57, which allows us to find the common difference \( d \). Since the second term is \( a + d \), we have \( 57 = 61 + d \).

\( S_n = \frac{n}{2} \times (2a + (n – 1)d) \)

\( n = \frac{n}{2} \times (2 \times 61 + (n – 1) \times (-4)) \)

\( n \): \( 0 \) and \( 31 \).

Since we are looking for a positive integer value of \( n \), the solution is \( n = 31 \).

**Question**

The first term of a progression is 4x and the second term is \(x^{2}\)**(i) For the case where the progression is arithmetic with a common difference of 12, find the possible values of x and the corresponding values of the third term.**

(ii) For the case where the progression is geometric with a sum to infinity of 8, find the third term.

**Answer/Explanation**

(i)\(x^{2}-4x=12\)

x=-2 or 6

Third term\(=(-2)^{2}+12=16\) or \(6^{2}+12=48\)

(ii)\(r^{2}=\frac{x^{2}}{4x}=\frac{x}{4}\)

\(\frac{4x}{1-\frac{x}{4}}=8\)

\(x=\frac{4}{3}\) or \(r=\frac{1}{3}\)

**ALT**

\(\frac{4x}{1-r}=8\rightarrow r=1-\frac{1}{2}x\) or \(\frac{4x}{1-r}=8\rightarrow x=2(1-r)\)

\(x^{2}=4x(1-\frac{1}{2}x)\) \(r=\frac{2(1-r)}{4}\)

\(x=\frac{4}{3}\) \(r=\frac{1}{3}\)

**Question**

(a) In an arithmetic progression the sum of the first ten terms is 400 and the sum of the next ten terms is 1000. Find the common difference and the first term. **(b) A geometric progression has first term a, common ratio r and sum to infinity 6. A second geometric progression has first term 2a, common ratio \(r^{2}\) and sum to infinity 7. Find the values of a and r.**

**Answer/Explanation**

(a)\(\frac{10}{2}(2a+9d)=400\)

\(\frac{20}{2}(2a+19d)=1400\)

\(\frac{10}{2}\left [ 2(a+10d)+9d \right ]=1000\)

d=6 a=13

(b)\(\frac{a}{1-r}=6 \) \(\frac{2a}{1-r^{2}}=7\)

\(\frac{12(1-r)}{1-r^{2}}=7\) or \(\frac{1-r^{2}}{1-r}=\frac{12}{7}\)

\(r=\frac{5}{7}\) or 0.714

\(a=\frac{12}{7}\) or 1.714

**Question**

(i) A geometric progression has first term a (a ≠ 0), common ratio r and sum to infinity S. A second geometric progression has first term a, common ratio 2r and sum to infinity 3S. Find the value of r.**(ii) An arithmetic progression has first term 7. The nth term is 84 and the (3n)th term is 245. Find the value of n.**

**Answer/Explanation**

(i)\(S=\frac{a}{1-r}\) ,\(3S=\frac{a}{1-2r}\)

1-r=3-6r

\(r=\frac{2}{5}\)

(ii)\(7+(n-1)d=84\) and/or \(7+(3n-1)d=245\)

\(\left [ (n-1)d=77,(3n-1)d=238,2nd=161 \right ]\)

\(\frac{n-1}{3n-1}=\frac{77}{238}\) (must be from the correct \(u_{n}\) formula)

n=23 \((d=\frac{77}{22}=3.5)\)

**Question.**

(a) An arithmetic progression has a first term of 32, a 5th term of 22 and a last term of −28. Find the

sum of all the terms in the progression.

(b) Each year a school allocates a sum of money for the library. The amount allocated each year

increases by 2.5% of the amount allocated the previous year. In 2005 the school allocated \($2000\).

Find the total amount allocated in the years 2005 to 2014 inclusive.

**Answer/Explanation**

(a) \(a=32,a+4d=22\rightarrow d=-2.5\)

\(a+(n-1)d=-28\rightarrow n=25\)

\(S_{25}=\frac{25}{2}(64-2.5\times 24)=50\)

(b)a=2000,r=1.025

\(S_{10}=2000\left ( \frac{1.025^{10}-1}{1.025-1} \right )=22400\) or a value which rounds to this

**Question.**

(a) The first term of a geometric progression in which all the terms are positive is 50. The third term is 32. Find the sum to infinity of the progression.

(b) The first three terms of an arithmetic progression are 2 sin x, 3 cos x and (sin x + 2 cos x) respectively, where x is an acute angle.

(i) Show that \(tan x = \frac {4}{3}\)

(ii) Find the sum of the first twenty terms of the progression.

**Answer/Explanation**

a=50,\(ar^{2}=32\)

\(\rightarrow r=\frac{4}{5}\) (allow \(\frac{4}{5}\) for M mark)

\(\rightarrow S_{\infty }=250\)

(b) (i) \(2\sin x,3\cos x,\left ( \sin x+2\cos x \right )\)

3c-2s=(s+2c)-3c

(or uses a,a+d,a+2d)

\(\rightarrow 4c=3s\rightarrow t=\frac{4}{3}\)

SC uses \(\rightarrow 4c=3s\rightarrow t=\frac{4}{3}\)

\(u_{1}=\frac{8}{5},u_{2}=\frac{9}{5},u_{3}=\frac{10}{5}\)

(ii)\(\rightarrow c=\frac{3}{5},s=\frac{4}{5}\) or calculator \( x=53.1^{\circ}\)

→a=1.6,d=0.2

\(S_{\infty }=70\)

**Question.**

The function f is defined for x≥ 0 by \(f(x) =(4x + 1)^\frac{3}{2}\).

(i) Find f′(x) and f′′(x).

The first, second and third terms of a geometric progression are respectively f(2), f′(2) and kf′′(2).**(ii) Find the value of the constant k.**

**Answer/Explanation**

(i) \(f{}’\left ( x \right )=\left [ \frac{3}{2}\left ( 4x+1 \right )^{\frac{1}{2}} \right ]\left [ 4 \right ]\)

\(f{}”\left ( x \right )=6\times \frac{1}{2}\times \left ( 4x+1 \right )^{-\frac{1}{2}}\times 4\)

(ii) f(2),\(f{}’\left ( 2 \right )\),\(kf{}”\left ( 2 \right )=27\),4k OR 12

\(\frac{27}{18}=\frac{18}{4k}\) oe OR \(kf{}”\left ( 2 \right )=12\Rightarrow k=3\)

**Question.**

The 12th term of an arithmetic progression is 17 and the sum of the first 31 terms is 1023. Find the 31st term.

**Answer/Explanation**

Ans: \(a+11d=17\)

\(\frac{31}{2}(2a+30d)=1023\)

on solving both ,

\(d=4\) , \(a=-27\)

31st term \(=93\)

**Question **

**(a) The third and fourth terms of a geometric progression are 48 and 32 respectively. Find the sum****to infinity of the progression.**

(b) Two schemes are proposed for increasing the amount of household waste that is recycled each

week.

Scheme A is to increase the amount of waste recycled each month by 0.16 tonnes.

Scheme B is to increase the amount of waste recycled each month by 6% of the amount recycled in the previous month.

The proposal is to operate the scheme for a period of 24 months. The amount recycled in the

first month is 2.5 tonnes.For each scheme, find the total amount of waste that would be recycled over the 24- month period.

**Answer/Explanation**

**Queastion**

**(i) The first and second terms of a geometric progression are p and 2p respectively, where p is a****positive constant. The sum of the first n terms is greater than 1000p. Show that 2n > 1001.**

(ii) In another case, p and 2p are the first and second terms respectively of an arithmetic progression.

The nth term is 336 and the sum of the first n terms is 7224. Write down two equations in n and

p and hence find the values of n and p.

**Answer/Explanation**

**(i) **

\(S_{n}=\frac{p(2^{n}-1)}{2-1} \)

\(p(2^{n}-1)> 1000p\rightarrow 2^{n}> 1001\)

**(ii) **

\( p + ( n−1)=336\)

\(\frac{n}{2}\left [ 2p=(n-1) p\right ]=7224\)

Eliminate n or p to an equation in one variable \(n = 42\) ,

\(p = 8\)

### Question

A geometric progression has first term *a*, common ratio *r* and sum to infinity S. A second geometric

progression has first term *a*, common ratio R and sum to infinity 2S.

(a) Show that r = 2R − 1. [3]

It is now given that the 3rd term of the first progression is equal to the 2nd term of the second

progression.

(b) Express *S* in terms of *a*. [4]

**Answer/Explanation**

Ans

8 (a) \(S=\frac{1}{1-r}, \ 2S=\frac{a}{1-R}\)

\(\frac{2a}{1-r}=\frac{a}{1-R}\)

\(2-2R=1-r\rightarrow r=2R-1\)

8 (b) \(ar^{2}=aR\rightarrow (a)(2R-1)^{2}=R(a)\)

\(4R^{2}-5R+1(=0\rightarrow )(4R-1)(R-1)(=0)\)

\(R=\frac{1}{4}\)

\(S=\frac{2a}{3}\)

**Alternative method for question 8(b)**

\(ar^{2}=aR\rightarrow (a)r^{2}=\frac{1}{2}(r+1)(a)\)

\(2r^{2}-r-1(=0)\rightarrow (2r+1)(r-1)(=0)\)

\(r=-\frac{1}{2}\)

\(S=\frac{2a}{3}\)

**Question**

**Question**

A woman’s basic salary for her first year with a particular company is $30000 and at the end of the year she also gets a bonus of $600.

(a) For her first year, express her bonus as a percentage of her basic salary.

At the end of each complete year, the woman’s basic salary will increase by 3% and her bonus will increase by $100.

(b) Express the bonus she will be paid at the end of her 24th year as a percentage of the basic salary paid during that year.

**Answer/Explanation**

**Ans:**

(a) 2%

(b) Bonus \(=600+23 \times 100 = 2900\)

Salary \(=30000 \times 1.03^{23}\)

= 59207.60

\(\frac{their2900}{their59200}\)

4.9(0)%

### Question

The sum of the first 20 terms of an arithmetic progression is 405 and the sum of the first 40 terms

is 1410.

Find the 60th term of the progression. [5]

**Answer/Explanation**

Ans

2 10(2a + 19d) = 405

20(2a + 39d) = 1410

Solving simultaneously two equations obtained from using the correct sum

formulae [a = 6, d = 1.5]

Using the correct formula for 60th term with their a and d

60th term = 94.5

*Question*

The first term of a geometric progression and the first term of an arithmetic progression are both equal to a.

The third term of the geometric progression is equal to the second term of the arithmetic progression.

The fifth term of the geometric progression is equal to the sixth term of the arithmetic progression.

Given that the terms are all positive and not all equal, find the sum of the first twenty terms of the arithmetic progression in terms of a.

**Answer/Explanation**

Ans:

\(ar^{2} = a + d\)

\(ar^{4} = a + 5d\)

\(a^{2} r^{4} = a (a+5d) leading to a^{2} + 5ad = (a+d)^{2}\)

\([3ad -d^{2} = 0 leading] d = 3a OR [r= 2 leading to] d = 3a\)

\(S_{20} = \frac{20}{2} [2a + 19 \times 3a]\)

590a

*Question*

*Question*

The first term of a progression is \(cos\theta\), where \(0<\theta<\frac{1}{2}\pi\).

(a) For the case where the progression is geometric, the sum to infinity is \(\frac{1}{cos\theta}\).

(i) Show that the second term is \(cos\theta sin^2 \theta\).

(ii) Find the sum of the first 12 terms when \(\theta=\frac{1}{3}\pi\), giving your answer correct to 4 significant figures.

(b) For the case where the progression is arithmetic, the first two terms are again \(cos \theta \) and \(cos \theta sin^2 \theta\) respectively.

Find the 85th term when \(\theta =\frac{1}{3}\pi\).

**Answer/Explanation**

**Ans:**

(a) (i) \(\frac{cos \theta}{1-r}=\frac{1}{cos \theta}\)

\(1-r=cos^2 \theta\) leading to \(r=1-cos^2 \theta\)

\(r=sin^2 \theta\) leading to 2nd term \(=cos \theta sin^2 \theta\)

(a)(ii) \(S_{12}=\frac{cos(\frac{\pi}{3})[1-(sin^2(\frac{\pi}{3}))^{12}]}{1-sin^2(\frac{\pi}{3})}\)=\(0.5[1-(0.75)^{12}]}{1-0.75}\)

1.937

(b) \([d=]cos\theta sin^2 \theta – cos \theta\)

\(-\frac{1}{8}\)

[85th term=] \(\frac{1}{2} + 84 \times – \frac{1}{8}\)

-10

### Question

The equation of a curve is \(y=2\sqrt{3x+4}-x\)

(a) Find the equation of the normal to the curve at the point (4, 4), giving your answer in the form

y = mx + c. [5]

(b) Find the coordinates of the stationary point. [3]

(c) Determine the nature of the stationary point. [2]

(d) Find the exact area of the region bounded by the curve, the x-axis and the lines x = 0 and x = 4. [4]

**Answer/Explanation**

Ans

11 (a) \(\frac{dx}{dy}=3(3x+4)^{-0.5}-1\)

Gradient of tangent \(=-\frac{1}{4}\) and Gradient of normal = 4

Equation of line is (y – 4) = 4(x – 4) or evaluate c

So y = 4x – 12

11 (b) 3(3x+4)^{-0.5} -1 = 0

Solving as far as x=

\(x=\frac{5}{3}, y=2\left ( 3\times \frac{5}{3}+4 \right )^{0.5}-\frac{5}{3}=\frac{13}{3}\)

At \(x=\frac{5}{3} \frac{d^{2}y}{dx^{2}}\) is negative so the point is a maximum

11 (d) \(Area-[f2(3x+4)^{0.5}-xdx=]\frac{4}{9}(3x+4)^{1.5}-\frac{1}{2}x^{2}\)

\(\left ( \frac{4}{9}(16)^{1.5}-\frac{1}{2}(4)^{2} \right )-\frac{4}{9}(4)^{1.5}=\frac{256}{9}-8-\frac{32}{9}\)

\(16\frac{8}{9}\)

### Question.

Each year the selling price of a diamond necklace increases by 5% of the price the year before. The selling price of the necklace in the year 2000 was \($36 000\).**(a)** Write down an expression for the selling price of the necklace n years later and hence find the

selling price in 2008.

**(b)** The company that makes the necklace only sells one each year. Find the total amount of money obtained in the ten-year period starting in the year 2000.

**Answer/Explanation**

**(a)** \($36 000 \times (1.05)^{n}\)

(**B1** for *r* = 1.05. **M1** method for *r*th term)

\( $53 200\) after 8 years.

** (b)** \(S_{10}=36000\frac{(1.05^{10}-1)}{(1.05-1)}\)

\( $453 000 \)

*Question*.

The first term of an arithmetic progression is a and the common difference is −4. The first term of a geometric progression is 5a and the common ratio is \(-\frac{1}{4}\). The sum to infinity of the geometric progression is equal to the sum of the first eight terms of the arithmetic progression.

**(a) **Find the value of a.

The *k*th term of the arithmetic progression is zero.

**(b)** Find the value of k

**Answer/Explanation**

**(a) ** \(\frac{5a}{1-\left ( \pm \frac{1}{4} \right )}\)

\(\frac{8}{2}\left [ 2a + 7(-4) \right ]\)

4a = 8a – 112 leading to a = [28]

a = 28

**(b) **their k

28 + ( k – 1) (-4_) = 0

[k = ] 8