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)\(ar^{2}=-108,ar^{5}=32\)

\(r^{3}=-\frac{32}{108}=\left (- \frac{8}{27} \right )\)

\(r=\left ( -\frac{2}{3} \right )\) or -0.666 or -0.667

(ii)a=-243

(iii)\(S_{\infty }=\frac{-243}{1+\frac{2}{3}}=-\frac{729}{5}or -145.8\)

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) \(S_{10}=\frac{10}{2\left [ 2+9\left ( \cos ^{2}x-1 \right ) \right ]}\)

\(S_{10}=5\left [ 2-9\sin ^{2}x \right ]\)

\(S_{10}=10-45\sin ^{2}x\)

(b)(i)\(0< \frac{1}{3}\tan ^{2}\Theta < 1\)

\(0< \Theta < \frac{\pi }{3}\)

(ii)\(S_{\infty }=\frac{1}{1-\frac{1}{3}\tan ^{2}\frac{\pi }{6}}\)

\(S_{\infty }=\frac{9}{8}\) or 1.125

Question

The sum of the 1st and 2nd terms of a geometric progression is 50 and the sum of 2nd and 3rd terms is 30.Find the sum to infinity.

Answer/Explanation

\(a(1+r)=50)\) or  \(\frac{a\left ( 1-r^{2} \right )}{1-r}=50\)

\(ar\left ( 1+r \right )=30\) or \(\frac{a\left ( 1-r^{3} \right )}{\left ( 1-r \right )}=30+a\)

Eliminating a or r

\(r=\frac{3}{5}\)

\(a=\frac{125}{4}\)

\(S=\frac{625}{8}\)

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

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

\(3+6r=1-r\)

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

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

n=91

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

(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

  The fifth, sixth and seventh terms of a geometric progression are 8k, −12 and 2k respectively.
    Given that k is negative, find the sum to infinity of the progression.                                                     [4]

Answer/Explanation

Ans

5 (−12)2 = 8k × 2k 
    k =−3 
    Using correct formula for S∞ [r = 0.5,a =−384]
    S∞ =−768

Alternative method for Question 5

    \(r^{2}=\frac{2k}{8k}\)

      r = ±[ ]0.5 A1
      Using correct formula for S∞ [r = 0.5,a =−384]
      S∞ =−768 

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

\(S_{\infty }=70\)