**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) Given:

\( a r^2 = -108 \)

\( a r^5 = 32 \)

Dividing both we get,

\( \frac{a r^5}{a r^2} = \frac{32}{-108} \)

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

\( r^3 = -\frac{8}{27} \)

\( r = \left( -\frac{8}{27} \right)^{\frac{1}{3}} \)

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

(ii) Using \( a r^2 = -108 \) and \( r = -\frac{2}{3} \):

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

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

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

\( a = -243 \)

(iii) The sum to infinity is given by:

\( S_{\infty} = \frac{a}{1 – r} \]

Since \( |r| < 1 \) in this case, the sum to infinity exists. Using \( a = -243 \) and \( r = -\frac{2}{3} \):

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

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

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

\( S_{\infty} = -\frac{729}{5} \) or approximately \( -145.8 \)

**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)The common difference \( d \) is the difference between the second and the first term:

\( d = \cos^2 x – 1 \)

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

\( S_n = \frac{n}{2} \left(2a_1 + (n – 1)d\right) \)

where \( a_1 \) is the first term, and \( d \) is the common difference.

For the first ten terms (\( n = 10 \)):

\( S_{10} = \frac{10}{2} \left(2 \cdot 1 + (10 – 1)(\cos^2 x – 1)\right) \)

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

\( S_{10} = 5 \left(2 + 9\cos^2 x – 9\right) \)

\( S_{10} = 5(9\cos^2 x – 7) \)

We know that \( \cos^2 x + \sin^2 x = 1 \), so \( \cos^2 x = 1 – \sin^2 x \).

Substituting this into the expression:

\( S_{10} = 5(9(1 – \sin^2 x) – 7) \)

\( S_{10} = 5(9 – 9\sin^2 x – 7) \)

\( S_{10} = 5(2 – 9\sin^2 x) \)

\( S_{10} = 10 – 45\sin^2 x \)

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

(b) (i) The common ratio \( r \) is given by:

\( r = \frac{a_2}{a_1} = \frac{1}{3} \tan^2 \Theta \)

For the progression to be convergent, \( |r| < 1 \):

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

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

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

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

Since \( 0 < \Theta < \frac{\pi}{2} \), \( \Theta \) is in the first quadrant where the tangent function is positive. Thus:

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

\( \Theta < \tan^{-1}(\sqrt{3}) \)

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

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

(ii) \( S_{\infty} = \frac{a_1}{1 – r} \)

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

Now, \( \tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3} \), so:

\( S_{\infty} = \frac{1}{1 – \frac{1}{3} \left(\frac{\sqrt{3}}{3}\right)^2} \)

\( S_{\infty} = \frac{1}{1 – \frac{1}{3} \cdot \frac{1}{3}} \)

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

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

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

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

**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**

(a)For the first GP, the first term is \( 3a \) and the common ratio is \( r \). The sum to infinity is:

\( S_1 = \frac{3a}{1 – r} \)

For the second GP, the first term is \( a \) and the common ratio is \( -2r \). The sum to infinity is:

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

Since the sums to infinity are equal, we set \( S_1 = S_2 \):

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

\( 3(1 + 2r) = 1 – r \)

\( 3 + 6r = 1 – r \)

\( 6r + r = 1 – 3 \)

\( 7r = -2 \)

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

So, the value of \( r \) is \( -\frac{2}{7} \).

(b) \( S_n = \frac{n}{2} (2a_1 + (n – 1)d) \),

where \( a_1 \) is the first term and \( d \) is the common difference.

For the first AP, the first term is 15 and the common difference is \( 19 – 15 = 4 \). The sum of the first \( n \) terms is:

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

For the second AP, the first term is 420 and the common difference is \( 415 – 420 = -5 \). The sum of the first \( n \) terms is:

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

Setting these sums equal:

\( \frac{n}{2} (30 + 4n – 4) = \frac{n}{2} (840 – 5n + 5) \)

\( n(30 + 4n – 4) = n(840 – 5n + 5) \)

\( 30n + 4n^2 – 4n = 840n – 5n^2 + 5n \)

\( 4n^2 – 4n = -5n^2 + 845n \)

\( 9n^2 – 849n = 0 \)

\( n(9n – 849) = 0 \)

This gives us two solutions, \( n = 0 \) and \( n = \frac{849}{9} \).

\( n = \frac{849}{9} \)

\( n = 94.33 \)

**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**

**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)We have two geometric progressions (GPs):

GP1 with first term \( a \) (where \( a \neq 0 \)), common ratio \( r \), and sum to infinity \( S \).

GP2 with first term \( a \), common ratio \( 2r \), and sum to infinity \( 3S \).

The formula for the sum to infinity of a GP is given by \( S = \frac{a}{1 – r} \), where \( a \) is the first term and \( r \) is the common ratio.

For GP1,

\( S = \frac{a}{1 – r} \)

For GP2,

\( 3S = \frac{a}{1 – 2r} \)

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

\( 1 – r = 3(1 – 2r) \)

\( 1 – r = 3 – 6r \)

\( 5r = 2 \)

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

Therefore, the common ratio \( r \) of the first GP is \( \frac{2}{5} \).

(ii) We have an arithmetic progression (AP) with:

First term \( a = 7 \).

\( n \)th term is 84.

\( 3n \)th term is 245.

The formula for the \( n \)th term of an AP is \( a_n = a + (n – 1)d \), where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the term number.

For the \( n \)th term, we have:

\( 84 = 7 + (n – 1)d \)

For the \( 3n \)th term, we have:

\( 245 = 7 + (3n – 1)d \)

\( 84 = 7 + (n – 1)d \)

\( 77 = (n – 1)d \)

\( d = \frac{77}{n – 1} \)—(1)

\( 245 = 7 + (3n – 1)d \)

\( 238 = (3n – 1)d \)

Substituting \( d \) from equation (1):

\( 238 = (3n – 1) \frac{77}{n – 1} \)

\( 238(n – 1) = 77(3n – 1) \)

\( 238n – 238 = 231n – 77 \)

\( 7n = 161 \)

\( n = \frac{161}{7} \)

\( n = 23 \)

Therefore, the value of \( n \) is 23.

### 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.

**▶️Answer/Explanation**

The fifth term of the GP is \( 8k \).

The sixth term of the GP is \( -12 \).

The seventh term of the GP is \( 2k \).

\( 8k \cdot r = -12 \) (sixth term)

\( -12 \cdot r = 2k \) (seventh term)

\( r = \frac{-12}{8k} \)

\( -12 \cdot \frac{-12}{8k} = 2k \)

\( \frac{144}{8k} = 2k \)

\( 18 = 2k^2 \)

\( k^2 = \frac{18}{2} \)

\( k^2 = 9 \)

\( k = -3 \)

\( a \cdot r^5 = -12 \)

\( 8k = 8(-3) = -24 \):

\( a \cdot r^4 = -24 \)

\( r = \frac{-12}{-24} = \frac{1}{2} \)

\( a \cdot (\frac{1}{2})^4 = -24 \)

\( a \cdot \frac{1}{16} = -24 \)

\( a = -24 \cdot 16 \)

\( a = -384 \)

\( S_{\infty} = \frac{a}{1 – r} \)

\( S_{\infty} = \frac{-384}{1 – \frac{1}{2}} \)

\( S_{\infty} = \frac{-384}{\frac{1}{2}} \)

\( S_{\infty} = -768 \)

So, the sum to infinity of the progression is \( -768 \).

**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)The first term (a) is 50.

The third term is 32.

The \( n \)th term of a GP is given by \( a \cdot r^{(n-1)} \), where \( r \) is the common ratio. For the third term, we have:

\( 50 \cdot r^2 = 32 \)

\( r^2 = \frac{32}{50} \)

\( r^2 = \frac{16}{25} \)

\( r = \sqrt{\frac{16}{25}} \)

\( r = \frac{4}{5} \)

The formula for the sum to infinity of a GP is:

\( S_{\infty} = \frac{a}{1 – r} \)

\( S_{\infty} = \frac{50}{1 – \frac{4}{5}} \)

\( S_{\infty} = \frac{50}{\frac{1}{5}} \)

\( S_{\infty} = 50 \times 5 \)

\( S_{\infty} = 250 \)

So, the sum to infinity of the progression is 250.

(b)Given an arithmetic progression (AP) with the first three terms as \( 2 \sin x \), \( 3 \cos x \), and \( \sin x + 2 \cos x \) respectively, where \( x \) is an acute angle.

(i)In an AP, the common difference between consecutive terms is constant.

\( 3 \cos x – 2 \sin x = d \)

\( (\sin x + 2 \cos x) – 3 \cos x = d \)

\( \sin x + 2 \cos x – 3 \cos x = 3 \cos x – 2 \sin x \)

\( \sin x – \cos x = 3 \cos x – 2 \sin x \)

\( 3 \sin x = 4 \cos x \)

\( \frac{\sin x}{\cos x} = \frac{4}{3} \)

\( \tan x = \frac{4}{3} \)

(ii) The formula for the sum of the first \( n \) terms of an AP is:

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

Here \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms. Since \( n = 20 \), we have:

\( S_{20} = \frac{20}{2}(2 \times 2 \sin x + (20-1)d) \)

\( d = 3 \cos x – 2 \sin x \)

\( S_{20} = 10(4 \sin x + 19(3 \cos x – 2 \sin x)) \)

\( S_{20} = 10(4 \sin x + 57 \cos x – 38 \sin x) \)

\( S_{20} = 10(57 \cos x – 34 \sin x) \)

Since \( \tan x = \frac{4}{3} \), we can express \( \sin x \) and \( \cos x \) in terms of \( \tan x \):

\( \sin x = \frac{\tan x}{\sqrt{1 + \tan^2 x}} \)

\(\cos x = \frac{1}{\sqrt{1 + \tan^2 x}} \)

\( \sin x = \frac{4/3}{\sqrt{1 + (4/3)^2}} \)

\( \cos x = \frac{1}{\sqrt{1 + (4/3)^2}} \)

\( \sin x = \frac{\tan x}{\sqrt{1 + \tan^2 x}} = \frac{\frac{4}{3}}{\sqrt{1 + (\frac{4}{3})^2}} \)

\( \cos x = \frac{1}{\sqrt{1 + \tan^2 x}} = \frac{1}{\sqrt{1 + (\frac{4}{3})^2}} \)

\( \sin x = \frac{\frac{4}{3}}{\sqrt{1 + \frac{16}{9}}} = \frac{\frac{4}{3}}{\sqrt{\frac{25}{9}}} = \frac{4}{3} \cdot \frac{3}{5} = \frac{4}{5} \)

\( \cos x = \frac{1}{\sqrt{\frac{25}{9}}} = \frac{3}{5} \)

\( S_{20} = 10(57 \cdot \frac{3}{5} – 34 \cdot \frac{4}{5}) \)

\( S_{20} = 10(57 \cdot \frac{3}{5} – 34 \cdot \frac{4}{5}) \)

\( S_{20} = 10 \left( \frac{171}{5} – \frac{136}{5} \right) \)

\( S_{20} = 10 \cdot \frac{35}{5} \)

\( S_{20} = 10 \cdot 7 \)

\( S_{20} = 70 \)

Therefore, the sum of the first twenty terms of the progression is \(70\).