If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it. - Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

Solved Examples on Geometric Sequences

Samuel Dominic Chukwuemeka (SamDom For Peace) Prerequisites:
(1.) Linear Systems
(2.) Exponents
(3.) Quadratic Equations

Formulas: Formulas
Calculators: Calculators

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
Use at least two methods whenever applicable.
Show all work

(1.) WAEC:BECE A sequence is given as $3, 6, 12, 24,...$
Find the:
(i) next two terms of the sequence;
(ii) sum of the two terms in (i)


$ 3, 6, 12, 24,... \\[3ex] 3 * 2 = 6 \\[3ex] 6 * 2 = 12 \\[3ex] 12 * 2 = 24 \\[3ex] (i) \\[3ex] 24 * 2 = 48 \\[3ex] 48 * 2 = 96 \\[3ex] The\:\:next\:\:two\:\:terms = 48, 96 \\[3ex] OR \\[3ex] The\:\:next\:\:two\:\:terms\:\:are\:\:the\:\:5th\:\:and\:\:6th\:\:terms \\[3ex] a = 3 \\[3ex] r = \dfrac{6}{3} = 2 \\[5ex] GS_n = ar^{n - 1} \\[3ex] GS_5 = 3 * 2^{5 - 1} \\[3ex] = 3 * 2^4 \\[3ex] = 3 * 16 \\[3ex] = 48 \\[3ex] GS_6 = 3 * 2^{6 - 1} \\[3ex] = 3 * 2^5 \\[3ex] = 3 * 32 \\[3ex] = 96 \\[3ex] (ii) \\[3ex] Sum\:\:of\:\:48\:\:and\:\:96 \\[3ex] = 48 + 96 \\[3ex] = 144 \\[3ex] OR \\[3ex] The\:\:next\:\:two\:\:terms\:\:are\:\:the\:\:5th\:\:and\:\:6th\:\:terms \\[3ex] They\:\:are\:\:the\:\:last\:\:two\:\:terms \\[3ex] Sum\:\:of\:\:the\:\:last\:\:two\:\:terms = Sum\:\:of\:\:the\:\:first\:\:six\:\:terms - Sum\:\:of\:\:the\:\:first\:\:four\:\:terms \\[3ex] Sum\:\:of\:\:the\:\:last\:\:two\:\:terms = SGS_6 - SGS_4 \\[3ex] r = 2 \implies 2 \gt 1 \\[3ex] SGS_n = \dfrac{a(r^{n} - 1)}{r - 1} \:\:for\:\: r \gt 1 \\[5ex] SGS_6 = \dfrac{3 * (2^{6} - 1)}{2 - 1} \\[5ex] = \dfrac{3 * (64 - 1)}{1} \\[5ex] = \dfrac{3 * 63}{1} \\[5ex] = \dfrac{189}{1} \\[5ex] = 189 \\[3ex] SGS_4 = \dfrac{3 * (2^{4} - 1)}{2 - 1} \\[5ex] = \dfrac{3 * (16 - 1)}{1} \\[5ex] = \dfrac{3 * 15}{1} \\[5ex] = \dfrac{45}{1} \\[5ex] = 45 \\[3ex] \therefore Sum\:\:of\:\:the\:\:last\:\:two\:\:terms \\[3ex] = 189 - 45 \\[3ex] = 144 $
(2.) ACT The first term is $1$ in the geometric sequence $1, -3, 9, -27,...$.
What is the SEVENTH term of the geometric sequence?

$ A.\:\: -243 \\[3ex] B.\:\: -30 \\[3ex] C.\:\: 81 \\[3ex] D.\:\: 189 \\[3ex] E.\:\: 729 \\[3ex] $

$ GS_n = ar^{n - 1} \\[3ex] a = 1 \\[3ex] r = \dfrac{-3}{1} = -3 \\[5ex] GS_7 = ar^{7 - 1} \\[3ex] GS_7 = ar^6 \\[3ex] GS_7 = 1 * (-3)^6 \\[3ex] GS_7 = 1 * 729 \\[3ex] GS_7 = 729 $
(3.) JAMB The sum of infinity of a geometric progression is $-\dfrac{1}{10}$ and the first is $-\dfrac{1}{8}$.
Find the common ratio of the progression.

$ A.\:\: -\dfrac{1}{5} \\[5ex] B.\:\: -\dfrac{1}{4} \\[5ex] C.\:\: -\dfrac{1}{3} \\[5ex] D.\:\: -\dfrac{1}{2} \\[5ex] $

$ S_{\infty} = -\dfrac{1}{10} \\[5ex] a = -\dfrac{1}{8} \\[5ex] r = ? \\[3ex] r = \dfrac{S_{\infty} - a}{S_{\infty}} \\[5ex] r = \dfrac{-\dfrac{1}{10} - \left(-\dfrac{1}{8}\right)}{-\dfrac{1}{10}} \\[7ex] -\dfrac{1}{10} - \left(-\dfrac{1}{8}\right) \\[5ex] = -\dfrac{1}{10} + \dfrac{1}{8} \\[5ex] = \dfrac{-8 + 10}{80} \\[5ex] = \dfrac{2}{80} \\[5ex] = \dfrac{1}{40} \\[5ex] \dfrac{1}{40} \div -\dfrac{1}{10} \\[5ex] = \dfrac{1}{40} * -\dfrac{10}{1} \\[5ex] = -\dfrac{1}{4} \\[5ex] r = -\dfrac{1}{4} $
(4.) ACT What is the fourth term in the geometric sequence $36, -12, 4, ...?$

$ F.\:\: -\dfrac{4}{3} \\[5ex] G.\:\: -1 \\[3ex] H.\:\: -\dfrac{3}{4} \\[5ex] J.\:\: \dfrac{3}{4} \\[5ex] K.\:\: \dfrac{4}{3} \\[5ex] $

$ GS_n = ar^{n - 1} \\[3ex] a = 36 \\[3ex] r = \dfrac{-12}{36} = -\dfrac{1}{3} \\[5ex] GS_4 = ar^3 \\[3ex] GS_4 = 36 * \left(-\dfrac{1}{3}\right)^3 \\[5ex] GS_4 = 36 * -\dfrac{1}{3} * -\dfrac{1}{3} * -\dfrac{1}{3} \\[5ex] GS_4 = 4 * -1 * -1 * -\dfrac{1}{3} \\[5ex] GS_4 = -\dfrac{4}{3} $
(5.) ACT The $1st$ and $2nd$ terms of a certain geometric sequence are $10$ and $-5$ respectively.
What is the $5th$ term of the geometric sequence?

$ F.\:\: -\dfrac{5}{8} \\[5ex] G.\:\: -\dfrac{5}{16} \\[5ex] H.\:\: \dfrac{5}{8} \\[5ex] J.\:\: \dfrac{5}{16} \\[5ex] K.\:\: \dfrac{5}{32} \\[5ex] $

$ GS_1 = a = 10 \\[3ex] GS_2 = -5 \\[3ex] r = -\dfrac{5}{10} \\[5ex] r = -\dfrac{1}{2} \\[5ex] n = 5 \\[3ex] GS_n = ar^{n - 1} \\[3ex] GS_5 = 10\left(-\dfrac{1}{2}\right)^{5 - 1} \\[5ex] GS_5 = 10\left(-\dfrac{1}{2}\right)^{4} \\[5ex] GS_5 = 10 * \dfrac{(-1)^4}{2^4} \\[5ex] GS_5 = 10 * \dfrac{1}{16} \\[5ex] GS_5 = \dfrac{5}{8} $
(6.) ACT What is the sixth term of the geometric sequence whose second term is $-4$ and whose fifth term is $32$?

$ F.\:\: -128 \\[3ex] G.\:\: -64 \\[3ex] H.\:\: 44 \\[3ex] J.\:\: 128 \\[3ex] K.\:\: 256 \\[3ex] $

$ GS_n = ar^{n - 1} \\[3ex] GS_2 = ar^{2 - 1} = ar = -4...eqn.(1) \\[3ex] GS_5 = ar^{5 - 1} = ar^4 = 32...eqn.(2) \\[3ex] eqn.(2) \div eqn.(1) \:\: gives \\[3ex] \dfrac{ar^4}{ar} = \dfrac{32}{-4} \\[5ex] r^3 = -8 \\[3ex] r = \sqrt[3]{-8} \\[3ex] r = -2 \\[3ex] From\:\: eqn.(1) \\[3ex] a = -\dfrac{4}{r} \\[5ex] a = \dfrac{-4}{-2} \\[5ex] a = 2 \\[3ex] GS_6 = ar^{6 - 1} = ar^5 \\[3ex] GS_6 = 2(-2^{5}) \\[3ex] GS_6 = 2(-32) \\[3ex] GS_6 = -64 \\[3ex] $ The sixth term is $-64$
(7.) WASSCE The difference between the third and first terms of a Geometric Progression ($G.P$) is $42$.
If the fourth term is greater than the second term by $168$, find the:
(i) first term
(ii) fourth term
of the progression.


$ GS_3 - GS_1 = 42 \\[3ex] GS_4 - GS_2 = 168 \\[3ex] GS_n = ar^{n - 1} \\[3ex] GS_1 = a \\[3ex] GS_2 = ar^{2 - 1} = ar \\[3ex] GS_3 = ar^{3 - 1} = ar^2 \\[3ex] GS_4 = ar^{4 - 1} = ar^3 \\[3ex] \implies ar^2 - a = 42 ...eqn.(1) \\[3ex] \implies ar^3 - ar = 168 ...eqn.(2) \\[3ex] From\:\: eqn(1),\:\: a(r^2 - 1) = 42 ...eqn.(3) \\[3ex] From\:\: eqn(2),\:\: a(r^3 - r) = 168 ...eqn.(4) \\[3ex] Divide\:\: eqn.(4)\:\: by\:\: eqn.(3) \\[3ex] \dfrac{a(r^3 - r)}{a(r^2 - 1)} = \dfrac{168}{42} \\[5ex] \dfrac{r^3 - r}{r^2 - 1} = 4 \\[5ex] LCD = (r^2 - 1) \\[3ex] Multiply\:\: both\:\: sides\:\: by\:\: the\:\: LCD \\[3ex] r^3 - r = 4(r^2 - 1) \\[3ex] r^3 - r = 4r^2 - 4 \\[3ex] r^3 - r - 4r^2 + 4 = 0 \\[3ex] Factor\:\: by\:\: Grouping \\[3ex] r(r^2 - 1) - 4(r^2 - 1) = 0 \\[3ex] (r^2 - 1)(r - 4) = 0 \\[3ex] r^2 - 1 = r^2 - 1^2 = (r + 1)(r - 1)...Difference \:\:of\:\: Two\:\: Squares \\[3ex] (r + 1)(r - 1)(r - 4) = 0 \\[3ex] r + 1 = 0 \:\:OR\:\: r - 1 = 0 \:\:OR\:\: r - 4 = 0 \\[3ex] r = -1 \:\:OR\:\: r = 1 \:\:OR\:\: r = 4 \\[3ex] From\:\: eqn(3),\:\: a = \dfrac{42}{r^2 - 1} ...eqn.(5) \\[5ex] Based\:\: on\:\: the\:\: denominator; r \ne 1, r\ne -1 \\[3ex] This\:\: is\:\: because \\[3ex] 1^2 - 1 = 1 - 1 = 0 \\[3ex] (-1)^2 - 1 = 1 - 1 = 0 \\[3ex] So,\:\: r = 4 \\[3ex] a = \dfrac{42}{r^2 - 1} \implies a = \dfrac{42}{(4)^2 - 1} \\[5ex] a = \dfrac{42}{16 - 1} \\[5ex] a = \dfrac{42}{15} \\[5ex] (i)\:\: a = \dfrac{14}{5} \\[5ex] GS_4 = ar^3 \\[3ex] GS_4 = \dfrac{14}{5} * (4)^3 \\[5ex] GS_4 = \dfrac{14}{5} * 64 \\[5ex] (ii)\:\: GS_4 = \dfrac{896}{5} $
(8.) NSC Given the following geometric sequence: $30\:\:;\:\:10\:\:;\:\:\dfrac{10}{3}\:\:;...$

(12.1.1) Determine $n$ if the $n^{th}$ term of the sequence is equal to $\dfrac{10}{729}$.

(12.1.2) Calculate: $30 + 10 + \dfrac{10}{3} +...$


$ a = 30 \\[3ex] r = \dfrac{10}{30} = \dfrac{1}{3} \\[5ex] (12.1.1) \\[3ex] GS_n = a * r^{n - 1} \\[3ex] GS_n = \dfrac{10}{729} \\[5ex] \rightarrow \dfrac{10}{729} = 30 * \left(\dfrac{1}{3}\right)^{n - 1} \\[5ex] 30 *\left(\dfrac{1}{3}\right)^{n - 1} = \dfrac{10}{729} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\: \dfrac{1}{30} \\[5ex] \dfrac{1}{30} * 30 * \left(\dfrac{1}{3}\right)^{n - 1} = \dfrac{1}{30} * \dfrac{10}{729} \\[5ex] \left(\dfrac{1}{3}\right)^{n - 1} = \dfrac{1}{3} * \dfrac{1}{729} \\[5ex] \left(\dfrac{1}{3}\right)^{n - 1} = \dfrac{1}{3 * 3^6} \\[5ex] \left(\dfrac{1}{3}\right)^{n - 1} = \dfrac{1}{3^7} \\[5ex] \left(\dfrac{1}{3}\right)^{n - 1} = \dfrac{1^7}{3^7} \\[5ex] \left(\dfrac{1}{3}\right)^{n - 1} = \left(\dfrac{1}{3}\right)^{7} \\[5ex] Base\:\:is\:\:the\:\:same \\[3ex] Equate\:\:the\:\:exponents \\[3ex] n - 1 = 7 \\[3ex] n = 7 + 1 \\[3ex] n = 8 \\[3ex] OR \\[3ex] \left(\dfrac{1}{3}\right)^{n - 1} = \dfrac{1}{3^7} \\[5ex] \dfrac{1^{n - 1}}{3^{n - 1}} = \dfrac{1}{3^7} \\[5ex] 1^{anything} = 1 \\[3ex] 1^{n - 1} = 1 \\[3ex] \rightarrow \dfrac{1}{3^{n - 1}} = \dfrac{1}{3^7} \\[5ex] Cross\:\:Multiply \\[3ex] 3^{n - 1} * 1 = 1 * 3^7 \\[3ex] 3^{n - 1} = 3^7 \\[3ex] Base\:\:is\:\:the\:\:same \\[3ex] Equate\:\:the\:\:exponents \\[3ex] n - 1 = 7 \\[3ex] n = 7 + 1 \\[3ex] n = 8 \\[3ex] OR \\[3ex] \dfrac{1}{3^{n - 1}} = \dfrac{1}{3^7} \\[5ex] 3^{-(n - 1)} = 3^{-7} \\[3ex] Base\:\:is\:\:the\:\:same \\[3ex] Equate\:\:the\:\:exponents \\[3ex] -(n - 1) = -7 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:-1 \\[3ex] n - 1 = \dfrac{-7}{-1} \\[5ex] n - 1 = 7 \\[3ex] n = 7 + 1 \\[3ex] n = 8 \\[3ex] (12.1.2) \\[3ex] r \lt 1...series\:\:converges \\[3ex] S_{\infty} = \dfrac{a}{1 - r} \\[5ex] 1 - r = 1 - \dfrac{1}{3} = \dfrac{3}{3} - \dfrac{1}{3} = \dfrac{3 - 1}{3} = \dfrac{2}{3} \\[5ex] \rightarrow S_{\infty} = a \div (1 - r) \\[3ex] S_{\infty} = 30 \div \dfrac{2}{3} \\[5ex] S_{\infty} = 30 * \dfrac{3}{2} \\[5ex] S_{\infty} = 15(3) \\[3ex] S_{\infty} = 45 $
(9.) WASSCE The product of the second and third terms of an exponential sequence (GP) is $972$.
If the first term is $\dfrac{3}{4}$, find the common ratio.

$ A.\:\: 8 \\[3ex] B.\:\: 10 \\[3ex] C.\:\: 11 \\[3ex] D.\:\: 12 \\[3ex] $

$ GS_n = ar^{n - 1} \\[3ex] GS_2 = ar^{2 - 1} = ar^1 = ar \\[3ex] GS_3 = ar^{3 - 1} = ar^2 \\[3ex] ar * ar^2 = 972 \\[3ex] a^2 * r^3 = 972 \\[3ex] a^2 = \left(\dfrac{3}{4}\right)^2 = \dfrac{3^2}{4^2} = \dfrac{9}{16} \\[5ex] \rightarrow \dfrac{9}{16} * r^3 = 972 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\: \dfrac{16}{9} \\[5ex] \dfrac{16}{9} * \dfrac{9}{16} * r^3 = \dfrac{16}{9} * 972 \\[5ex] r^3 = 16 * 108 \\[3ex] r^3 = 1728 \\[3ex] r = \sqrt[3]{1728} \\[3ex] r = 12 $
(10.) WASSCE The $3^{rd}$, $4^{th}$ and $5^{th}$ terms of an exponential sequence are $x$, $y$, and $z$ respectively.
Which of the following statements is true?

$ A.\:\: 3x = 20yz \\[3ex] B.\:\: x^2 = yz \\[3ex] C.\:\: y^2 = xz \\[3ex] D.\:\: z^2 = xy \\[3ex] $

$ 3rd = GS_3 = ar^2 = x \\[3ex] 4th = GS_4 = ar^3 = y \\[3ex] 5th = GS_5 = ar^4 = z \\[3ex] Let\:\:us\:\:check\:\:each\:\:option \\[3ex] $ Check
$ \underline{LHS} \\[3ex] 3x \\[3ex] 3(ar^2) \\[3ex] 3ar^2 $ $ \underline{RHS} \\[3ex] 20yz \\[3ex] 20(ar^3)(ar^4) \\[3ex] 20a^2r^7 $
NO
$ x^2 \\[3ex] (ar^2)^2 \\[3ex] a^2r^4 $ $ yz \\[3ex] (ar^3)(ar^4) \\[3ex] a^2r^7 $
NO
$ y^2 \\[3ex] (ar^3)^2 \\[3ex] a^2r^6 \checkmark $ $ xz \\[3ex] (ar^2)(ar^4) \\[3ex] a^2r^6 \checkmark $
YES
$ z^2 \\[3ex] (ar^4)^2 \\[3ex] a^2r^8 $ $ xy \\[3ex] (ar^2)(ar^3) \\[3ex] a^2r^5 $
NO
(11.) WASSCE:FM Find the third term of the exponential sequence (G.P)

$(\sqrt{2} - 1),\:\:(3 - 2\sqrt{2}),\:\:...$


$ a = \sqrt{2} - 1 \\[3ex] r = \dfrac{3 - 2\sqrt{2}}{\sqrt{2} - 1} \\[5ex] GS_3 = ar^2 \\[3ex] = (\sqrt{2} - 1) * \left(\dfrac{3 - 2\sqrt{2}}{\sqrt{2} - 1}\right)^2 \\[5ex] = (\sqrt{2} - 1) * \left(\dfrac{3 - 2\sqrt{2}}{\sqrt{2} - 1}\right) * \left(\dfrac{3 - 2\sqrt{2}}{\sqrt{2} - 1}\right) \\[5ex] = (3 - 2\sqrt{2}) * \dfrac{3 - 2\sqrt{2}}{\sqrt{2} - 1} \\[5ex] = \dfrac{(3 - 2\sqrt{2})(3 - 2\sqrt{2})}{\sqrt{2} - 1} \\[5ex] = \dfrac{9 - 6\sqrt{2} - 6\sqrt{2} + (2\sqrt{2})^2}{\sqrt{2} - 1} \\[5ex] = \dfrac{9 - 12\sqrt{2} + 4(2)}{\sqrt{2} - 1} \\[5ex] = \dfrac{9 - 12\sqrt{2} + 8}{\sqrt{2} - 1} \\[5ex] = \dfrac{17 - 12\sqrt{2}}{\sqrt{2} - 1} \\[5ex] Rationalize \\[3ex] = \dfrac{17 - 12\sqrt{2}}{\sqrt{2} - 1} * \dfrac{\sqrt{2} + 1}{\sqrt{2} + 1} \\[5ex] = \dfrac{17\sqrt{2} + 17 - 12(2) - 12\sqrt{2}}{(\sqrt{2})^2 - 1^2} \\[5ex] = \dfrac{17\sqrt{2} + 17 - 24 - 12\sqrt{2}}{2 - 1} \\[5ex] = \dfrac{5\sqrt{2} - 7}{1} \\[5ex] = 5\sqrt{2} - 7 $
(12.) WASSCE If $(3 - x), 6, (7 - 5x)$ are consecutive terms of a Geometric Progression with constant ratio $r \gt 0$, find the:
(i) values of $x$;
(ii) common ratio


$ (3 - x), 6, (7 - 5x) \\[3ex] r = \dfrac{6}{3 - x} \\[5ex] r = \dfrac{7 - 5x}{6} \\[5ex] r = r \\[3ex] \implies \dfrac{6}{3 - x} = \dfrac{7 - 5x}{6} \\[5ex] (3 - x)(7 - 5x) = 6(6) \\[3ex] 21 - 15x - 7x + 5x^2 = 36 \\[3ex] 21 - 22x + 5x^2 = 36 \\[3ex] 5x^2 - 22x + 21 = 36 \\[3ex] 5x^2 - 22x + 21 - 36 = 0 \\[3ex] 5x^2 - 22x - 15 = 0 \\[3ex] 5x^2 + 3x - 25x - 15 = 0 \\[3ex] x(5x + 3) - 5(5x + 3) = 0 \\[3ex] 5x + 3 = 0 \:\:OR\:\: x - 5 = 0 \\[3ex] 5x = -3 \:\:OR\:\: x = 5 \\[3ex] (i)\:\: x = -\dfrac{3}{5} \:\:OR\:\: x = 5 \\[5ex] r = \dfrac{7 - 5x}{6} \\[5ex] when\:\: x = -\dfrac{3}{5} \\[5ex] 5x = 5 * -\dfrac{3}{5} = -3 \\[5ex] r = \dfrac{7 - (-3)}{6} \\[5ex] r = \dfrac{7 + 3}{6} = \dfrac{10}{6} = \dfrac{5}{3} \\[5ex] when\:\: x = 5 \\[3ex] r = \dfrac{7 - 5(5)}{6} \\[5ex] r = \dfrac{7 - 25}{6} = -\dfrac{18}{6} = -3 \\[5ex] Because\:\: r \gt 0 \\[3ex] (ii)\:\: r = \dfrac{5}{3} $
(13.) JAMB Three consecutive terms of a geometric progression are given as $n - 2$, $n$, and $n + 3$.
Find the common ratio.

$ A.\:\: \dfrac{3}{2} \\[5ex] B.\:\: \dfrac{2}{3} \\[5ex] C.\:\: \dfrac{1}{2} \\[5ex] D.\:\: \dfrac{1}{4} \\[5ex] $

$ n - 2,\:\:n,\:\:n + 3 \\[3ex] r = \dfrac{n}{n - 2} \\[5ex] r = \dfrac{n + 3}{n} \\[5ex] \rightarrow \dfrac{n}{n - 2} = \dfrac{n + 3}{n} \\[5ex] Cross\:\:multiply \\[3ex] n(n) = (n - 2)(n + 3) \\[3ex] n^2 = n^2 + 3n - 2n - 6 \\[3ex] n^2 = n^2 + n - 6 \\[3ex] n^2 + n - 6 = n^2 \\[3ex] n = n^2 - n^2 + 6 \\[3ex] n = 6 \\[3ex] r = \dfrac{n}{n - 2} \\[5ex] r = \dfrac{6}{6 - 2} \\[5ex] r = \dfrac{6}{4} \\[5ex] r = \dfrac{3}{2} $
(14.) ACT Let a, b, c, and d be distinct positive integers.
What is the 4th term of the geometric sequence below?
$bcd, abc^2d, a^2bc^3d, ...$

$ F.\:\: a^3bc^4d \\[3ex] G.\:\: a^3b^2c^3d \\[3ex] H.\:\: a^3b^2c^4d^2 \\[3ex] J.\:\: a^4bc^6d \\[3ex] K.\:\: a^4bc^9d \\[3ex] $

$ r = \dfrac{2nd\:\: term}{1st\:\: term} \\[5ex] r = \dfrac{abc^2d}{bcd} = \dfrac{a * b * c * c * d}{b * c * d} = ac \\[5ex] 4th\:\:term = 3rd\:\: term * r \\[3ex] 4th\:\:term = a^2bc^3d * ac = a^2 * a * b * c^3 * c * d = a^3bc^4d \\[3ex] 4th\:\:term = a^3bc^4d $
(15.) JAMB Find to infinity, the sum of the sequence

$ 1,\:\:\dfrac{9}{10},\:\:\left(\dfrac{9}{10}\right)^2,\:\:\left(\dfrac{9}{10}\right)^3,... \\[5ex] A.\:\: 10 \\[3ex] B.\:\: 9 \\[3ex] C.\:\: \dfrac{10}{9} \\[5ex] D.\:\: \dfrac{9}{10} \\[5ex] $

$ S_{\infty} = \dfrac{a}{1 - r} \\[5ex] a = 1 \\[3ex] r = \dfrac{9}{10} \div 1 \\[5ex] = \dfrac{9}{10} \\[5ex] 1 - r \\[3ex] = 1 - \dfrac{9}{10} \\[5ex] = \dfrac{10}{10} - \dfrac{9}{10} \\[5ex] = \dfrac{10 - 9}{10} \\[5ex] = \dfrac{1}{10} \\[5ex] S_{\infty} = a \div (1 - r) \\[3ex] = 1 \div \dfrac{1}{10} \\[5ex] = 1 * \dfrac{10}{1} \\[5ex] = 1 * 10 \\[3ex] = 10 \\[3ex] S_{\infty} = 10 $
(16.) ACT The $1st$ term in the geometric sequence below is $-6$.
If it can be determined, what is the $6th$ term?
$-6, 12, -24, 48, -96, ...$

$ F.\:\: 192 \\[3ex] G.\:\: 144 \\[3ex] H.\:\: -144 \\[3ex] J.\:\: -192 \\[3ex] K.\:\:Cannot\:\:be\:\:determined\:\:from\:\:the\:\:given\:\:information \\[3ex] $

$ GS_n = ar^{n - 1} \\[3ex] a = -6 \\[3ex] r = \dfrac{12}{-6} = -2 \\[5ex] GS_6 = ar^{6 - 1} \\[3ex] GS_6 = ar^5 \\[3ex] GS_6 = -6 * (-2)^5 \\[3ex] GS_6 = -6 * -32 \\[3ex] GS_6 = 192 $
(17.) JAMB If the sum of the first two terms of a G.P is $3$, and the sum of the second and the third terms is $-6$, find the sum of the first term and the common ratio.

$ A.\:\: -3 \\[3ex] B.\:\: -5 \\[3ex] C.\:\: 5 \\[3ex] D.\:\: -2 \\[3ex] $

The first two terms are the first term and the second term.

$ GS_n = ar^n \\[3ex] first\:\:term = a \\[3ex] second\:\:term = GS_2 = ar \\[3ex] a + ar = 3 \\[3ex] a(1 + r) = 3...eqn.(1) \\[3ex] third\:\:term = GS_3 = ar^2 \\[3ex] ar + ar^2 = -6 \\[3ex] ar(1 + r) = -6...eqn.(2) \\[3ex] eqn.(2) \div eqn.(1)\:\:gives \\[3ex] \dfrac{ar(1 + r)}{a(1 + r)} = \dfrac{-6}{3} \\[5ex] r = -\dfrac{6}{3} \\[5ex] r = -2 \\[3ex] From\:\:eqn.(1) \\[3ex] a = \dfrac{3}{1 + r} \\[5ex] a = \dfrac{3}{1 + (-2)} \\[5ex] a = \dfrac{3}{1 - 2} \\[5ex] a = \dfrac{3}{-1} \\[5ex] a = -3 \\[3ex] a + r \\[3ex] = -3 + (-2) \\[3ex] = -3 - 2 \\[3ex] = -5 $
(18.) ACT What is the next term after $-\dfrac{1}{4}$ in the geometric sequence $16, -4, 1, -\dfrac{1}{4}, ...$?


$ 16, -4, 1, -\dfrac{1}{4}, ... \\[5ex] r = \dfrac{-4}{16} = -\dfrac{1}{4} \\[5ex] Next\:\: term = -\dfrac{1}{4} * -\dfrac{1}{4} \\[5ex] Next\:\: term = \dfrac{1}{16} $
(19.) JAMB Given that the first and fourth terms of a G.P are $6$ and $162$ respectively, find the sum of the first three terms of the progression.

$ A.\:\: 48 \\[3ex] B.\:\: 78 \\[3ex] C.\:\: 27 \\[3ex] D.\:\: 8 \\[3ex] $

$ a = 6 \\[3ex] GS_n = ar^{n - 1} \\[3ex] GS_4 = ar^3 \\[3ex] 162 = 6 * r^3 \\[3ex] 6 * r^3 = 162 \\[3ex] r^3 = \dfrac{162}{6} \\[5ex] r^3 = 27 \\[3ex] r = \sqrt[3]{27} \\[3ex] r = 3 \\[3ex] r \gt 3 \\[3ex] SGS_n = \dfrac{a(r^{n} - 1)}{r - 1} \:\:for\:\: r \gt 1 \\[5ex] SGS_3 = \dfrac{6 * (3^{3} - 1)}{3 - 1} \\[5ex] = \dfrac{6 * (27 - 1)}{2} \\[5ex] = \dfrac{6 * 26}{2} \\[5ex] = 3 * 26 \\[3ex] = 78 \\[3ex] SGS_3 = 78 $
(20.) WASSCE:FM The $3rd$ and $6th$ terms of a geometric progression (G.P) are $2$ and $54$ respectively.
Find the:
(a.) common ratio
(b.) first term
(c.) sum of the first ten terms, correct to the nearest whole number.


$ (a.) \\[3ex] GP_n = ar^{n - 1} \\[3ex] GP_3 = ar^{3 - 1} = ar^2 = 2...eqn.(1) \\[3ex] GP_6 = ar^{6 - 1} = ar^5 = 54...eqn.(2) \\[3ex] eqn.(2) \div eqn.(1) \\[3ex] \implies \dfrac{ar^5}{ar^2} = \dfrac{54}{2} \\[5ex] r^3 = 27 \\[3ex] r = \sqrt[3]{27} \\[3ex] r = 3 \\[3ex] (b.) \\[3ex] From\:\:eqn.(1) \\[3ex] ar^2 = 2 \\[3ex] a(3^2) = 2 \\[3ex] a(9) = 2 \\[3ex] a = \dfrac{2}{9} \\[5ex] (c.) \\[3ex] r \gt 1 \\[3ex] \therefore SGP_n = \dfrac{a(r^{n} - 1)}{r - 1} \:\:for\:\: r \gt 1 \\[5ex] SGP_n = \dfrac{a(r^{n} - 1)}{r - 1} \\[5ex] SGP_{10} = \dfrac{\dfrac{2}{9}\left(3^{10} - 1\right)}{3 - 1} \\[7ex] = \dfrac{\dfrac{2}{9}\left(59049 - 1\right)}{2} \\[7ex] = \dfrac{\dfrac{2}{9}(59048)}{2} \\[7ex] = \dfrac{2}{9}(59048) \div \dfrac{2}{1} \\[5ex] = \dfrac{2}{9}(59048) * \dfrac{1}{2} \\[5ex] = \dfrac{59048}{9} \\[5ex] = 6560.888889 \\[3ex] SGP_{10} \approx 6561 $




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(21.) JAMB Find the sum to infinity of the series $\dfrac{1}{2},\;\;\dfrac{1}{6},\;\;\dfrac{1}{18},...$

$ A.\;\; \dfrac{2}{3} \\[5ex] B.\;\; 1 \\[3ex] C.\;\; \dfrac{1}{3} \\[5ex] D.\;\; \dfrac{3}{4} \\[5ex] $

$ a = \dfrac{1}{2} \\[5ex] r = \dfrac{1}{6} \div \dfrac{1}{2} \\[5ex] = \dfrac{1}{6} * \dfrac{2}{1} \\[5ex] = \dfrac{1}{3} \\[5ex] r \lt 1 \implies the\:\:series\:\:converges\:\:and\:\: S_{\infty} = \dfrac{a}{1 - r} \\[5ex] 1 - r \\[3ex] = 1 - \dfrac{1}{3} \\[5ex] = \dfrac{3}{3} - \dfrac{1}{3} \\[5ex] = \dfrac{2}{3} \\[5ex] S_{\infty} = \dfrac{1}{2} \div \dfrac{2}{3} \\[5ex] = \dfrac{1}{2} * \dfrac{3}{2} \\[5ex] = \dfrac{3}{4} $
(22.) ACT A geometric sequence is a sequence of numbers in which each term is multiplied by a constant to obtain the following term.
What is the $4th$ term in the geometric sequence with first $3$ terms $4$, $6$, and $9?$

$ F.\:\: 10.5 \\[3ex] G.\:\: 12 \\[3ex] H.\:\: 13 \\[3ex] J.\:\: 13.5 \\[3ex] K.\:\: 15 \\[3ex] $

$ Let\:\:the\:\:4th\:\:term = p \\[3ex] GS:\:\: 4, 6, 9, p \\[3ex] r = \dfrac{6}{4} = \dfrac{3}{2} \\[5ex] p = 9 * \dfrac{3}{2} \\[5ex] p = 9(1.5) \\[3ex] p = 13.5 \\[3ex] $ The $4th$ term is $13.5$
(23.) JAMB The sum to infinity of the series $1 + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + ...$ is

$ A.\;\; \dfrac{10}{3} \\[5ex] B.\;\; \dfrac{11}{3} \\[5ex] C.\;\; \dfrac{3}{2} \\[5ex] D.\;\; \dfrac{5}{2} \\[5ex] $

$ a = 1 \\[3ex] r = \dfrac{1}{3} \div 1 \\[5ex] = \dfrac{1}{3} \\[5ex] r \lt 1 \implies the\:\:series\:\:converges\:\:and\:\: S_{\infty} = \dfrac{a}{1 - r} \\[5ex] 1 - r \\[3ex] = 1 - \dfrac{1}{3} \\[5ex] = \dfrac{3}{3} - \dfrac{1}{3} \\[5ex] = \dfrac{2}{3} \\[5ex] S_{\infty} = 1 \div \dfrac{2}{3} \\[5ex] = 1 * \dfrac{3}{2} \\[5ex] = \dfrac{3}{2} $
(24.) ACT The $1st$ term in the geometric sequence below is $-4$.
If it can be determined, what is the $6th$ term?
$-4, 8, -16, 32, -64, ...$

$ F.\:\: 128 \\[3ex] G.\:\: 96 \\[3ex] H.\:\: -96 \\[3ex] J.\:\: -128 \\[3ex] K.\:\:Cannot\:\:be\:\:determined\:\:from\:\:the\:\:given\:\:information \\[3ex] $

$ GS_n = ar^{n - 1} \\[3ex] a = -4 \\[3ex] r = \dfrac{8}{-4} = -2 \\[5ex] GS_6 = ar^{6 - 1} \\[3ex] GS_6 = ar^5 \\[3ex] GS_6 = -4 * (-2)^5 \\[3ex] GS_6 = -4 * -32 \\[3ex] GS_6 = 128 $
(25.) JAMB Find the sum of infinity of the following series.

$ 0.5 + 0.05 + 0.005 + 0.0005 + ... \\[3ex] A.\;\; \dfrac{5}{8} \\[5ex] B.\;\; \dfrac{5}{7} \\[5ex] C.\;\; \dfrac{5}{11} \\[5ex] D.\;\; \dfrac{5}{9} \\[5ex] $

$ a = 0.5 \\[3ex] = \dfrac{5}{10} \\[5ex] = \dfrac{1}{2} \\[5ex] r = 0.05 \div 0.5 \\[5ex] = \dfrac{0.05}{0.5} \\[5ex] = \dfrac{5}{50} \\[5ex] = \dfrac{1}{10} \\[5ex] r \lt 1 \implies the\:\:series\:\:converges\:\:and\:\: S_{\infty} = \dfrac{a}{1 - r} \\[5ex] 1 - r \\[3ex] = 1 - \dfrac{1}{10} \\[5ex] = \dfrac{10}{10} - \dfrac{1}{10} \\[5ex] = \dfrac{9}{10} \\[5ex] S_{\infty} = \dfrac{1}{2} \div \dfrac{9}{10} \\[5ex] = \dfrac{1}{2} * \dfrac{10}{9} \\[5ex] = \dfrac{5}{9} $
(26.) ACT The $1st$ term in the geometric sequence below is $-12$.
If it can be determined, what is the $6th$ term?
$-12, 24, -48, 96, -192, ...$

$ A.\:\: -384 \\[3ex] B.\:\: -288 \\[3ex] C.\:\: 288 \\[3ex] D.\:\: 384 \\[3ex] E.\:\:Cannot\:\:be\:\:determined\:\:from\:\:the\:\:given\:\:information \\[3ex] $

$ GS_n = ar^{n - 1} \\[3ex] a = -12 \\[3ex] r = \dfrac{24}{-12} = -2 \\[5ex] GS_6 = ar^{6 - 1} \\[3ex] GS_6 = ar^5 \\[3ex] GS_6 = -12 * (-2)^5 \\[3ex] GS_6 = -12 * -32 \\[3ex] GS_6 = 384 $
(27.) JAMB The second term of a geometric series is $4$ while the fourth term is $16$
Find the sum of the first five terms.

$ A.\;\; 62 \\[3ex] B.\;\; 60 \\[3ex] C.\;\; 54 \\[3ex] D.\;\; 64 \\[3ex] $

$ GS_n = ar^{n - 1} \\[3ex] GS_2 = ar \\[3ex] GS_2 = 4 \\[3ex] \implies ar = 4 ...eqn.(1) \\[3ex] GS_4 = ar^3 \\[3ex] GS_4 = 16 \\[3ex] ar^3 = 16 ...eqn.(2) \\[3ex] eqn.(2) \div eqn.(1) \implies \\[3ex] \dfrac{ar^3}{ar} = \dfrac{16}{4} \\[5ex] r^2 = 4 \\[3ex] r = \pm \sqrt{4} \\[3ex] r = 2 \\[3ex] Substitute\;\; r = 2 \;\;into\;\;eqn.(1) \\[3ex] a(2) = 4 \\[3ex] a = \dfrac{4}{2} \\[5ex] a = 2 \\[3ex] Based\;\;on\;\;the\;\;answer\;\;options;\;\; r = 2 \\[3ex] SGS_n = \dfrac{a(r^{n} - 1)}{r - 1} \:\:for\:\: r \gt 1 \\[5ex] SGS_5 = \dfrac{2(2^{5} - 1)}{2 - 1} \\[5ex] = \dfrac{2(32 - 1)}{1} \\[5ex] = 2(31) \\[3ex] = 62 $
(28.) ACT The first 3 terms of a geometric sequence are 4, 10, and 25.
What is the next term in the sequence?

$ A.\:\: 35 \\[3ex] B.\:\: 40 \\[3ex] C.\:\: 55 \\[3ex] D.\:\: 62.5 \\[3ex] E.\:\: 70 \\[3ex] $

$ GS:\:\: 4, 10, 25 \\[3ex] r = \dfrac{10}{4} = \dfrac{5}{2} \\[5ex] GS_3 = 25 \\[3ex] GS_4 = GS_3 * r \\[3ex] GS_4 = 25 * \dfrac{5}{2} \\[5ex] GS_4 = 25(2.5) \\[3ex] GS_4 = 62.5 $
(29.) WASSCE (a.) The third and sixth terms of a Geometric Progression (G.P) are $\dfrac{1}{4}$ and $\dfrac{1}{32}$ respectively.
Find the:
(i) first term and the common ratio
(ii) seventh term


$ (i) \\[3ex] GS_n = ar^{n - 1} \\[3ex] GS_3 = ar^{3 - 1} \\[3ex] \dfrac{1}{4} = ar^2 \\[5ex] ar^2 = \dfrac{1}{4}...eqn.(1) \\[5ex] GS_6 = ar^{6 - 1} \\[3ex] \dfrac{1}{32} = ar^5 \\[5ex] ar^5 = \dfrac{1}{32}...eqn.(2) \\[5ex] eqn.(2) \div eqn.(1) \implies \\[3ex] \dfrac{ar^5}{ar^2} = \dfrac{1}{32} \div \dfrac{1}{4} \\[5ex] r^3 = \dfrac{1}{32} * \dfrac{4}{1} \\[5ex] r^3 = \dfrac{1}{8} \\[5ex] r = \sqrt[3]{\dfrac{1}{8}} \\[5ex] r = \dfrac{1}{2} \\[5ex] From\;\;eqn.(1) \\[3ex] ar^2 = \dfrac{1}{4} \\[5ex] a = \dfrac{1}{4} \div r^2 \\[5ex] a = \dfrac{1}{4} \div \left(\dfrac{1}{2}\right)^2 \\[5ex] a = \dfrac{1}{4} \div \dfrac{1}{4} \\[5ex] a = 1 \\[3ex] (ii) \\[3ex] GS_7 = ar^6 \\[3ex] = 1 * \left(\dfrac{1}{2}\right)^6 \\[5ex] = \dfrac{1}{64} $
(30.) ACT If 4 is the first term and 256 is the fourth term of a geometric progression, which of the following is the second term?

$ A.\;\; 8 \\[3ex] B.\;\; 12 \\[3ex] C.\;\; 16 \\[3ex] D.\;\; 32 \\[3ex] E.\;\; 64 \\[3ex] $

$ GS_n = ar^{n - 1} \\[3ex] a = 4 \\[3ex] GS_4 = 256 \\[3ex] GS_4 = ar^{4 - 1} \\[3ex] GS_4 = ar^3 \\[3ex] \implies \\[3ex] ar^3 = 256 \\[3ex] 4 * r^3 = 256 \\[3ex] r^3 = \dfrac{256}{4} \\[5ex] r^3 = 64 \\[3ex] r = \sqrt[3]{64} \\[3ex] r = 4 \\[3ex] GS_2 = ar^{2 - 1} \\[3ex] GS_2 = ar \\[3ex] GS_2 = 4 * 4 \\[3ex] GS_2 = 16 $
(31.) JAMB The sum of the first three terms of a geometric progression is half its sum to infinity.
Find the positive common ratio of the progression.

$ A.\;\; \dfrac{1}{4} \\[5ex] B.\;\; \sqrt{\dfrac{3}{2}} \\[5ex] C.\;\; \dfrac{1}{\sqrt{3}} \\[5ex] D.\;\; \dfrac{1}{\sqrt{2}} \\[5ex] $

For there to be a sum to infinity, the series must converge
For the series to converge, the common ratio must be less than $1$
So, we have to use the sum of the first $n$ terms of a Geometric Sequence when the common ratio is less than $1$

$ If\:\:r \lt 1,\:\:the\:\:series\:\:converges\:\:and\:\: S_{\infty} = \dfrac{a}{1 - r} \\[5ex] Because\;\;r \lt 1: \\[3ex] SGS_n = \dfrac{a(1 - r^{n})}{1 - r} \:\:for\:\: r \lt 1 \\[5ex] n = 3 \\[3ex] SGS_3 = \dfrac{a(1 - r^{3})}{1 - r} \\[5ex] SGS_3 = \dfrac{1}{2} * S_{\infty} \\[5ex] \implies \\[3ex] \dfrac{a(1 - r^{3})}{1 - r} = \dfrac{1}{2} * \dfrac{a}{1 - r} \\[5ex] Multiply\;\;both\;\;sides\;\;by\;\; \dfrac{1-r}{a} \\[5ex] \dfrac{1 - r}{a} * \dfrac{a(1 - r^{3})}{1 - r} = \dfrac{1}{2} * \dfrac{1 - r}{a} * \dfrac{a}{1 - r} \\[5ex] 1 - r^3 = \dfrac{1}{2} \\[5ex] 1 - \dfrac{1}{2} = r^3 \\[5ex] r^3 = 1 - \dfrac{1}{2} \\[5ex] r^3 = \dfrac{1}{2} \\[5ex] r = \sqrt[3]{\dfrac{1}{2}} \\[5ex] = \dfrac{\sqrt[3]{1}}{\sqrt[3]{2}} \\[5ex] = \dfrac{1}{\sqrt[3]{2}} \\[5ex] = \dfrac{1}{\sqrt[3]{2}} * \dfrac{\sqrt[3]{4}}{\sqrt[3]{4}} \\[5ex] = \dfrac{\sqrt[3]{4}}{\sqrt[3]{8}} \\[5ex] = \dfrac{\sqrt[3]{4}}{2} \\[5ex] $ No correct option
(32.) JAMB $S_n$ is the sum of the first $n$ terms of a series given by $S_n = n^2 - 1$
Find the $nth$ term

$ A.\;\; 4n + 1 \\[3ex] B.\;\; 4n - 1 \\[3ex] C.\;\; 2n + 1 \\[3ex] D.\;\; 2n - 1 \\[3ex] $

The first term of the $nth$ term of a sequence is the same as the sum of the first $1$ term of the sequence

$ Let\;\;Tn = nth\;\;term\;\;of\;\;the\;\;sequence \\[3ex] a = T_1 = S_1 \\[3ex] S_n = n^2 - 1 \\[3ex] S_1 = 1^2 - 1 \\[3ex] = 1 - 1 \\[3ex] = 0 \\[3ex] \implies a = T_1 = S_1 = 0 \\[3ex] Test\;\;the\;\;options \\[3ex] (A.)\;\; T_n = 4n + 1 \\[3ex] T_1 = 4(1) + 1 \\[3ex] = 4 + 1 = 5 \\[3ex] 5 \ne 0...No \\[3ex] (B.)\;\; T_n = 4n -1 \\[3ex] T_1 = 4(1) - 1 \\[3ex] = 4 - 1 = 3 \\[3ex] 3 \ne 0...No \\[3ex] (C.)\;\; T_n = 2n + 1 \\[3ex] T_1 = 2(1) + 1 \\[3ex] = 2 + 1 = 3 \\[3ex] 3 \ne 0 ...No \\[3ex] (D.)\;\; T_n = 2n - 1 \\[3ex] T_1 = 2(1) - 1 \\[3ex] = 2 - 1 = 1 \\[3ex] 1 \ne 0...No \\[3ex] $ No correct option
(33.)

(34.) JAMB The first term of a geometrical progression is twice its common ratio.
Find the sum of the first two terms of the progression if its sum to infinity is $8$

$ A.\;\; \dfrac{8}{25} \\[5ex] B.\;\; \dfrac{8}{3} \\[5ex] C.\;\; \dfrac{72}{25} \\[5ex] D.\;\; \dfrac{56}{9} \\[5ex] $

For there to be a sum to infinity, the series must converge
For the series to converge, the common ratio must be less than $1$
So, we have to use the sum of the first $n$ terms of a Geometric Sequence when the common ratio is less than $1$

$ a = 2 * r \\[3ex] a = 2r...eqn.(1) \\[3ex] If\:\:r \lt 1,\:\:the\:\:series\:\:converges\:\:and\:\: S_{\infty} = \dfrac{a}{1 - r} \\[5ex] S_{\infty} = 8 \\[3ex] \implies \\[3ex] \dfrac{a}{1 - r} = 8 \\[5ex] a = 8(1 - r)...eqn.(2) \\[3ex] a = a \implies eqn.(1) = eqn.(2) \\[3ex] 2r = 8(1 - r) \\[3ex] 2r = 8 - 8r \\[3ex] 2r + 8r = 8 \\[3ex] 10r = 8 \\[3ex] r = \dfrac{8}{10} \\[5ex] r = \dfrac{4}{5} \\[5ex] From\;\;eqn.(1): \\[3ex] a = 2r \\[3ex] a = 2\left(\dfrac{4}{5}\right) \\[5ex] a = \dfrac{8}{5} \\[5ex] Because\;\;r \lt 1: \\[3ex] SGS_n = \dfrac{a(1 - r^{n})}{1 - r} \:\:for\:\: r \lt 1 \\[5ex] n = 2 \\[3ex] SGS_2 = \dfrac{a(1 - r^{2})}{1 - r} \\[5ex] 1 - r \\[3ex] = 1 - \dfrac{4}{5} \\[5ex] = \dfrac{5}{5} - \dfrac{4}{5} \\[5ex] = \dfrac{1}{5} \\[5ex] 1 - r^2 \\[3ex] = 1 - \left(\dfrac{4}{5}\right)^2 \\[5ex] = 1 - \dfrac{16}{25} \\[5ex] = \dfrac{25}{25} - \dfrac{16}{25} \\[5ex] = \dfrac{9}{25} \\[5ex] a * (1 - r^2) \\[3ex] = \dfrac{8}{5} * \dfrac{9}{25} \\[5ex] SGS_2 = a(1 - r^2) \div (1 - r) \\[3ex] = \left(\dfrac{8}{5} * \dfrac{9}{25}\right) \div \dfrac{1}{5} \\[5ex] = \dfrac{8}{5} * \dfrac{9}{25} * \dfrac{5}{1} \\[5ex] = \dfrac{72}{25} $
(35.)

(36.) ACT The sum of an infinite geometric series with first term a and common ratio r < 1 is given by $\dfrac{a}{1 - r}$.
The sum of a given infinite geometric series is 200, and the common ratio is 0.15
What is the second term of this series?

$ F.\;\; 25.5 \\[3ex] G.\;\; 30 \\[3ex] H.\;\; 169.85 \\[3ex] J.\;\; 170 \\[3ex] K.\;\; 199.85 \\[3ex] $

$ S_{\infty} = 200 \\[3ex] r = 0.15 \\[3ex] S_{\infty} = \dfrac{a}{1 - r} \\[5ex] 200 = \dfrac{a}{1 - 0.15} \\[5ex] \dfrac{a}{0.85} = 200 \\[5ex] a = 200(0.85) \\[3ex] a = 170 \\[3ex] GS_n = ar^{n - 1} \\[3ex] GS_2 = 170 * 0.15^(2 - 1) \\[3ex] GS_2 = 170 * 0.15^1 \\[3ex] GS_2 = 170 * 0.15 \\[3ex] GS_2 = 25.5 \\[3ex] $ The second term of the series is 25.5
(37.)

(38.)

(39.)

(40.)





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(41.)

(42.)

(42.)

(44.)

(45.) JAMB Find the sum of the infinity of the following series: $3 + 2 + \dfrac{4}{3} + \dfrac{8}{9} + \dfrac{16}{27} + ...$

$ A.\;\; 1270 \\[3ex] B.\;\; 190 \\[3ex] C.\;\; 18 \\[3ex] D.\;\; 9 \\[3ex] $

$ a = 3 \\[3ex] r = 2 \div 3 \\[3ex] = \dfrac{2}{3} \\[5ex] r \lt 1 \implies the\:\:series\:\:converges\:\:and\:\: S_{\infty} = \dfrac{a}{1 - r} \\[5ex] 1 - r \\[3ex] = 1 - \dfrac{2}{3} \\[5ex] = \dfrac{3}{3} - \dfrac{2}{3} \\[5ex] = \dfrac{1}{3} \\[5ex] S_{\infty} = 3 \div \dfrac{1}{3} \\[5ex] = 3 * \dfrac{3}{1} \\[5ex] = 9 $
(46.) ACT What is the $7th$ term of the geometric sequence $1, -2, 4, -8, ...?$

$ A.\:\: -32 \\[3ex] B.\:\: -10 \\[3ex] C.\:\: 16 \\[3ex] D.\:\: 56 \\[3ex] E.\:\: 64 \\[3ex] $

$ GS_n = ar^{n - 1} \\[3ex] a = 1 \\[3ex] r = \dfrac{-2}{1} = -2 \\[5ex] GS_7 = ar^{7 - 1} \\[3ex] GS_7 = ar^6 \\[3ex] GS_7 = 1 * (-2)^6 \\[3ex] GS_7 = 1 * 64 \\[3ex] GS_7 = 64 $
(47.) WASSCE:FM The sum of the second and third terms of a Geometric Progression (G.P) is $48$
If the sum of the third and fourth terms is $144$, find the first term of the progression.


$ GS_n = ar^{n - 1} \\[3ex] GS_2 = ar^{2 - 1} = ar^1 = ar \\[3ex] GS_3 = ar^{3 - 1} = ar^2 \\[3ex] GS_4 = ar^{4 - 1} = ar^3 \\[3ex] GS_2 + GS_3 = 48 \\[3ex] \implies ar + ar^2 = 48 \\[3ex] \implies ar(1 + r) = 48...eqn.(1) \\[3ex] GS_3 + GS_4 = 144 \\[3ex] \implies ar^2 + ar^3 = 144 \\[3ex] \implies ar^2(1 + r) = 144...eqn.(2) \\[3ex] eqn.(2) \div eqn.(1) \implies \\[3ex] \dfrac{ar^2(1 + r)}{ar(1 + r)} = \dfrac{144}{48} \\[5ex] r = 3 \\[3ex] From\;\;eqn.(1) \\[3ex] ar(1 + r) = 48 \\[3ex] a(3)(1 + 3) = 48 \\[3ex] a(3)(4) = 48 \\[3ex] 12a = 48 \\[3ex] a = \dfrac{48}{12} \\[5ex] a = 4 $
(48.) WASSCE:FM The fourth and sixth terms of a Geometric Progression (G.P) are $54$ and $486$ respectively.
If $r \gt 0$, find the third term.


$ GS_n = ar^{n - 1} \\[3ex] GS_4 = ar^{4 - 1} = ar^3 = 54...eqn.(1) \\[3ex] GS_6 = ar^{6 - 1} = ar^5 = 486...eqn.(2) \\[3ex] eqn.(2) \div eqn.(1) \\[3ex] \implies \dfrac{ar^5}{ar^3} = \dfrac{486}{54} \\[5ex] r^2 = 9 \\[3ex] r = \pm \sqrt{9} \\[3ex] r = \pm 3 \\[3ex] But:\;\; r \gt 0...from\;\;the\;\;question \\[3ex] \therefore r = 3 \\[3ex] From\;\;eqn.(1) \\[3ex] ar^3 = 54 \\[3ex] a = \dfrac{54}{r^3} \\[5ex] a = \dfrac{54}{3^3} \\[5ex] a = \dfrac{54}{27} \\[5ex] a = 2 \\[3ex] \underline{Third\;\;term} \\[3ex] GP_3 \\[3ex] = ar^{3 - 1} \\[3ex] = ar^2 \\[3ex] = 2 * (3)^2 \\[3ex] = 2(9) \\[3ex] = 18 $
(49.) JAMB Find the sum to infinity of the following sequence

$ 1,\:\:\dfrac{9}{10},\:\:\left(\dfrac{9}{10}\right)^2,\:\:\left(\dfrac{9}{10}\right)^3 \\[5ex] A.\:\: \dfrac{1}{10} \\[5ex] B.\:\: \dfrac{9}{10} \\[5ex] C.\:\: \dfrac{10}{9} \\[5ex] D.\:\: 10 \\[3ex] $

$ S_{\infty} = \dfrac{a}{1 - r} \\[5ex] a = 1 \\[3ex] r = \dfrac{9}{10} \div 1 \\[5ex] = \dfrac{9}{10} \\[5ex] 1 - r \\[3ex] = 1 - \dfrac{9}{10} \\[5ex] = \dfrac{10}{10} - \dfrac{9}{10} \\[5ex] = \dfrac{10 - 9}{10} \\[5ex] = \dfrac{1}{10} \\[5ex] S_{\infty} = a \div (1 - r) \\[3ex] = 1 \div \dfrac{1}{10} \\[5ex] = 1 * \dfrac{10}{1} \\[5ex] = 1 * 10 \\[3ex] = 10 \\[3ex] S_{\infty} = 10 $
(50.) ACT The 1st term of a geometric sequence is 27, and the 4th term is 64.
In terms of n, what is the nth term of the sequence?

$ A.\:\: 27\left(\dfrac{3}{4}\right)^{n - 1} \\[5ex] B.\:\: 27\left(\dfrac{3}{4}\right)^n \\[5ex] C.\:\: 27\left(\dfrac{4}{3}\right)^{n - 1} \\[5ex] D.\:\: 27\left(\dfrac{4}{3}\right)^n \\[5ex] E.\:\: 27\left(\dfrac{4}{3}\right)n \\[5ex] $

$ GS_n = ar^{n - 1} \\[3ex] a = 27 \\[3ex] GS_4 = 27 * r^{4 - 1} \\[3ex] GS_4 = 64 \\[3ex] 64 = 27 * r^3 27r^3 = 64 \\[3ex] r^3 = \dfrac{64}{27} \\[5ex] r = \sqrt[3]{\dfrac{64}{27}} \\[5ex] r = \dfrac{4}{3} \\[5ex] \therefore GS_n = 27 * \left(\dfrac{4}{3}\right)^{n - 1} \\[5ex] GS_n = 27\left(\dfrac{4}{3}\right)^{n - 1} $