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Sequences and Series

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$ (1.)\:\: AS_n = a + d(n - 1) \\[5ex] (2.)\:\: AS_n = vn + w \:\:where\:\: v = d \:\:and\:\: w = a - d \\[5ex] (3.)\:\: p = a + d(n - 1) \\[5ex] (4.)\:\: SAS_n = \dfrac{n}{2}(a + AS_n) \\[7ex] (5.)\:\: SAS_n = \dfrac{n}{2}(a + p) \\[7ex] (6.)\:\: SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[7ex] (7.)\:\: n = \dfrac{2 * SAS_n}{a + p} \\[7ex] (8.)\:\: n = \dfrac{p - a + d}{d} \\[7ex] (9.)\:\: n = \dfrac{-(2a - d) \pm \sqrt{(2a - d)^2 + 8d*SAS_n}}{2d} \\[7ex] (10.)\;\; d = \dfrac{(p - a)(p + a)}{2 * SAS_n - p - a} $


$ (1.)\:\: GS_n = ar^{n - 1} \\[5ex] (2.)\:\: SGS_n = \dfrac{a(r^{n} - 1)}{r - 1} \:\:for\:\: r \gt 1 \\[7ex] (3.)\:\: SGS_n = \dfrac{a(1 - r^{n})}{1 - r} \:\:for\:\: r \lt 1 \\[7ex] (4.)\:\: n = \dfrac{\log{\left[\dfrac{SGS_n(r - 1)}{a} + 1\right]}}{\log r} \\[7ex] (5.)\:\: If\:\:r \lt 1,\:\:the\:\:series\:\:converges\:\:and\:\: S_{\infty} = \dfrac{a}{1 - r} \\[7ex] (6.)\:\: If\:\:r \gt 1,\:\:the\:\:series\:\:diverges \\[5ex] (7.)\:\: If\:\:r = 1,\:\:S_{\infty}\:\:DNE \\[5ex] (8.)\:\: r = \dfrac{S_{\infty} - a}{S_{\infty}} \\[7ex] (9.)\:\: a = S_{\infty}(1 - r) $


$ QS = 1st,\:\:\:\:2nd,\:\:\:3rd,\:\:\:4th,... \\[5ex] QS_n = an^2 + bn + c \\[5ex] (1.)\:\: a = \dfrac{1st + 3rd - 2(2nd)}{2} \\[7ex] (2.)\:\: b = \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} \\[7ex] (3.)\:\: c = 3(1st) - 3(2nd) + 3rd \\[5ex] (4.)\:\: \therefore QS_n = \dfrac{1st + 3rd - 2(2nd)}{2} * n^2 + \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} * n + 3(1st) - 3(2nd) + 3rd \\[7ex] The\:\:Left\:\:Hand\:\:Side\:\:must\;\;be\:\:equal\:\:to\;\;the\:\:Right\:\:Hand\:\:Side \\[5ex] (5.)\:\: a + b + c = 1st \\[5ex] (6.)\:\: 4a + 2b + c = 2nd \\[5ex] (7.)\:\: 9a + 3b + c = 3rd \\[5ex] (8.)\:\: 3a + b = 2nd - 1st \\[5ex] (9.)\:\: 8a + 2b = 3rd - 1st $


$ \underline{Triangular\;\;Number\;\;Sequence} \\[3ex] (1.)\:\: TS_n = \dfrac{n(n + 1)}{2} \\[7ex] (2.)\;\; n = \dfrac{\sqrt{8 * TS_n + 1} - 1}{2} \\[7ex] (3.)\:\: TS_n = C(n + 1, 2)...Combinatorics\;\;symbol\;\;for\;\;Combinations \\[5ex] (4.)\;\; STS_n = \dfrac{n(n + 1)(n + 2)}{6} \\[7ex] (5.)\:\: STS_n = C(n + 2, 3)...Combinatorics\;\;symbol\;\;for\;\;Combinations \\[7ex] \underline{Square\;\;Number\;\;Sequence} \\[3ex] (1.)\:\: SS_n = n^2 \\[5ex] (2.)\;\; n = \sqrt{SS_n} \\[5ex] (3.)\;\; SSS_n = \dfrac{n(n + 1)(2n + 1)}{6} \\[7ex] \underline{Cube\;\;Number\;\;Sequence} \\[3ex] (1.)\:\: CS_n = n^3 \\[5ex] (2.)\;\; n = \sqrt[3]{CS_n} \\[5ex] (3.)\;\; SCS_n = \left[\dfrac{n(n + 1)}{2}\right]^2 \\[7ex] (4.)\;\; n = \dfrac{\sqrt{8\sqrt{SCS_n} + 1} - 1}{2} \\[7ex] $


$ \underline{First-Order\;\;Linear\;\;Recurrence\;\;Relation} \\[3ex] (1.)\:\: RS_{n + 1} = r * RS_{n} + a \\[5ex] (2.)\:\: RS_{n + 1} = \dfrac{RS_1 * r^n(r - 1) + a(r^n - 1)}{r - 1} \;\;\;for\;\;r \gt 1 \\[7ex] (3.)\;\; RS_{n + 1} = \dfrac{RS_1 * r^n(1 - r) + a(1 - r^n)}{1 - r} \;\;\;for\;\;r \lt 1 \\[7ex] \underline{Fibonacci\;\;Sequence} \\[3ex] (1.)\;\; \phi = \dfrac{1 + \sqrt{5}}{2} \\[7ex] (2.)\;\; FS_n = \dfrac{\sqrt{5}}{5}\left[\left(\dfrac{1 + \sqrt{5}}{2}\right)^n - \left(\dfrac{1 - \sqrt{5}}{2}\right)^n\right] \\[7ex] (3.)\;\; FS_n = \dfrac{\sqrt{5}}{5}\left[\phi^n - \left(\dfrac{1 - \sqrt{5}}{2}\right)^n\right] \\[7ex] (4.)\;\; SFS_n = \dfrac{\sqrt{5}}{5}\left[\dfrac{\phi^3(\phi^{n - 1} - 1) + [(\phi - 1)(1 - \phi^2)][(1 - \phi)^{n - 1} - 1]}{\phi(\phi - 1)}\right] + 1 $

(1.) 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 $
(2.) 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 $
(3.) 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 $
(4.) The fourth and sixth terms of a Geometric Progression (G.P) are 54 and 486 respectively.
If r > 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 $
(5.) The sum of the first twelve terms of an Arithmetic Progression is 168
If the third term is 7, find the values of the common difference and the first term.


$ SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] SAS_{12} = \dfrac{12}{2}[2a + d(12 - 1)] \\[5ex] 168 = 6[2a + 11d] \\[3ex] \dfrac{168}{6} = 2a + 11d \\[3ex] 28 = 2a + 11d \\[3ex] 2a + 11d = 28...eqn.(1) \\[3ex] AS_n = a + d(n - 1) \\[3ex] AS_3 = a + d(3 - 1) \\[3ex] 7 = a + 2d \\[3ex] a + 2d = 7...eqn.(2) \\[3ex] Multiply\;\;eqn.(2)\;\;by\;\;2 \\[3ex] 2 * eqn.(2) \implies 2(a + 2d) = 2(7) \\[3ex] 2a + 4d = 14...eqn.(3) \\[3ex] eqn.(1) - eqn.(3) \implies \\[3ex] (2a + 11d) - (2a + 4d) = 28 - 14 \\[3ex] 2a + 11d - 2a - 4d = 14 \\[3ex] 7d = 14 \\[3ex] d = \dfrac{14}{7} \\[5ex] d = 2 \\[3ex] From\;\;eqn.(2) \\[3ex] a + 2d = 7 \\[3ex] a = 7 - 2d \\[3ex] a = 7 - 2(2) \\[3ex] a = 7 - 4 \\[3ex] a = 3 $
(6.) How many terms of the series $-3 - 1 + 1 + ...$ add up to 165?


$ a = -3 \\[3ex] d = -1 - (-3) \\[3ex] d = -1 + 3 \\[3ex] d = 2 \\[3ex] SAS_n = 165...because\;\;of\;\;"add\;\;up\;\;to" \\[3ex] n = ? \\[3ex] SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] 165 = \dfrac{n}{2}[2(-3) + 2(n - 1)] \\[5ex] 165 = \dfrac{n}{2}[-6 + 2n - 2] \\[5ex] 165 = \dfrac{n}{2}[2n - 8] \\[5ex] 2(165) = n(2n - 8) \\[3ex] 330 = 2n^2 - 8n \\[3ex] 2n^2 - 8n = 330 \\[3ex] 2n^2 - 8n - 330 = 0 \\[3ex] Divide\;\;all\;\;terms\;\;by\;\;2...to\;\;simplify \\[3ex] Work\;\;with\;\;smaller\;\;numbers\;\;when\;\;possible \\[3ex] n^2 - 4n - 165 = 0 \\[3ex] Factor\;\;Quadratic\;\;Trinomial \\[3ex] (n + 11)(n - 15) = 0 \\[3ex] n + 11 = 0 \;\;OR\;\; n - 15 = 0 \\[3ex] n = -11 \;\;OR\;\; n = 15 \\[3ex] number\;\;of\;\;terms\;\;cannot\;\;be\;\;negative \\[3ex] \implies n = 15 $
(7.) The $nth$ term of an Arithmetic Progression (A.P) is given by $U_n = n^2\log 5$
Find the sum of the first n terms of the Arithmetic Progression in terms of $\log 5$


$ U_n = n^2\log 5 \\[3ex] \underline{first\;\;term} \\[3ex] \implies n = 1 \\[3ex] U_1 = 1^2\log 5 \\[3ex] = 1 * \log 5 \\[3ex] = \log 5 \\[3ex] U_1 = a = \log 5 \\[3ex] \underline{second\;\;term} \\[3ex] Necessary\;\;so\;\;we\;\;can\;\;the\;\;common\;\;difference \\[3ex] U_2 = 2^2 * \log 5 \\[3ex] U_2 = 4\log 5 \\[3ex] \underline{common\;\;difference} \\[3ex] d = U_2 - U_1 \\[3ex] d = 4\log 5 - \log 5 \\[3ex] d = 3\log 5 \\[3ex] \underline{Sum\;\;of\;\;the\;\;first\;\;n\;\;terms} \\[5ex] SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] = \dfrac{n}{2}[2(\log 5) + 3\log 5(n - 1)] \\[5ex] = \dfrac{n}{2}[2\log 5 + 3n\log 5 - 3\log 5] \\[5ex] = \dfrac{n}{2}[3n\log 5 - \log 5] \\[5ex] = \dfrac{n\log 5}{2}[3n - 1] $
(8.) If the sixth term of an Arithmetic Progression (A.P) is 37 and the sum of the first six terms is 147, find the:
(a.) first term
(b.) sum of the first fifteen terms


$ AS_n = a + d(n - 1) \\[3ex] AS_6 = a + d(6 - 1) \\[3ex] AS_6 = a + 5d \\[3ex] AS_6 = 37 \\[3ex] \implies a + 5d = 37...eqn.(1) \\[3ex] SAS_n = \dfrac{n}{2}(a + AS_n) \\[5ex] SAS_6 = \dfrac{6}{2}(a + AS_6) \\[5ex] SAS_6 = 3(a + 37) \\[5ex] SAS_6 = 147 \\[3ex] \implies 3(a + 37) = 147 \\[3ex] Divide\;\;both\;\;sides\;\;by\;\;3...to\;\;simplify \\[3ex] a + 37 = 49...eqn.(2) \\[5ex] (a.) \\[3ex] a = 49 - 37 \\[3ex] a = 12 \\[5ex] (b.) \\[3ex] Subst.\;\;a = 12 \;\;into\;\;eqn.(1) \\[3ex] From\;\;eqn.(1) \\[3ex] 12 + 5d = 37 \\[3ex] 5d = 37 - 12 \\[3ex] 5d = 25 \\[3ex] d = \dfrac{25}{5} \\[5ex] d = 5 \\[3ex] AS_{15} = a + 14d \\[3ex] = 12 + 14(5) \\[3ex] = 12 + 70 \\[3ex] = 82 \\[3ex] SAS_{15} = \dfrac{n}{2}(a + AS_{15}) \\[5ex] = \dfrac{15}{2}(12 + 82) \\[5ex] = \dfrac{15}{2}(94) \\[5ex] = 15(47) \\[3ex] = 705 $
(9.) The sum of the first n terms of a sequence is given by $S_n = \dfrac{5n^2}{2} + \dfrac{5n}{2}$

Write down the first four terms of the sequence and find an exprsssion for the nth term.


We can solve this question in at least two ways.
Use any method you prefer.

We are given the sum of the first $n$ terms
However, we do not know the kind of sequence
So, let us try to determine the kind of sequence by testing some numbers
Let us find the sum of the first one term; sum of the first two terms; sum of the first three terms; sum of the first four terms
Then, we can find the four terms of the sequence
And determine the fifth term and the sum of the first five terms
And verify that sum by finding the sum of the first five terms.

$ \underline{First\:\:Method} \\[3ex] Let\:\:the\:\:terms\:\:be\:\:a, b, c, d, e \\[3ex] S_1 = sum\:\:of\:\:the\:\:first\:\:1\:\:term \\[3ex] S_2 = sum\:\:of\:\:the\:\:first\:\:2\:\:terms \\[3ex] S_3 = sum\:\:of\:\:the\:\:first\:\:3\:\:terms \\[3ex] S_4 = sum\:\:of\:\:the\:\:first\:\:4\:\:terms \\[3ex] S_5 = sum\:\:of\:\:the\:\:first\:\:5\:\:terms \\[3ex] S_n = \dfrac{5n^2}{2} + \dfrac{5n}{2} \\[5ex] S_1 = \dfrac{5(1)^2}{2} + \dfrac{5(1)}{2} \\[5ex] = \dfrac{5(1)}{2} + \dfrac{5}{2} \\[5ex] = \dfrac{5}{2} + \dfrac{5}{2} \\[5ex] = \dfrac{5 + 5}{2} \\[5ex] = \dfrac{10}{2} \\[5ex] S_1 = 5 \\[3ex] \implies \\[3ex] a = 5 \\[3ex] S_2 = \dfrac{5(2)^2}{2} + \dfrac{5(2)}{2} \\[5ex] = \dfrac{5(4)}{2} + \dfrac{10}{2} \\[5ex] = \dfrac{20}{2} + 5 \\[5ex] = 10 + 5 \\[3ex] S_2 = 15 \\[3ex] \implies \\[3ex] a + b = 15 \\[3ex] 5 + b = 15 \\[3ex] b = 15 - 5 \\[3ex] b = 10 \\[3ex] S_3 = \dfrac{5(3)^2}{2} + \dfrac{5(3)}{2} \\[5ex] = \dfrac{5(9)}{2} + \dfrac{15}{2} \\[5ex] = \dfrac{45}{2} + \dfrac{15}{2} \\[5ex] = \dfrac{45 + 15}{2} \\[5ex] = \dfrac{60}{2} \\[5ex] S_3 = 30 \\[3ex] \implies \\[3ex] a + b + c = 30 \\[3ex] 15 + c = 30 \\[3ex] c = 30 - 15 \\[3ex] c = 15 \\[3ex] a, b, c = 5, 10, 15...is\:\:an\:\:arithmetic\:\:sequence \\[3ex] But,\:\:we\:\:can\:\:still\:\:explore\:\:more \\[3ex] S_4 = \dfrac{5(4)^2}{2} + \dfrac{5(4)}{2} \\[5ex] = \dfrac{5(16)}{2} + \dfrac{20}{2} \\[5ex] = 5(8) + 10 \\[3ex] = 40 + 10 \\[3ex] S_4 = 50 \\[3ex] \implies \\[3ex] a + b + c + d = 50 \\[3ex] 30 + d = 50 \\[3ex] d = 50 - 30 \\[3ex] d = 20 \\[3ex] a, b, c, d = 5, 10, 15, 20 \\[3ex] a, b, c, d, e = 5, 10, 15, 20, 25 \\[3ex] a + b + c + d + e = 5 + 10 + 15 + 20 + 25 = 75 \\[3ex] Verify \\[3ex] S_5 = \dfrac{5(5)^2}{2} + \dfrac{5(5)}{2} \\[5ex] = \dfrac{5(25)}{2} + \dfrac{25}{2} \\[5ex] = \dfrac{125}{2} + \dfrac{25}{2} \\[5ex] = \dfrac{125 + 25}{2} \\[5ex] = \dfrac{150}{2} \\[5ex] S_5 = 75...verified \\[3ex] \therefore AP:\:\: 5, 10, 15, 20 \\[3ex] a = 5 \\[3ex] d = 10 - 5 = 5 \\[3ex] AP_n = a + d(n - 1) \\[3ex] AP_n = 5 + 5(n - 1) \\[3ex] AP_n = 5 + 5n - 5 \\[3ex] AP_n = 5n \\[3ex] \underline{Second\:\:Method} \\[3ex] S_n = \dfrac{5n^2}{2} + \dfrac{5n}{2} \\[5ex] S_n = \dfrac{n}{2}(5n + 5) \\[5ex] S_n = \dfrac{n}{2}(5 + 5n) \\[5ex] Compare\:\:to\:\:the\:\:Sum\:\:of\:\:the\:\:first\:\:n\:\:terms\:\:of\:\:an\:\:A.P \\[3ex] S_n = \dfrac{n}{2}(a + AP_n) \\[5ex] \therefore a = 5 \\[3ex] \therefore AP_n = 5n $
(10.) Find the sum of all natural numbers between 403 and 603 which are divisible by 7.


The closest number after 403 which is divisble by 7 is 406
The closest number before 603 which is divisible by 7 is 602
We get the sequence by adding 7 to subsequent terms after 406 until we get to 602
This implies that:
The sequence is an arithmetic sequence
The first term, a = 406
The common difference, d = 7
The last term, p = $AS_n$ = 602
So, the sequence looks like this:
406, 406 + 7 = 413, 413 + 7 = 420, ..., 602
406, 413, 420, ..., 602
We do not know the number of terms, n so we need to find it so we can find the sum of the sequence, $SAS_n$

$ AS_n = a + d(n - 1) \\[3ex] 602 = 406 + 7(n - 1) \\[3ex] 406 + 7(n - 1) = 602 \\[3ex] 7(n - 1) = 602 - 406 \\[3ex] 7(n - 1) = 196 \\[3ex] n - 1 = \dfrac{196}{7} \\[5ex] n - 1 = 28 \\[3ex] n = 28 + 1 \\[3ex] n = 29 \\[5ex] SAS_n = \dfrac{n(a + AS_n)}{2} \\[5ex] = \dfrac{29(406 + 602)}{2} \\[5ex] = \dfrac{29 * 1008}{2} \\[5ex] = 14616 $
(11.) (a.) If $3, x, y, 18$ are in Arithmetic Progression (A.P), find the values of x and y
(b.) (i.) The sum of the second and third terms of a geometric progression is six times the fourth term.
Find the two possible values of the common ratio.
(ii.) If the second term is 8 and the common ratio is positive, find the first six terms.


$ (a.) \\[3ex] AP:\:\: 3, x, y, 18 \\[3ex] d = x - 3 \\[3ex] d = y - x \\[3ex] d = 18 - y \\[3ex] d = d \\[3ex] \implies \\[3ex] x - 3 = y - x \\[3ex] 2x - 3 = y \\[3ex] y = 2x - 3...eqn.(1) \\[3ex] \implies \\[3ex] y - x = 18 - y \\[3ex] 2y - x = 18 \\[3ex] Substitute\:\: (2x - 3)\:\:for\:\:y \\[3ex] 2(2x - 3) - x = 18 \\[3ex] 4x - 6 - x = 18 \\[3ex] 3x - 6 = 18 \\[3ex] 3x = 18 + 6 \\[3ex] 3x = 24 \\[3ex] x = \dfrac{24}{3} \\[5ex] x = 8 \\[3ex] Substitute\:\:x = 8\:\:in\:\:eqn.(1) \\[3ex] y = 2x - 3 \\[3ex] y = 2(8) - 3 \\[3ex] y = 16 - 3 \\[3ex] y = 13 \\[3ex] AP:\:\: 3, 8, 13, 18 \\[5ex] (b.) \\[3ex] (i.) \\[3ex] GP_n = ar^{n - 1} \\[3ex] GP_2 = ar \\[3ex] GP_3 = ar^2 \\[3ex] GP_4 = ar^3 \\[3ex] GP_2 + GP_3 = 6 * GP_4 \\[3ex] ar + ar^2 = 6(ar^3) \\[3ex] a(r + r^2) = 6ar^3 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:a \\[3ex] r + r^2 = 6r^3 \\[3ex] 0 = 6r^3 - r^2 - r \\[3ex] 6r^3 - r^2 - r = 0 \\[3ex] r(6r^2 - r - 1) = 0 \\[3ex] Factor\:\: 6r^2 - r - 1 \\[3ex] 6r^2 + 2r - 3r - 1 \\[3ex] 2r(3r + 1) - 1(3r + 1) \\[3ex] (3r + 1)(2r - 1) \\[3ex] \implies \\[3ex] r(3r + 1)(2r - 1) = 0 \\[3ex] r = 0 \:\:OR \\[3ex] 3r + 1 = 0 \\[3ex] 3r = 0 - 1 \\[3ex] 3r = -1 \\[3ex] r = -\dfrac{1}{3} \:\:OR \\[5ex] 2r - 1 = 0 \\[3ex] 2r = 0 + 1 \\[3ex] 2r = 1 \\[3ex] r = \dfrac{1}{2} \\[5ex] r = -\dfrac{1}{3},0,\dfrac{1}{2} \\[5ex] r \ne 0 \\[3ex] \therefore r = -\dfrac{1}{3},\dfrac{1}{2} \\[5ex] (ii.) \\[3ex] r \:\:is\:\:positive \\[3ex] \implies \\[3ex] r = \dfrac{1}{2} \\[5ex] GP_2 = ar = 8 \\[3ex] a = 8 \div r \\[3ex] a = 8 \div \dfrac{1}{2} \\[5ex] a = 8 * \dfrac{2}{1} \\[5ex] a = 16 \\[3ex] GP_3 = ar^2 = ar * r = GP_2 * r = 8 * \dfrac{1}{2} = 4 \\[5ex] GP_4 = GP_3 * r = 4 * \dfrac{1}{2} = 2 \\[5ex] GP_5 = GP_4 * r = 2 * \dfrac{1}{2} = 1 \\[5ex] GP_6 = GP_5 * r = 1 * \dfrac{1}{2} = \dfrac{1}{2} \\[5ex] GP:\:\: 16, 8, 4, 2, 1, \dfrac{1}{2} $
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