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Sequences and Series

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For the Classic ACT exam:
The ACT Mathematics test is a timed exam...60 questions in 60 minutes
This implies that you have to solve each question in one minute.
Each of the first 20 questions (less challenging) will typically take less than a minute a solve.
Each of the next 20 questions (medium challenging) may take about a minute to solve.
Each of the last 20 questions (more challenging) may take more than a minute to solve.
The goal is to maximize your time.
You use the time saved on the questions you solve in less than a minute to solve 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.
Also: please note that unless specified otherwise, geometric figures are drawn to scale. So, you can figure out the correct answer by eliminating the incorrect options.
Other suggestions are listed in the solutions/explanations as applicable.

These are the solutions to the ACT past questions on Sequences.
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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.) On the first day of school, Mr. Thibodeaux gave his third-grade students 9 new spelling words to learn.
On each day of school after that, he gave the students 6 new spelling words.
How many new spelling words had he given the students by the end of the 30th day of school?

$ F.\;\; 174 \\[3ex] G.\;\; 180 \\[3ex] H.\;\; 183 \\[3ex] J.\;\; 189 \\[3ex] K.\;\; 195 \\[3ex] $

1st day: 9 new spelling words
2nd day: 6 new spelling words, total = 9 + 6 = 15 new spelling words
3rd day: 6 new spelling words, total = 9 + 6(2) = 9 + 12 = 21 new spelling words
4th day: 6 new spelling words, total = 9 + 6(3) = 9 + 18 = 27 new spelling words
This is an Arithmetic Sequence

$ a = 9 \\[3ex] d = 6 \\[3ex] n = 30 \\[3ex] AS_n = a + d(n - 1) \\[5ex] AS_{30} = 9 + 6(30 - 1) \\[5ex] = 9 + 6(29) \\[3ex] = 9 + 174 \\[3ex] = 183 \\[3ex] $ By the end of the 30th day of school, Mr. Thibodeaux would have given his third-grade students 183 new spelling words.
Calculator 1
(2.) In the triangular arrangement of fractions below, the first and last fraction in row n is $\dfrac{1}{n}$.
Any other entry is the sum of the 2 fractions on either side of that entry in the row directly beneath it.
What is the 3rd fraction in the 5th row?

Number 2

$ F.\;\; \dfrac{1}{4} \\[5ex] G.\;\; \dfrac{1}{15} \\[5ex] H.\;\; \dfrac{1}{25} \\[5ex] J.\;\; \dfrac{1}{30} \\[5ex] K.\;\; \dfrac{1}{100} \\[5ex] $

$ \text{Just to confirm} \\[3ex] \dfrac{1}{12} + \dfrac{1}{12} \\[5ex] = \dfrac{2}{12} \\[5ex] = \dfrac{1}{6} \\[5ex] Similarly: \\[3ex] \dfrac{1}{20} + ? = \dfrac{1}{12} \\[5ex] ? = \dfrac{1}{12} - \dfrac{1}{20} \\[5ex] = \dfrac{5 - 3}{60} \\[5ex] = \dfrac{2}{60} \\[5ex] = \dfrac{1}{30} $
(3.) Some consecutive terms of a geometric sequence are below.

$ ..., 64, 48, 36, 27, ... \\[3ex] $ What is the common ratio of this sequence?

$ F.\;\; -16 \\[3ex] G.\;\; -12 \\[3ex] H.\;\; -9 \\[3ex] J.\;\; \dfrac{3}{4} \\[5ex] K.\;\; \dfrac{4}{3} \\[5ex] $

$ \text{common ratio}, r = \dfrac{48}{64} \\[5ex] = \dfrac{48 \div 16}{64 \div 16} \\[5ex] = \dfrac{3}{4} $

Calculator 3
(4.) A sequence is defined for all positive integers by $s_n = 2s_{(n - 1)} + n + 1$ and $s_1 = 3$
What is $s_4?$

$ F.\:\: 9 \\[3ex] G.\:\: 18 \\[3ex] H.\:\: 22 \\[3ex] J.\:\: 49 \\[3ex] K.\:\: 111 \\[3ex] $

Recursive Seqeunce

$ s_n = 2s_{(n - 1)} + n + 1 \\[3ex] s_1 = 3 \\[3ex] s_2 = 2s_{(2 - 1)} + 2 + 1 \\[3ex] s_2 = 2s_1 + 2 + 1 \\[3ex] s_2 = 2(3) + 2 + 1 \\[3ex] s_2 = 6 + 2 + 1 \\[3ex] s_2 = 9 \\[3ex] s_3 = 2s_{(3 - 1)} + 3 + 1 \\[3ex] s_3 = 2s_2 + 3 + 1 \\[3ex] s_3 = 2(9) + 3 + 1 \\[3ex] s_3 = 18 + 3 + 1 \\[3ex] s_3 = 22 \\[3ex] s_4 = 2s_{(4 - 1)} + 4 + 1 \\[3ex] s_4 = 2s_3 + 4 + 1 \\[3ex] s_4 = 2(22) + 4 + 1 \\[3ex] s_4 = 44 + 4 + 1 \\[3ex] s_4 = 49 $
(5.) The first 3 terms of an arithmetic sequence are $2\dfrac{1}{6}$, $3\dfrac{1}{3}$, and $4\dfrac{1}{2}$ in that order.
What is the fourth term of the sequence?

$ A.\:\: 4\dfrac{5}{6} \\[5ex] B.\:\: 5\dfrac{1}{6} \\[5ex] C.\:\: 5\dfrac{1}{3} \\[5ex] D.\:\: 5\dfrac{2}{3} \\[5ex] E.\:\: 6 \\[3ex] $

$ \text{Let the fourth term } = p \\[3ex] 2\dfrac{1}{6}, \;\; 3\dfrac{1}{3}, \;\; 4\dfrac{1}{2}, \;\; p \\[5ex] 2\dfrac{1}{6} = \dfrac{6 * 2 + 1}{6} = \dfrac{12 + 1}{6} = \dfrac{13}{6} \\[5ex] 3\dfrac{1}{3} = \dfrac{3 * 3 + 1}{3} = \dfrac{9 + 1}{3} = \dfrac{10}{3} \\[5ex] d = \dfrac{10}{3} - \dfrac{13}{6} \\[5ex] = \dfrac{20}{6} - \dfrac{13}{6} \\[5ex] = \dfrac{20 - 13}{6} \\[5ex] d = \dfrac{7}{6} \\[5ex] 3rd\:\:term + common\:\:difference = fourth\:\:term \\[3ex] p = 4\dfrac{1}{2} + d \\[5ex] 4\dfrac{1}{2} = \dfrac{2 * 4 + 1}{2} = \dfrac{8 + 1}{2} = \dfrac{9}{2} \\[5ex] p = \dfrac{9}{2} + \dfrac{7}{6} \\[5ex] = \dfrac{27}{6} + \dfrac{7}{6} \\[5ex] = \dfrac{27 + 7}{6} \\[5ex] = \dfrac{34}{6} \\[5ex] = \dfrac{17}{3} \\[5ex] = 5\dfrac{2}{3} \\[5ex] $ The fourth term is $5\dfrac{2}{3}$
(6.) 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 $
(7.) The degree measures of the interior angles of $\triangle ABC$, shown below, form an arithmetic sequence with common difference 10°.
What is the first term of the sequence?

Number 7

$ F.\:\: 80^\circ \\[3ex] G.\:\: 60^\circ \\[3ex] H.\:\: 50^\circ \\[3ex] J.\:\: 40^\circ \\[3ex] K.\:\: 30^\circ \\[3ex] $

$ d = 10^\circ \\[3ex] Let\:\:the\:\:smallest\:\:angle = first\:\:term = a \\[3ex] Second\:\:angle = second\:\:term = a + d = a + 10 \\[3ex] Third\:\:angle = third\:\:term = a + 2d = a + 2(10) = a + 20 \\[3ex] a + (a + d) + (a + 2d) = 180...sum\:\:of\:\:\angle s\:\:of\:\:a\:\:\triangle \\[3ex] a + (a + 10) + (a + 20) = 180 \\[3ex] a + a + 10 + a + 20 = 180 \\[3ex] 3a + 30 = 180 \\[3ex] 3a = 180 - 30 \\[3ex] 3a = 150 \\[3ex] a = \dfrac{150}{3} \\[5ex] a = 50^\circ $
(8.) The recursive formula for a sequence is given below, where $a_n$ is the value of the nth term.

$ a_1 = 10 \\[3ex] a_n = a_{n - 1} + 5 \\[3ex] $ Which of the following equations is an explicit formula for this sequence?

$ A.\:\: a_n = -5n + 10 \\[3ex] B.\:\: a_n = 5n + 5 \\[3ex] C.\:\: a_n = 5n + 10 \\[3ex] D.\:\: a_n = 10n - 5 \\[3ex] E.\:\: a_n = 10n + 5 \\[3ex] $

Recursive Seqeunce

$ a_1 = 10 \\[3ex] a_n = a_{n - 1} + 5 \\[3ex] a_2 = a_{2 - 1} + 5 \\[3ex] a_2 = a_1 + 5 = 10 + 5 = 15 \\[3ex] a_3 = a_{3 - 1} + 5 \\[3ex] a_3 = a_2 + 5 = 15 + 5 = 20 \\[3ex] a_1, a_2, a_3 = 10, 15, 20...arithmetic\:\:sequence \\[3ex] a = a_1 = 10 \\[3ex] d = 15 - 10 = 5 \\[3ex] a_n = a + d(n - 1) \\[3ex] a_n = 10 + 5(n - 1) \\[3ex] a_n = 10 + 5n - 5 \\[3ex] a_n = 5n + 5 $
(9.) 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} $
(10.) The 3rd and 4th terms of an arithmetic sequence are 13 and 18, respectively.
What is the 50th term of the sequence?

$ A.\:\: 248 \\[3ex] B.\:\: 250 \\[3ex] C.\:\: 253 \\[3ex] D.\:\: 258 \\[3ex] E.\:\: 263 \\[3ex] $

$ 3rd\:\:term = AS_3 = a + 2d = 13...eqn.(1) \\[3ex] 4th\:\:term = AS_4 = a + 3d = 18...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \rightarrow 3d - 2d = 18 - 13 \\[3ex] d = 5 \\[3ex] From\:\:eqn.(1);\:\: a = 13 - 2d \\[3ex] a = 13 - 2(5) \\[3ex] a = 13 - 10 \\[3ex] a = 3 \\[3ex] AS_{50} = a + 49d \\[3ex] = 3 + 49(5) \\[3ex] = 3 + 245 \\[3ex] AS_{50} = 248 $
(11.) The function below is defined for constants a and b and for all positive integers n. $$ r(n) = ab^n $$
It is known that $r(1) = \dfrac{1}{2},\;\;r(2) = \dfrac{3}{2},\;\;r(3) = \dfrac{9}{2}$, and $r(4) = \dfrac{27}{2}$.

Which of the following functions is equivalent to $r(n)$?

$ F.\;\; f(n) = 3\left(\dfrac{1}{2}\right)^{n - 1} \\[6ex] G.\;\; g(n) = 3\left(\dfrac{1}{2}\right)^n \\[6ex] H.\;\; h(n) = \dfrac{1}{2}\left(\dfrac{3}{2}\right)^{n - 1} \\[6ex] J.\;\; j(n) = \dfrac{1}{2}(3)^{n - 1} \\[6ex] K.\;\; k(n) = \dfrac{1}{2}(3)^n \\[6ex] $

$ r(n) = ab^n \\[4ex] r(1) = ab^1 = \dfrac{1}{2} \\[5ex] ab = \dfrac{1}{2}...eqn.(1) \\[7ex] r(2) = ab^2 = \dfrac{3}{2} \\[5ex] ab^2 = \dfrac{3}{2}...eqn.(2) \\[7ex] eqn.(2) \div eqn.(1) \implies \\[3ex] \dfrac{ab^2}{ab} = \dfrac{3}{2} \div \dfrac{1}{2} \\[6ex] b = \dfrac{3}{2} \cdot \dfrac{2}{1} \\[5ex] b = 3 \\[3ex] Substitute\;\;b = 3\;\;into\;\; eqn.(1) \\[3ex] a \cdot 3 = \dfrac{1}{2}...eqn.(1) \\[5ex] a = \dfrac{1}{2} \cdot \dfrac{1}{3} \\[5ex] a = \dfrac{1}{6} \\[5ex] \implies \\[3ex] r(n) = \dfrac{1}{6} \cdot 3^n \\[6ex] = \dfrac{1}{2 \cdot 3} \cdot 3^n \\[6ex] = \dfrac{1}{2} \cdot \dfrac{1}{3} \cdot 3^n \\[6ex] = \dfrac{1}{2}\left(\dfrac{3^n}{3^1}\right) ...Law\;2...Exp \\[6ex] = \dfrac{1}{2}(3)^{n - 1} $
(12.) 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.\;\; \text{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 $
(13.) The first 5 terms of a sequence are given in the table below.
The sequence is defined by setting $a_1 = 9$ and $a_n = a_{n - 1} + (n - 1)^2$ for $n \ge 2$
What is the sixth term, $a_6$, of this sequence?

$a_1$ $a_2$ $a_3$ $a_4$ $a_5$ $a_6$
$9$ $10$ $14$ $23$ $39$ $?$

$ A.\:\: 62 \\[3ex] B.\:\: 64 \\[3ex] C.\:\: 76 \\[3ex] D.\:\: 78 \\[3ex] E.\:\: 95 \\[3ex] $

Recursive Seqeunce

$ a_n = a_{n - 1} + (n - 1)^2 \\[3ex] a_6 = a_{6 - 1} + (6 - 1)^2 \\[3ex] a_6 = a_{5} + 5^2 \\[3ex] a_6 = 39 + 25 \\[3ex] a_6 = 64 $
(14.) 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} $
(15.) The 13th, 14th, and 15th terms of an arithmetic sequence are 61, 65, and 69 respectively.
What are the first $2$ terms of the sequence?

$ A.\:\: 4, 8 \\[3ex] B.\:\: 9, 11 \\[3ex] C.\:\: 9, 13 \\[3ex] D.\:\: 13, 15 \\[3ex] E.\:\: 13, 17 \\[3ex] $

$ AS_n = a + d(n - 1) \\[3ex] AS_{13} = a + 12d = 61...eqn.(1) \\[3ex] AS_{14} = a + 13d = 65...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \\[3ex] a + 13d - (a + 12d) = 65 - 61 \\[3ex] a + 13d - a - 12d = 4 \\[3ex] d = 4 \\[3ex] From\:\:eqn.(1) \\[3ex] a + 12d = 61 \\[3ex] a = 61 - 12d \\[3ex] a = 61 - 12(4) \\[3ex] a = 61 - 48 \\[3ex] a = 13 \\[3ex] 2nd\:\:term = a + d = 13 + 4 = 17 \\[3ex] First\:\:two\:\:terms = 13, 17 $
(16.) The second term of an arithmetic sequence is 12, and the third term is 6.
What is the first term?
(Note: In an arithmetic sequence, consecutive terms differ by the same amount.)

$ A.\:\: -12 \\[3ex] B.\:\: -6 \\[3ex] C.\:\: \dfrac{1}{12} \\[5ex] D.\:\: 6 \\[3ex] E.\:\: 18 \\[3ex] $

The ACT is a timed test: a question should typically take a minute to solve
We shall do it two ways
The first method is much faster. It is recommended for the ACT

$ \underline{\text{First Method: Faster}} \\[3ex] d = 3rd\:\:term - 2nd\:\: term \\[3ex] d = 6 - 12 = -6 \\[3ex] Also,\:\: d = 2nd\:\: term - 1st\:\: term \\[3ex] \rightarrow -6 = 12 - a \\[3ex] a = 12 + 6 \\[3ex] a = 18 \\[5ex] \underline{\text{Second Method: Longer}} \\[3ex] AS_n = a + d(n - 1) \\[3ex] AS_2 = a + d = 12 ...eqn.(1) \\[3ex] AS_3 = a + 2d = 6 ...eqn.(2) \\[3ex] 2 * eqn.(1) \implies 2(a + d) = 2(12) \\[3ex] 2 * eqn.(1) \implies 2a + 2d = 24...eqn.(3) \\[3ex] eqn.(3) - eqn.(2) \implies \\[3ex] (2a + 2d) - (a + 2d) = 24 - 6 \\[3ex] 2a + 2d - a - 2d = 18 \\[3ex] a = 18 \\[3ex] $ The first term is 18
(17.) The nth term of an arithmetic progression is given by the formula $a_n = a_1 + (n - 1)d$, where d is the common difference and $a_1$ is the first term.
If the third term of an arithmetic progression is $\dfrac{5}{2}$ and the sixth term is $\dfrac{1}{4}$, what is the seventh term?

$ A.\:\: -\dfrac{1}{2} \\[5ex] B.\:\: 0 \\[3ex] C.\:\: \dfrac{1}{2} \\[5ex] D.\:\: \dfrac{3}{4} \\[5ex] E.\:\: 1 \\[3ex] $

$ Let\:\:the\:\:seventh\:\:term = p \\[3ex] 3rd\:\:term = a + 2d = \dfrac{5}{2}...eqn.(1) \\[5ex] 6th\:\:term = a + 5d = \dfrac{1}{4}...eqn.(2) \\[5ex] eqn.(2) - eqn.(1) \implies (a - a) + (5d - 2d) = \dfrac{1}{4} - \dfrac{5}{2} \\[5ex] 3d = \dfrac{1}{4} - \dfrac{10}{4} \\[5ex] 3d = \dfrac{1 - 10}{4} \\[5ex] 3d = -\dfrac{9}{4} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\: \dfrac{1}{3} \\[5ex] \dfrac{1}{3} * 3d = \dfrac{1}{3} * -\dfrac{9}{4} \\[5ex] d = -\dfrac{3}{4} \\[5ex] 6th\:\:term + common\:\:difference = seventh\:\:term \\[3ex] p = \dfrac{1}{4} + - \dfrac{3}{4} \\[5ex] = \dfrac{1}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{1 - 3}{4} \\[5ex] = \dfrac{-2}{4} \\[5ex] p = -\dfrac{1}{2} \\[5ex] $ The seventh term is $-\dfrac{1}{2}$
(18.) 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 $-64$
(19.) A finite arithmetic sequence has 7 terms, and the first term is $\dfrac{3}{4}$.
What is the difference between the mean and the median of the 7 terms?

$ A.\:\: 0 \\[3ex] B.\:\: \dfrac{3}{4} \\[5ex] C.\:\: \dfrac{4}{3} \\[5ex] D.\:\: 3 \\[3ex] E.\:\: 4 \\[3ex] $

$ n = 7 \\[3ex] a = \dfrac{3}{4} \\[5ex] AS_n = a + d(n - 1) \\[3ex] AS_7 = a + 6d \\[3ex] Mean, \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \Sigma x = SAS_7 \\[3ex] SAS_7 = \dfrac{n}{2}[a + AS_7] \\[5ex] = \dfrac{7}{2} * [a + a + 6d] \\[5ex] = \dfrac{7}{2} * [2a + 6d] \\[5ex] = 7a + 21d \\[3ex] \Sigma x = SAS_7 = 7a + 21d \\[3ex] \implies Mean = \dfrac{7a + 21d}{7} \\[5ex] = \dfrac{7(a + 3d)}{7} \\[5ex] = a + 3d \\[3ex] Mean = a + 3d \\[3ex] Because\:\:the\:\:sample\:\:size\:\:is\:\:odd,\:\:n = 7 \\[3ex] Median = 4th\:\:term \\[3ex] Median = AS_4 \\[3ex] AS_4 = a + 3d \\[3ex] \implies Median = a + 3d \\[3ex] Mean - Median \\[3ex] = (a + 3d) - (a + 3d) \\[3ex] = 0 $
(20.) 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 $




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(21.) The table below shows the first $5$ terms of an arithmetic sequence.
Which of the following is a general expression for the $nth$ term?

Term position (n) nth term
1 1
2 5
3 9
4 13
5 17

$ F.\;\; 2n - 1 \\[3ex] G.\;\; 3n - 2 \\[3ex] H.\;\; 4n - 3 \\[3ex] J.\;\; 5n - 4 \\[3ex] K.\;\; 6n - 5 \\[3ex] $

$ AS_n = a + d(n - 1) \\[3ex] From\;\;the\;\;table \\[3ex] n = 1,\;\; AS_1 = 1 \\[3ex] \implies a = 1 \;\;because\;\;AS_1 = 1 \\[3ex] n = 2,\;\; AS_2 = 5 \\[3ex] AS_2 = a + d \\[3ex] 5 = 1 + d \\[3ex] 5 = 1 + d \\[3ex] 5 - 1 = d \\[3ex] 4 = d \\[3ex] d = 4 \\[3ex] Confirm\;\;for\;\; n = 3 \\[3ex] n = 3,\;\; AS_3 = 9 \\[3ex] AS_3 = a + 2d \\[3ex] 9 = 1 + 2d \\[3ex] 9 - 1 = 2d \\[3ex] 8 = 2d \\[3ex] 2d = 8 \\[3ex] d = \dfrac{8}{2} \\[5ex] d = 4 \\[3ex] Three\;\;terms\;\;is\;\;enough \\[3ex] AS_n = a + d(n - 1) \\[3ex] AS_n = 1 + 4(n - 1) \\[3ex] AS_n = 1 + 4n - 4 \\[3ex] AS_n = 4n - 3 $
(22.) What is the sum of the first 4 terms of the arithmetic sequence in which the 6th term is 8 and the 10th term is 13?

$ F.\:\: 10.5 \\[3ex] G.\:\: 14.5 \\[3ex] H.\:\: 18 \\[3ex] J.\:\: 21.25 \\[3ex] K.\:\: 39.5 \\[3ex] $

$ AS_n = a + d(n - 1) \\[3ex] AS_{6} = a + 5d = 8...eqn.(1) \\[3ex] AS_{10} = a + 9d = 13...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \rightarrow 4d = 5 \\[3ex] d = \dfrac{5}{4} \\[5ex] d = 1.25 \\[3ex] From\:\:eqn.(1);\:\: a = 8 - 5d \\[3ex] a = 8 - 5(1.25) \\[3ex] a = 8 - 6.25 \\[3ex] a = 1.75 \\[3ex] Two\:\:ways\:\:to\:\:calculate\:\:SAS_{4} \\[3ex] Choose\:\:your\:\:preference \\[3ex] \underline{First\:\:Method} \\[3ex] SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] SAS_{4} = \dfrac{4}{2}[2(1.75) + 1.25(4 - 1)] \\[5ex] = 2[3.5 + 1.25(3)] \\[3ex] = 2[3.5 + 3.75] \\[3ex] = 2(7.25) \\[3ex] SAS_{4} = 14.5 \\[3ex] \underline{Second\:\:Method} \\[3ex] SAS_n = \dfrac{n}{2}(a + p) \\[5ex] SAS_n = \dfrac{n}{2}(a + AS_{n}) \\[5ex] AS_{4} = a + 3d \\[3ex] AS_{4} = 1.75 + 3(1.25) \\[3ex] AS_{4} = 1.75 + 3.75 \\[3ex] AS_{4} = 5.5 \\[3ex] SAS_{4} = \dfrac{4}{2}(1.75 + 5.5) \\[5ex] SAS_{4} = 2(7.25) \\[3ex] SAS_{4} = 14.5 $
(23.) In an arithmetic series, the terms of the series are equally spread out.
For example, in 1 + 5 + 9 + 13 + 17, consecutive terms are 4 apart.
If the first term in an arithmetic series is 3, the last term is 136, and the sum is 1,390, what are the first 3 terms?

$ A.\;\; 3, 10, 17 \\[3ex] B.\;\; 3, 23, 43 \\[3ex] C.\;\; 3, 36\dfrac{1}{3}, 70 \\[5ex] D.\;\; 3, 69\dfrac{1}{2}, 136 \\[5ex] E.\;\; 3, 139, 1,251 \\[3ex] $

$ a = 3 \\[3ex] p = 136 \\[3ex] SAS_n = 1390 \\[3ex] d = ? \\[3ex] d = \dfrac{(p - a)(p + a)}{2 * SAS_n - p - a} \\[5ex] d = \dfrac{(136 - 3)(136 + 3)}{2 * 1390 - 136 - 3} \\[5ex] d = \dfrac{133 * 139}{2780 - 136 - 3} \\[5ex] d = \dfrac{18487}{2641} \\[5ex] d = 7 \\[3ex] 1st\;\;term = 3 \\[3ex] 2nd\;\;term = 3 + 7 = 10 \\[3ex] 3rd\;\;term = 10 + 7 = 17 \\[3ex] First\;\;three\;\;terms = 3, 10, 17 $
(24.) 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\;\; \text{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 $
(25.) Which of the following statements describes the total number of dots in the first n rows of the triangular arrangement illustrated below?

Number 25

A. This total is always equal to 25 regardless of the number of rows.
B. This total is equal to twice the number of rows.
C. This total is equal to 5 times the number of rows.
D. This total is equal to the square of the number of rows.
E. There is no consistent relationship between this total and the number of rows.


1st row: 1 dot
2nd row: 3 dots
3rd row: 5 dots
4th row: 7 dots
5th row: 9 dots
This is an arithmetic sequence
What is the sum of the first n terms of the arithmetic sequence?

$ a = 1 \\[3ex] d = 3 - 1 = 2 \\[3ex] SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] SAS_n = \dfrac{n}{2}[2(1) + 2(n - 1)] \\[5ex] SAS_n = \dfrac{n}{2}[2 + 2n - 2] \\[5ex] SAS_n = \dfrac{n}{2}(2n) \\[3ex] SAS_n = n(n) \\[3ex] SAS_n = n^2 \\[3ex] $ This total is equal to the square of the number of rows.
(26.) 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.\;\; \text{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.) The sum of a sequence of consecutive odd numbers, where the smallest term is 1, is always a perfect square.
For example, $1 + 3 = 2^2$ and $1 + 3 + 5 + 7 = 4^2$
One of the sequences described above has a sum of 144
What is the largest odd number in the sequence?

$ F.\:\: 11 \\[3ex] G.\:\: 13 \\[3ex] H.\:\: 15 \\[3ex] J.\:\: 23 \\[3ex] K.\:\: 73 \\[3ex] $

Notice the pattern

$ 1 + 3\:\:are\:\:2\:\:numbers \\[3ex] 1 + 3 = 4 = 2^2 \\[3ex] 1 + 3 + 5\:\:are\:\:3\:\:numbers \\[3ex] 1 + 3 + 5 = 9 = 3^3 \\[3ex] 1 + 3 + 5 + 7 \:\:are\:\:4\:\:numbers \\[3ex] 1 + 3 + 5 + 7 = 16 = 4^2 \\[3ex] 1 + 3 + 5 + 7 + 9 \:\:are\:\:5\:\:numbers \\[3ex] 1 + 3 + 5 + 7 + 9 = 5^2 \\[3ex] Similarly \\[3ex] For\:\:144 \\[3ex] 144 = 12^2 \\[3ex] We\:\:need\:\:12\:\:numbers \\[3ex] 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 \:\:are\:\:12\:\:numbers \\[3ex] 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12^2 \\[3ex] Highest\:\:odd\:\:number = 23 $
(28.) 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] $

$ \text{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
(29.) There is a pattern when adding the cubes of the first c consecutive counting numbers, as illustrated below.

$ ~~~~~~~~~~ 1^3 + 2^3 = 9 = (1 + 2)^2 \\[3ex] ~~~~~~~~~~ 1^3 + 2^3 + 3^3 = 36 = (1 + 2 + 3)^2 \\[3ex] $ Which of the following is an expression for the sum of the cubes of the first c consecutive counting numbers?

$ F.\;\; (c + 1)^3 \\[3ex] G.\;\; (c + 1)^2 \\[3ex] H.\;\; (1 + 2 + ... + c)^c \\[3ex] J.\;\; (1 + 2 + ... + c)^3 \\[3ex] K.\;\; (1 + 2 + ... + c)^2 \\[3ex] $

$ ~~~~~~~~~~ 1^3 + 2^3 = 9 = (1 + 2)^2 \\[3ex] ~~~~~~~~~~ 1^3 + 2^3 + 3^3 = 36 = (1 + 2 + 3)^2 \\[3ex] Similarly: \\[3ex] ~~~~~~~~~~ 1^3 + 2^3 + ... + c^3 = (1 + 2 + ... + c)^2 \\[3ex] $ The correct answer is Option K.
(30.) The second term of an arithmetic sequence is −11, and the third term is −38.
What is the first term?
(Note: In an arithmetic sequence, consecutive terms differ by the same amount.)

$ F.\:\: -27 \\[3ex] G.\:\: \dfrac{1}{11} \\[5ex] H.\:\: 11 \\[3ex] J.\:\: 16 \\[3ex] K.\:\: 27 \\[3ex] $

The ACT is a timed test: a question should typically take a minute to solve
We shall do it two ways
The first method is much faster. It is recommended for the ACT

$ \underline{\text{First Method: Faster}} \\[3ex] d = 3rd\:\:term - 2nd\:\: term \\[3ex] d = -38 - (-11) = -38 + 11 = -27 \\[3ex] Also,\:\: d = 2nd\:\: term - 1st\:\: term \\[3ex] \rightarrow -27 = -11 - a \\[3ex] a = -11 + 27 \\[3ex] a = 16 \\[5ex] \underline{\text{Second Method: Longer}} \\[3ex] AS_n = a + d(n - 1) \\[3ex] AS_2 = a + d = -11 ...eqn.(1) \\[3ex] AS_3 = a + 2d = -38 ...eqn.(2) \\[3ex] From\:\: eqn.(1);\:\: d = -11 - a \\[3ex] Substitute\:\: (-11 - a)\:\:for\:\:d\:\:in\:\:eqn.(2) \\[3ex] a + 2(-11 - a) = -38 \\[3ex] a - 22 - 2a = -38 \\[3ex] -a = -38 + 22 \\[3ex] -a = -16 \\[3ex] a = \dfrac{-16}{-1} \\[5ex] a = 16 \\[3ex] $ The first term is 16
(31.) The first $4$ elements of a pattern are shown below.
Each element is composed of small squares that are $18$ inches wide and $18$ inches long.
Each element is a square with both dimensions $18$ inches less than the dimensions of the next element.
What is the perimeter, in feet, of the 5th element?

Number 31

$ F.\;\; 6 \\[3ex] G.\;\; 7.5 \\[3ex] H.\;\; 20 \\[3ex] J.\;\; 25 \\[3ex] K.\;\; 30 \\[3ex] $

Notice the pattern:

$ 1\;\;square = 1^2 \\[3ex] 4\;\; squares = 2^2 \\[3ex] 9\;\; squares = 3^2 \\[3ex] 16\;\; squares = 4^2 \\[3ex] $ This is a Square sequence
Next pattern would be $25$ squares because of $5^2$
Side of each square = $18$ inches
Length of next pattern = 5 square sides (outer) = (5 * 18) inches
Width of next pattern = 5 squares sides (outer) = (5 * 18) inches

$ Perimeter\;\;of\;\;next\;\;pattern \\[3ex] = (2 * Length) + (2 * Width) \\[3ex] = (2 * 5 * 18) + (2 * 5 * 18) \\[3ex] = 180 + 180 \\[3ex] = 360\;inches \\[3ex] 12\;inches = 1\;\;foot \\[3ex] \therefore 360\;inches \\[3ex] = \dfrac{360}{12} \;feet \\[5ex] = 30\;feet $
(32.) 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 $
(33.) A sequence of 5 numbers has 6 as its first term and 32 as its last term.
The first 3 numbers are an arithmetic sequence.
The last 3 numbers are a geometric sequence with a common ratio of 2.
What is the common difference among the first 3 terms?

$ F.\:\: 0 \\[3ex] G.\:\: 1 \\[3ex] H.\:\: 61 \\[3ex] J.\:\: 67 \\[3ex] K.\:\: 72 \\[3ex] $

Let us set up the sequence...have a visual representation

$ .....~~~~.....~~~~.....~~~~.....~~~~.....~~~~..... \\[3ex] |~~~~~~~~~~~~~~~~AS~~~~~~~~~~~~~~~~|~~~~~~~~~~~~~~GS~~~~~~~~~~~~~~| \\[3ex] For\:\:the\:\:AS: \\[3ex] First\:\:term = a = 6 \\[3ex] Second\:\:term = a + d = 6 + d \\[3ex] Third\:\:term = a + 2d = 6 + 2d \\[3ex] For the GS: \\[3ex] Fourth\:\:term = r * (6 + 2d) = r(6 + 2d) \\[3ex] Fifth\:\:term = r * r * (6 + 2d) = r^2(6 + 2d) \\[3ex] Fifth\:\:term\:\:is\:\:also\:\:32 \\[3ex] 6,~~~~6 + d,~~~~6 + 2d,~~~~r(6 + d),~~~~r^2(6 + d) \\[3ex] 6,~~~~6 + d,~~~~6 + 2d,~~~~r(6 + d),~~~~32 \\[3ex] \rightarrow r^2(6 + 2d) = 32 \\[3ex] r = 2...given \\[3ex] 6 + 2d = \dfrac{32}{r^2} \\[5ex] 6 + 2d = \dfrac{32}{2^2} \\[5ex] 6 + 2d = \dfrac{32}{4} \\[5ex] 6 + 2d = 8 \\[3ex] 2d = 8 - 6 \\[3ex] 2d = 2 \\[3ex] d = \dfrac{2}{2} \\[5ex] d = 1...correct\:\:answer \\[3ex] Explore\:\:more \\[3ex] \underline{Check} \\[3ex] 6 + d = 6 + 1 = 7 \\[3ex] 6 + 2d = 6 + 2(1) = 6 + 2 = 8 \\[3ex] r(6 + d) = 2(7) = 14 \\[3ex] r^2(6 + d) = 32...given \\[3ex] Sequence\:\:is\:\: 6, 7, 8, 16, 32 \\[3ex] 6, 7, 8\:\:is\:\:an\:\:AS \\[3ex] 8, 16, 32\:\:is\:\:a\:\:GS $
(34.) 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 $
(35.)


(36.) What is the sum of the first $80$ terms of the arithmetic sequence $1, 2, 3, ...?$

$ F.\:\: 160 \\[3ex] G.\:\: 320 \\[3ex] H.\:\: 800 \\[3ex] J.\:\: 3,240 \\[3ex] K.\:\: 6,400 \\[3ex] $

$ AS:\:\: 1, 2, 3, ... \\[3ex] 80th\:\:term = 80...easily\:\:recognizable \\[3ex] a = 1 \\[3ex] last\:\:term = p = 80 \\[3ex] n = 80 \\[3ex] SAS_n = \dfrac{n}{2}(a + p) \\[5ex] SAS_{80} = \dfrac{80}{2}(1 + 80) \\[5ex] SAS_{80} = 40(81) \\[3ex] SAS_{80} = 3240 $
(37.)


(38.) Consecutive terms of a certain arithmetic sequence have a positive common difference.
The sum of the first 3 terms of the sequence is 120
Which of the following values CANNOT be the first term of the arithmetic sequence?

$ A.\:\: 20 \\[3ex] B.\:\: 24 \\[3ex] C.\:\: 30 \\[3ex] D.\:\: 39 \\[3ex] E.\:\: 44 \\[3ex] $

$ SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] SAS_3 = \dfrac{3}{2}[2a + d(3 - 1)] \\[5ex] 120 = \dfrac{3}{2}[2a + 2d] \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\: \dfrac{2}{3} \\[5ex] \dfrac{2}{3} * 120 = \dfrac{2}{3} * \dfrac{3}{2}[2a + 2d] \\[5ex] 2(40) = 2a + 2d \\[3ex] 80 = 2a + 2d \\[3ex] 2a + 2d = 80 \\[3ex] 2(a + d) = 80 \\[3ex] a + d = \dfrac{80}{2} \\[5ex] a + d = 40 \\[3ex] d = 40 - a \\[3ex] Condition:\:\: d\:\:is\:\:positive \\[3ex] \implies d \gt 0 \\[3ex] \implies a \lt 40 \\[3ex] a \ne 44 $
(39.) What is the 7th term in this sequence of "triangular" numbers, defined by the figures below: $1, 3, 6, 10,...?$

Number 39

$ F.\:\: 7 \\[3ex] G.\:\: 22 \\[3ex] H.\:\: 25 \\[3ex] J.\:\: 28 \\[3ex] K.\:\: 40 \\[3ex] $

$ TS_n = \dfrac{n(n + 1)}{2} \\[5ex] TS_7 = \dfrac{7(7 + 1)}{2} \\[5ex] TS_7 = \dfrac{7 * 8}{2} \\[5ex] TS_7 = 7(4) \\[3ex] TS_7 = 28 $
(40.) 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




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(41.)


(42.) What is the fifth term of the arithmetic sequence $8, 6, 4, ...?$

$ A.\:\: -2 \\[3ex] B.\:\: 0 \\[3ex] C.\:\: 4 \\[3ex] D.\:\: 8 \\[3ex] E.\:\: 16 \\[3ex] $

$ 8, 6, 4, ...? \\[3ex] a = 8 \\[3ex] d = 6 - 8 = -2 \\[3ex] AS_n = a + d(n - 1) \\[3ex] AS_5 = a + d(5 - 1) \\[3ex] AS_5 = a + 4d \\[3ex] AS_5 = 8 + 4(-2) \\[3ex] AS_5 = 8 - 8 \\[3ex] AS_5 = 0 $
(43.)


(44.) 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 $
(45.)


(46.) Which of the following statements is NOT true about the arithmetic sequence $17, 12, 7, 2, ...?$

A. The fifth term is −3
B. The sum of the first 5 terms is 35
C. The eighth term is −18
D. The common difference of consecutive terms is −5
E. The common ratio of consecutive terms is −5


Let us analyze the options.
Beginning with the process of elimination, the obvious answer is Option E.
For an arithmetic sequence, we do not use the term, 'common ratio'.
'Common ratio' is used for geometric sequence.
Because the ACT is a timed test, there is no need to review other options.
Option E. is the correct option to the question.

However, for the sake of knowledge, let us review the other options

$ 17, 12, 7, 2, ...? \\[3ex] d = 12 - 17 = -5...Option\:D\:\:is\:\:true \\[3ex] a = 17 \\[3ex] AS_n = a + d(n - 1) \\[3ex] AS_5 = a + 4d \\[3ex] AS_5 = 17 + 4(-5) \\[3ex] AS_5 = 17 - 20 \\[3ex] AS_5 = -3...Option\:A\:\:is\:\:true \\[3ex] AS_8 = a + 7d \\[3ex] AS_8 = 17 + 7(-5) \\[3ex] AS_8 = 17 - 35 \\[3ex] AS_8 = -18...Option\:C\:\:is\:\:true \\[3ex] SAS_n = \dfrac{n}{2}(a + AS_n) \\[5ex] SAS_5 = \dfrac{5}{2}(17 + AS_5) \\[5ex] SAS_5 = \dfrac{5}{2}(17 + -3) \\[5ex] SAS_5 = \dfrac{5}{2}(17 - 3) \\[5ex] SAS_5 = \dfrac{5}{2}(14) \\[5ex] SAS_5 = 5(7) \\[3ex] SAS_5 = 35...Option\:B\:\:is\:\:true $
(47.)


(48.) 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} $
(49.)


(50.) The second term of an arithmetic sequence is −14, and the third term is −34.
What is the first term?
(Note: In an arithmetic sequence, consecutive terms differ by the same amount.)

$ A.\:\: \dfrac{1}{14} \\[5ex] B.\:\: 6 \\[3ex] C.\:\: 14 \\[3ex] D.\:\: 20 \\[3ex] E.\:\: -20 \\[3ex] $

The ACT is a timed test: a question should typically take a minute to solve
We shall do it two ways
The first method is much faster. It is recommended for the ACT

$ \underline{\text{First Method: Faster}} \\[3ex] d = 3rd\:\:term - 2nd\:\: term \\[3ex] d = -34 - (-14) = -34 + 14 = -20 \\[3ex] Also,\:\: d = 2nd\:\: term - 1st\:\: term \\[3ex] \rightarrow -20 = -14 - a \\[3ex] -20 + a = -14 \\[3ex] a = -14 + 20 \\[3ex] a = 6 \\[5ex] \underline{\text{Second Method: Longer}} \\[3ex] AS_n = a + d(n - 1) \\[3ex] AS_2 = a + d = -14 ...eqn.(1) \\[3ex] AS_3 = a + 2d = -34 ...eqn.(2) \\[3ex] 2 * eqn.(1) \implies \\[3ex] (a + d) = 2(-14) \\[3ex] 2 * eqn.(1) \implies 2a + 2d = -28...eqn.(3) \\[3ex] eqn.(3) - eqn.(2) \implies \\[3ex] (2a + 2d) - (a + 2d) = -28 - (-34) \\[3ex] 2a + 2d - a - 2d = -28 + 34 \\[3ex] a = 6 \\[3ex] $ The first term is $6$
(51.)


(52.) How many terms are there between 13 and 37, exclusive of 13 and 37, in the arithmetic sequence below?
$4, 7, 10, 13, ..., 37$

$ A.\:\: 0 \\[3ex] B.\:\: 7 \\[3ex] C.\:\: 8 \\[3ex] D.\:\: 28 \\[3ex] E.\:\: 36 \\[3ex] $

$ 4, 7, 10, 13, ..., 37 \\[3ex] a = 4 \\[3ex] d = 7 - 4 = 3 \\[3ex] 13 = AS_4 \\[3ex] AS_5 = 13 + 3 \\[3ex] AS_5 = 16 \\[3ex] 37 = AS_n \\[3ex] AS_n = a + d(n - 1) \\[3ex] 37 = 4 + 3(n - 1) \\[3ex] 37 - 4 = 3(n - 1) \\[3ex] 33 = 3(n - 1) \\[3ex] 3(n - 1) = 33 \\[3ex] n - 1 = \dfrac{33}{3} \\[5ex] n - 1 = 11 \\[3ex] n = 11 + 1 \\[3ex] n = 12 \\[3ex] 37 = AS_{12} \\[3ex] From\:\:13\:\:to\:\:37...exclusive\:\:of\:\:13\:\:and\:\:37 \\[3ex] \rightarrow From\:\:AS_4\:\:to\:\:AS_{12}...exclusive\:\:of\:\:AS_4\:\:and\:\:AS_{12} \\[3ex] \rightarrow From\:\:AS_5\:\:to\:\:AS_{11}...inclusive\:\:of\:\:AS_5\:\:and\:\:AS_{11} \\[3ex] = (11 - 5) + 1 \\[3ex] = 6 + 1 \\[3ex] = 7\:\:terms \\[3ex] OR\:\:list\:\:and\:\:count \\[3ex] AS_5, AS_6, AS_7, AS_8, AS_9, AS_{10}, AS_{11} \\[3ex] = 7\:\:terms \\[3ex] $ There are 7 terms between 13 and 37, exclusive of 13 and 37
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