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 Arithmetic Sequences

Samuel Dominic Chukwuemeka (SamDom For Peace) Prerequisites:
(1.) Linear Systems
(2.) 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 any wrong answer.

For JAMB and CMAT 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.) GCSE The $nth$ term of a sequence is $5n - 2$
Work out the $3rd$ term.
Circle your answer.
$51\:\:\:\:\:\:\:5\:\:\:\:\:\:\:123\:\:\:\:\:\:\:13$


$ AS_n = 5n - 2 \\[3ex] AS_3 = 5(3) - 2 \\[3ex] = 15 - 2 \\[3ex] AS_3 = 13 $
(2.) GCSE The first four terms of a linear sequence are
$7\:\:\:\:\:\:\:11\:\:\:\:\:\:\:15\:\:\:\:\:\:\:19$
Circle the expression for the $nth$ term.
$n + 6\:\:\:\:\:\:\:4n + 3\:\:\:\:\:\:\:7n + 4\:\:\:\:\:\:\:n + 4$


$ a = 7 \\[3ex] d = 11 - 7 \\[3ex] d = 4 \\[3ex] AS_n = a + d(n - 1) \\[3ex] AS_n = 7 + 4(n - 1) \\[3ex] = 7 + 4n - 4 \\[3ex] = 3 + 4n \\[3ex] AS_n = 4n + 3 $
(3.) GCSE The $nth$ term of a sequence is $12n - 5$
Work out the numbers in the sequence that have two digits and are not prime.


$ AS_n = 12n - 5 \\[3ex] AS_1 = 12(1) - 5 = 12 - 5 = 7 \\[3ex] AS_2 = 12(2) - 5 = 24 - 5 = 19 \\[3ex] d = 19 - 7 = 12 \\[3ex] AS_3 = 19 + 12 = 31 \\[3ex] AS_4 = 31 + 12 = 43 \\[3ex] AS_5 = 43 + 12 = 55 \\[3ex] AS_6 = 55 + 12 = 67 \\[3ex] AS_7 = 67 + 12 = 79 \\[3ex] AS_8 = 79 + 12 = 91 \\[3ex] $ The numbers in the sequence that have two digits and are not prime are $55$ and $91$
(4.) GCSE The first $4$ terms of a linear sequence are
$3\:\:\:\:\:\:\:11\:\:\:\:\:\:\:19\:\:\:\:\:\:\:27$
Circle the expression for the $nth$ term.
$8 - 5n\:\:\:\:\:\:\:n + 8\:\:\:\:\:\:\:8n + 3\:\:\:\:\:\:\:8n - 5$


$ a = 3 \\[3ex] d = 11 - 3 \\[3ex] d = 8 \\[3ex] AS_n = a + d(n - 1) \\[3ex] AS_n = 3 + 8(n - 1) \\[3ex] = 3 + 8n - 8 \\[3ex] = -5 + 8n \\[3ex] AS_n = 8n - 5 $
(5.) WASSCE Given the Arithmetic Sequence $-6, -2\dfrac{1}{2}, 1, ..., 71,$
find the:
(i) common difference;
(ii) number of terms of the sequence.


$ a = -6 \\[3ex] p = 71 \\[3ex] d = 2nd\:\: term - a \\[3ex] OR \\[3ex] d = 3rd\:\: term - 2nd\:\: term \\[3ex] = -2\dfrac{1}{2} - (-6) \\[5ex] = -\dfrac{5}{2} + 6 \\[5ex] = -\dfrac{5}{2} + \dfrac{12}{2} \\[5ex] \implies d = \dfrac{-5 + 12}{2} = \dfrac{7}{2} \\[5ex] OR \\[3ex] d = 1 - \left(-2\dfrac{1}{2}\right) \\[5ex] = 1 + \left(2\dfrac{1}{2}\right) \\[5ex] = 1 + \dfrac{5}{2} \\[5ex] = \dfrac{2}{2} + \dfrac{5}{2} \\[5ex] \implies d = \dfrac{2 + 5}{2} = \dfrac{7}{2} \\[5ex] n = \dfrac{p - a + d}{d} \\[5ex] n = (p - a + d) \div d \\[3ex] p - a + d = 71 - (-6) + \dfrac{7}{2} \\[5ex] = 71 + 6 + \dfrac{7}{2} \\[5ex] = 77 + \dfrac{7}{2} \\[5ex] = \dfrac{154}{2} + \dfrac{7}{2} \\[5ex] = \dfrac{154 + 7}{2} = \dfrac{161}{2} \\[5ex] \implies p - a + d = \dfrac{161}{2} \\[5ex] (p - a + d) \div d = \dfrac{161}{2} \div \dfrac{7}{2} \\[5ex] = \dfrac{161}{2} * \dfrac{2}{7} \\[5ex] = 23 \\[3ex] \implies n = 23 $
(6.) GCSE A linear sequence starts
$a + 2b\:\:\:\:\:\:\:a + 6b\:\:\:\:\:\:\:a + 10b\:\:\:\:\:\:\:......\:\:\:\:\:\:\:......$
The $2nd$ term has value $8$
The $5th$ term has value $44$
Work out the values of $a$ and $b$


$ 1st\:\:term = AS_1 = a + 2b \\[3ex] 2nd\:\:term = AS_2 = a + 6b = 8 \\[3ex] d = (a + 6b) - (a + 2b) \\[3ex] = a + 6b - a - 2b \\[3ex] = 4b \\[3ex] d = 4b \\[3ex] AS_3 = a + 10b \\[3ex] AS_4 = a + 10b + 4b \\[3ex] AS_4 = a + 14b \\[3ex] AS_5 = a + 14b + 4b \\[3ex] AS_5 = a + 18b = 44 \\[3ex] \implies a + 6b = 8...eqn.(1) \\[3ex] a + 18b = 44...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \implies 18b - 6b = 44 - 8 \\[3ex] 12b = 36 \\[3ex] b = \dfrac{36}{12} \\[5ex] b = 3 \\[3ex] From\:\:eqn.(1);\:\: a = 8 - 6b \\[3ex] a = 8 - 6(3) \\[3ex] a = 8 - 18 \\[3ex] a = -10 \\[3ex] a = -10,\:\:\: b = 3 $
(7.) JAMB The sixth term of an arithmetical progression is half of its twelfth term.
The first term is equal to
A. zero
B. half of the common difference
C. double the common difference
D. the common difference


$ AS_6 = a + 5d \\[3ex] AS_{12} = a + 11d \\[3ex] AS_6 = \dfrac{1}{2} * AS_{12} \\[5ex] \implies a + 5d = \dfrac{1}{2}(a + 11d) \\[5ex] LCD = 2 \\[3ex] 2(a + 5d) = 2 * \dfrac{1}{2}(a + 11d) \\[5ex] 2a + 10d = a + 11d \\[3ex] 2a - a = 11d - 10d \\[3ex] a = d \\[3ex] $ The first term is equal to the common difference
(8.) JAMB The $3rd$ term of an arithmetic progression is $-9$ and the $7th$ term is $-29$.
Find the $10th$ term of the progression.

$ A.\:\: -44 \\[3ex] B.\:\: -165 \\[3ex] C.\:\: 165 \\[3ex] D.\:\: 44 \\[3ex] $

$ AS_3 = a + 2d = -9 ...eqn.(1) \\[3ex] AS_7 = a + 6d = -29 ...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \\[3ex] \implies (a + 6d) - (a + 2d) = -29 - (-9) \\[3ex] a + 6d - a - 2d = -29 + 9 \\[3ex] 4d = - 20 \\[3ex] d = -\dfrac{20}{4} \\[5ex] d = -5 \\[3ex] Subst.\:\: d = -5 \:\:into\:\: eqn.(1) \\[3ex] a + 2(-5) = -9 \\[3ex] a - 10 = -9 \\[3ex] a = -9 + 10 \\[3ex] a = 1 \\[3ex] AS_{10} = a + 9d \\[3ex] AS_{10} = 1 + 9(-5) \\[3ex] AS_{10} = 1 - 45 \\[3ex] AS_{10} = -44 $
(9.) WASSCE The second, fourth and sixth terms of an Arithmetic Progression ($AP$) are $x - 1$, $x + 1$, and $7$ respectively.
Find the
(i) common difference;
(ii) first term;
(iii) value of $x$


$ AS_n = a + d(n - 1) \\[3ex] AS_2 = a + d = x - 1...eqn.(1) \\[3ex] AS_4 = a + 3d = x + 1...eqn.(2) \\[3ex] AS_6 = a + 5d = 7...eqn.(3) \\[3ex] eqn.(2) - eqn.(1) \implies (a + 3d) - (a + d) = (x + 1) - (x - 1) \\[3ex] a + 3d - a - d = x + 1 - x + 1 \\[3ex] 2d = 2 \\[3ex] d = \dfrac{2}{2} \\[5ex] (i)\:\: d = 1 \\[3ex] From\:\: eqn.(3) \\[3ex] a + 5d = 7 \\[3ex] a = 7 - 5d \\[3ex] Subst.\:\: d = 1 \\[3ex] a = 7 - 5(1) \\[3ex] a = 7 - 5 \\[3ex] (ii)\:\: a = 2 \\[3ex] From\:\: eqn.(1) \\[3ex] a + d = x - 1 \\[3ex] x - 1 = a + d \\[3ex] x = a + d + 1 \\[3ex] Subst.\:\: a = 2 \:\:and\:\: d = 1 \\[3ex] x = 2 + 1 + 1 \\[3ex] (iii)\:\: x = 4 $
(10.) NSC The first three terms of an arithmetic sequence are $-1\:\:;2\:\:and\:\:5$
(10.1) Determine the $n^{th}$ term, $T_n$, of the sequence.

(10.2) Calculate $T_{43}$

(10.3) Evaluate $\sum\limits_{k=1}^n T_k$ in terms of $n$


$ a = -1 \\[3ex] d = 2 - (-1) = 2 + 1 = 3 \\[3ex] (10.1) \\[3ex] T_n = a + d(n - 1) \\[3ex] T_n = -1 + 3(n - 1) \\[3ex] T_n = -1 + 3n - 3 \\[3ex] T_n = 3n - 4 \\[3ex] (10.2) \\[3ex] T_n = 3n - 4 \\[3ex] T_{43} = 3(43) - 4 \\[3ex] T_{43} = 129 - 4 \\[3ex] T_{43} = 125 \\[3ex] (10.3) \\[3ex] \sum\limits_{k=1}^n T_k \\[5ex] k = 1 \rightarrow T_k = T_1 = a = -1 \\[3ex] k = 2 \rightarrow T_k = T_2 \\[3ex] k = n \rightarrow T_k = T_n \\[3ex] \sum\limits_{k=1}^n T_k = SAS_n \\[5ex] = T_1 + T_2 + ... + T_n \\[3ex] = \dfrac{n}{2}(a + T_n) \\[5ex] = \dfrac{n}{2}(-1 + 3n - 4) \\[5ex] = \dfrac{n}{2}(3n - 5) $
(11.) WASSCE Find the sum of the Arithmetic Progression ($AP$) $1, 3, 5, ..., 101$


This is a sequence of odd numbers
So, we are asked to find the sum of the first $n$ odd numbers
We need to find the number of terms first
We can find $n$ in two ways.

$ \underline{Mentally} \\[3ex] There\:\: are\:\: 5 \:\:odd\:\:numbers\:\: from\:\: 1 - 10 \:\:(1, 3, 5, 7, 9) \\[3ex] There\:\: are\:\: 6 \:\:odd\:\:numbers\:\: from\:\: 1 - 11 \:\:(1, 3, 5, 7, 9, 11) \\[3ex] Therefore,\:\: there\:\: are\:\: 50 \:\:odd\:\:numbers\:\: from\:\: 1 - 100 \\[3ex] So,\:\: there\:\: are\:\: 51 \:\:odd\:\:numbers\:\: from\:\: 1 - 101 \\[3ex] n = 51 \\[3ex] \underline{By\:\: Formula} \\[3ex] a = 1 \\[3ex] p = 101 \\[3ex] d = 3 - 1 = 2 \\[3ex] n = \dfrac{p - a + d}{d} \\[5ex] n = \dfrac{101 - 1 + 2}{2} \\[5ex] n = \dfrac{102}{2} \\[5ex] n = 51 \\[3ex] $ We can also find $SAS_n$ in two ways.
You may use any of the formulas for $SAS_n$

$ SAS_n = \dfrac{n}{2}(a + p) \\[5ex] SAS_n = \dfrac{51}{2}(1 + 101) \\[5ex] = \dfrac{51}{2}(102) \\[5ex] = 51 * 51 \\[3ex] SAS_n = 2601 \\[3ex] OR \\[3ex] SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] SAS_n = \dfrac{51}{2}[2(1) + 2(51 - 1)] \\[5ex] = \dfrac{51}{2}[2 + 2(50)] \\[5ex] = \dfrac{51}{2}[2 + 100] \\[5ex] = \dfrac{51}{2}(102) \\[5ex] = 51 * 51 \\[3ex] SAS_n = 2601 $
(12.) WAEC-BECE Given that $s = \dfrac{n}{2}[2a + (n - 1)d]$, $a = 3$, $d = 4$ and $n = 10$, find the value of $s$.


$ s = \dfrac{n}{2}[2a + (n - 1)d] \\[3ex] a = 3 \\[3ex] d = 4 \\[3ex] n = 10 \\[3ex] s = \dfrac{10}{2}[2(3) + (10 - 1)4] \\[5ex] = 5[6 + 9(4)] \\[3ex] = 5[6 + 36] \\[3ex] = 5[42] \\[3ex] = 210 \\[3ex] s = 210 $
(13.) Determine the first term and the common difference for an arithmetic sequence if the second term is $40 - 5n$ and the fourth term is $20m + n$


$ AS_2 = a + d = 40 - 5n ...eqn.(1) \\[3ex] AS_4 = a + 3d = 20m + n ...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \:\:gives \\[3ex] (a + 3d) - (a + d) = (20m + n) - (40 - 5n) \\[3ex] \rightarrow a + 3d - a - d = 20m + n - 40 + 5n \\[3ex] 2d = 20m + 6n - 40 \\[3ex] 2d = 2(10m + 3n - 20) \\[3ex] d = \dfrac{2(10m + 3n - 20)}{2} \\[5ex] d = 10m + 3n - 20 \\[3ex] eqn.(2) - 3 * eqn.(1) \:\:gives \\[3ex] (a + 3d) - [3(a + d)] = (20m + n) - [3(40 - 5n)] \\[3ex] \rightarrow a + 3d - (3a + 3d) = 20m + n - (120 - 15n) \\[3ex] a + 3d - 3a - 3d = 20m + n - 120 + 15n \\[3ex] -2a = 20m + 16n - 120 \\[3ex] -2a = 2(10m + 8n - 60) \\[3ex] a = \dfrac{2(10m + 8n - 60)}{-2} \\[3ex] a = -(10m + 8n - 60) \\[3ex] a = -10m - 8n + 60 $
(14.) NSC Given the arithmetic sequence: $35;\:\:\:28;\:\:\:21;...$
Calculate which term of the sequence will have a value of $-140$


$ AS_n = a + d(n - 1) \\[3ex] a = 35 \\[3ex] d = 28 - 35 = -7 \\[3ex] AS_n = -140 \\[3ex] n = ? \\[3ex] -140 = 35 + -7(n - 1) \\[3ex] -140 = 35 - 7(n - 1) \\[3ex] -140 = 35 - 7n + 7 \\[3ex] -140 = 42 - 7n \\[3ex] 7n = 42 + 140 \\[3ex] 7n = 182 \\[3ex] n = \dfrac{182}{7} \\[5ex] n = 26 \\[3ex] $ The $26th$ term of the sequence will have the value of $-140$
(15.) WASSCE If $\dfrac{1}{2}$, $\dfrac{1}{x}$, $\dfrac{1}{3}$ are successive terms of an arithmetic progression (A.P), show that $\dfrac{2 - x}{x - 3} = \dfrac{2}{3}$


$ AP:\:\: \dfrac{1}{2},\dfrac{1}{x},\dfrac{1}{3} \\[5ex] d = \dfrac{1}{x} - \dfrac{1}{2} \\[5ex] d = \dfrac{1}{3} - \dfrac{1}{x} \\[5ex] d = d \\[3ex] \rightarrow \dfrac{1}{x} - \dfrac{1}{2} = \dfrac{1}{3} - \dfrac{1}{x} \\[5ex] \dfrac{2}{2x} - \dfrac{x}{2x} = \dfrac{x}{3x} - \dfrac{3}{3x} \\[5ex] \dfrac{2 - x}{2x} = \dfrac{x - 3}{3x} \\[5ex] Cross\:\:Multiply \\[3ex] 3x(2 - x) = 2x(x - 3) \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:(x - 3) \\[3ex] \dfrac{3x(2 - x)}{x - 3} = \dfrac{2x(x - 3)}{x - 3} \\[5ex] \dfrac{3x(2 - x)}{x - 3} = \dfrac{2x}{1} \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:\dfrac{1}{3x} \\[5ex] \dfrac{1}{3x} * \dfrac{3x(2 - x)}{x - 3} = \dfrac{1}{3x} * \dfrac{2x}{1} \\[5ex] \dfrac{2 -x}{x - 3} = \dfrac{2x}{3x} \\[5ex] \dfrac{2 - x}{x - 3} = \dfrac{2}{3} $
(16.) WASSCE The $6^{th}$ and $12^{th}$ terms of a linear sequence (A.P) are $17$ and $41$ respectively.
Find the sum of the first $20$ terms.


$ AS_n = a + d(n - 1) \\[3ex] AS_{6} = a + 5d = 17...eqn.(1) \\[3ex] AS_{12} = a + 11d = 41...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \rightarrow 6d = 24 \\[3ex] d = \dfrac{24}{6} \\[5ex] d = 4 \\[3ex] From\:\:eqn.(1);\:\: a = 17 - 5d \\[3ex] a = 17 - 5(4) \\[3ex] a = 17 - 20 \\[3ex] a = -3 \\[3ex] Two\:\:ways\:\:to\:\:calculate\:\:SAS_{20} \\[3ex] Choose\:\:your\:\:preference \\[3ex] \underline{First\:\:Method} \\[3ex] SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] SAS_{20} = \dfrac{20}{2}[2(-3) + 4(20 - 1)] \\[5ex] = 10[-6 + 4(19)] \\[3ex] = 10[-6 + 76] \\[3ex] = 10(70) \\[3ex] SAS_{20} = 700 \\[3ex] \underline{Second\:\:Method} \\[3ex] SAS_n = \dfrac{n}{2}(a + p) \\[5ex] SAS_n = \dfrac{n}{2}(a + AS_{n}) \\[5ex] AS_{20} = a + 19d \\[3ex] AS_{20} = -3 + 19(4) \\[3ex] AS_{20} = -3 + 76 \\[3ex] AS_{20} = 73 \\[3ex] SAS_{20} = \dfrac{20}{2}(-3 + 73) \\[5ex] SAS_{20} = 10(70) \\[3ex] SAS_{20} = 700 $
(17.) ACT 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] $

$ 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 $5\dfrac{2}{3}$
(18.) ACT 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 $
(19.) ACT 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 $-\dfrac{1}{2}$
(20.) ACT 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{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 \\[3ex] \underline{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 $18$




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(21.) ACT 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 $
(22.) ACT 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.) ACT The degree measures of the interior angles of $\triangle ABC$, shown below, form an arithmetic sequence with common difference $10^\circ$.
What is the first term of the sequence?
AS - Question 23

$ 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 $
(24.) JAMB The sum of the first $n$ terms of an arithmetic progression is $252$.
If the first term is $-16$ and the last term is $72$, find the number of terms in the series.

$ A.\:\: 9 \\[3ex] B.\:\: 8 \\[3ex] C.\:\: 7 \\[3ex] D.\:\: 6 \\[3ex] $

$ a = -16 \\[3ex] p = 72 \\[3ex] SAS_n = 252 \\[3ex] n = \dfrac{2 * SAS_n}{a + p} \\[5ex] n = \dfrac{2 * 252}{-16 + 72} \\[5ex] n = \dfrac{504}{56} \\[5ex] n = 9 $
(25.) JAMB



$ AS_6 = a + 5d \\[3ex] AS_{12} = a + 11d \\[3ex] AS_6 = \dfrac{1}{2} * AS_{12} \\[5ex] \implies a + 5d = \dfrac{1}{2}(a + 11d) \\[5ex] LCD = 2 \\[3ex] 2(a + 5d) = 2 * \dfrac{1}{2}(a + 11d) \\[5ex] 2a + 10d = a + 11d \\[3ex] 2a - a = 11d - 10d \\[3ex] a = d \\[3ex] $ The first term is equal to the common difference
(26.) JAMB The sum of the first $n$ terms of the arithmetic progression $5,\:\: 11,\:\: 17,\:\: 23,\:\: 29,\:\: 35,...$ is

$ A.\:\: n(3n - 0.5) \\[3ex] B.\:\: n(3n + 2) \\[3ex] C.\:\: n(3n + 2.5) \\[3ex] D.\:\: n(3n + 5) \\[3ex] $

$ SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] a = 5 \\[3ex] d = 11 - 5 = 6 \\[3ex] SAS_n = \dfrac{n}{2}[2(5) + 6(n - 1)] \\[5ex] = \dfrac{n}{2}[10 + 6n - 6] \\[5ex] = \dfrac{n}{2}[6n + 4] \\[5ex] = \dfrac{n}{2} * 6n + \dfrac{n}{2} * 4 \\[5ex] = n(3n) + n(2) \\[3ex] = n(3n + 2) $
(27.) ACT 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 $
(28.) ACT 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{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 \\[3ex] \underline{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 $16$
(29.) JAMB



$ AS_6 = a + 5d \\[3ex] AS_{12} = a + 11d \\[3ex] AS_6 = \dfrac{1}{2} * AS_{12} \\[5ex] \implies a + 5d = \dfrac{1}{2}(a + 11d) \\[5ex] LCD = 2 \\[3ex] 2(a + 5d) = 2 * \dfrac{1}{2}(a + 11d) \\[5ex] 2a + 10d = a + 11d \\[3ex] 2a - a = 11d - 10d \\[3ex] a = d \\[3ex] $ The first term is equal to the common difference
(30.) JAMB If the $9th$ term of an A.P is five times the $5th$ term, find the relationship betyween $a$ and $d$

$ A.\:\: 3a + 5d = 0 \\[3ex] B.\:\: a + 2d = 0 \\[3ex] C.\:\: 2a + d = 0 \\[3ex] D.\:\: a + 3d = 0 \\[3ex] $

$ AS_n = a + d(n - 1) \\[3ex] AS_9 = a + 8d \\[3ex] AS_5 = a + 4d \\[3ex] AS_9 = 5 * AS_5 \\[3ex] \rightarrow a + 8d = 5(a + 4d) \\[3ex] a + 8d = 5a + 20d \\[3ex] 5a + 20d = a + 8d \\[3ex] 5a - a + 20d - 8d = 0 \\[3ex] 4a + 12d = 0 \\[3ex] 4(a + 3d) = 0 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\: 4 \\[3ex] \dfrac{4(a + 3d)}{4} = \dfrac{0}{4} \\[5ex] a + 3d = 0 $
(31.) WASSCE (a.) How many numbers between $75$ and $500$ are divisible by $7$?

(b.) The $8th$ term of an Arithmetic Progression (A.P.) is $5$ times the third term while the $7th$ term is $9$ greater than the $4th$ term.
Write the first five terms of the A.P.


How many numbers between $75$ and $500$ are divisible by $7$?
Between $75$ and $500$ typically means that $7$ and $500$ are not included
Even if they are, both numbers are NOT divisible by $7$
So, we need to find the number immediately greater than $75$ that is divisible by $7$
That number is $77$.
This implies that $77$ is the first term.
Also, we need to find the number immediately less than $500$ that is divisible by $7$
That number is $497$.
This implies that $497$ is the last term.
Because the numbers are divisible by $7$, the common difference is $7$
The count of the numbers divisible by $7$ is the number of terms

$ (a.) \\[3ex] a = 77 \\[3ex] AP_n = 497 \\[3ex] d = 7 \\[3ex] n = ? \\[3ex] AP_n = a + d(n - 1) \\[3ex] 497 = 77 + 7(n - 1) \\[3ex] 497 - 77 = 7(n - 1) \\[3ex] 420 = 7(n - 1) \\[3ex] 7(n - 1) = 420 \\[3ex] n - 1 = \dfrac{420}{7} \\[5ex] n - 1 = 60 \\[3ex] n = 60 + 1 \\[3ex] n = 61 \\[3ex] $ There are $61$ numbers between $75$ and $500$ that are divisible by $7$

$ (b.) \\[3ex] AP_n = a + d(n - 1) \\[3ex] AP_8 = a + 7d \\[3ex] AP_3 = a + 2d \\[3ex] AP_7 = a + 6d \\[3ex] AP_4 = a + 3d \\[3ex] AP_8 = 5 * AP_3 \\[3ex] a + 7d = 5(a + 2d) \\[3ex] a + 7d = 5a + 10d \\[3ex] 5a + 10d = a + 7d \\[3ex] 5a - a + 10d - 7d = 0 \\[3ex] 4a + 3d = 0 ...eqn.(1) \\[3ex] Also: \\[3ex] AP_7 = 9 + AP_4 \\[3ex] a + 6d = 9 + (a + 3d) \\[3ex] a + 6d = 9 + a + 3d \\[3ex] a - a + 6d - 3d = 9 \\[3ex] 3d = 9 \\[3ex] d = \dfrac{9}{3} \\[5ex] d = 3 \\[3ex] Substitute \:\:d = 3\:\:into\:\:eqn.(1) \\[3ex] 4a + 3d = 0 \\[3ex] 4a = -3d \\[3ex] 4a = -3(3) \\[3ex] 4a = -9 \\[3ex] a = -\dfrac{9}{4} \\[5ex] AP_2 = a + d = -\dfrac{9}{4} + 3 = -\dfrac{9}{4} + \dfrac{12}{4} = \dfrac{-9 + 12}{4} = \dfrac{3}{4} \\[5ex] AP_3 = a + 2d = AP_2 + d = \dfrac{3}{4} + 3 = \dfrac{3}{4} + \dfrac{12}{4} = \dfrac{3 + 12}{4} = \dfrac{15}{4} \\[5ex] AP_4 = a + 3d = AP_3 + d = \dfrac{15}{4} + 3 = \dfrac{15}{4} + \dfrac{12}{4} = \dfrac{15 + 12}{4} = \dfrac{27}{4} \\[5ex] AP_5 = a + 2d = AP_4 + d = \dfrac{27}{4} + 3 = \dfrac{27}{4} + \dfrac{12}{4} = \dfrac{27 + 12}{4} = \dfrac{39}{4} \\[5ex] AP:\:\: -\dfrac{9}{4}, \dfrac{3}{4}, \dfrac{15}{4}, \dfrac{27}{4}, \dfrac{39}{4} $
(32.) WASSCE The first term of an Arithmetic Progression (A.P) is $31$ and the common difference is $9$
Show that the $nth$ term is $9n + 22$
Hence, find the $20th$ term


$ a = 31 \\[3ex] d = 9 \\[3ex] AP_n = a + d(n - 1) \\[3ex] AP_n = 31 + 9(n - 1) \\[3ex] AP_n = 31 + 9n - 9 \\[3ex] AP_n = 9n + 22 \\[3ex] AP_{20} = 9(20) + 22 \\[3ex] AP_{20} = 180 + 22 \\[3ex] AP_{20} = 202 $
(33.)


$ AS_6 = a + 5d \\[3ex] AS_{12} = a + 11d \\[3ex] AS_6 = \dfrac{1}{2} * AS_{12} \\[5ex] \implies a + 5d = \dfrac{1}{2}(a + 11d) \\[5ex] LCD = 2 \\[3ex] 2(a + 5d) = 2 * \dfrac{1}{2}(a + 11d) \\[5ex] 2a + 10d = a + 11d \\[3ex] 2a - a = 11d - 10d \\[3ex] a = d \\[3ex] $ The first term is equal to the common difference
(34.) ACT 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 $
(35.) JAMB



$ AS_6 = a + 5d \\[3ex] AS_{12} = a + 11d \\[3ex] AS_6 = \dfrac{1}{2} * AS_{12} \\[5ex] \implies a + 5d = \dfrac{1}{2}(a + 11d) \\[5ex] LCD = 2 \\[3ex] 2(a + 5d) = 2 * \dfrac{1}{2}(a + 11d) \\[5ex] 2a + 10d = a + 11d \\[3ex] 2a - a = 11d - 10d \\[3ex] a = d \\[3ex] $ The first term is equal to the common difference
(36.) ACT 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 $
(37.)


$ AS_6 = a + 5d \\[3ex] AS_{12} = a + 11d \\[3ex] AS_6 = \dfrac{1}{2} * AS_{12} \\[5ex] \implies a + 5d = \dfrac{1}{2}(a + 11d) \\[5ex] LCD = 2 \\[3ex] 2(a + 5d) = 2 * \dfrac{1}{2}(a + 11d) \\[5ex] 2a + 10d = a + 11d \\[3ex] 2a - a = 11d - 10d \\[3ex] a = d \\[3ex] $ The first term is equal to the common difference
(38.) ACT 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 $
(39.)


$ AS_6 = a + 5d \\[3ex] AS_{12} = a + 11d \\[3ex] AS_6 = \dfrac{1}{2} * AS_{12} \\[5ex] \implies a + 5d = \dfrac{1}{2}(a + 11d) \\[5ex] LCD = 2 \\[3ex] 2(a + 5d) = 2 * \dfrac{1}{2}(a + 11d) \\[5ex] 2a + 10d = a + 11d \\[3ex] 2a - a = 11d - 10d \\[3ex] a = d \\[3ex] $ The first term is equal to the common difference
(40.) ACT 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 $




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(41.)

$ 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 $-\dfrac{1}{2}$
(42.) ACT 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{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 \\[3ex] \underline{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 $6$
(43.)


$ AS_6 = a + 5d \\[3ex] AS_{12} = a + 11d \\[3ex] AS_6 = \dfrac{1}{2} * AS_{12} \\[5ex] \implies a + 5d = \dfrac{1}{2}(a + 11d) \\[5ex] LCD = 2 \\[3ex] 2(a + 5d) = 2 * \dfrac{1}{2}(a + 11d) \\[5ex] 2a + 10d = a + 11d \\[3ex] 2a - a = 11d - 10d \\[3ex] a = d \\[3ex] $ The first term is equal to the common difference
(44.) ACT 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 $37$