3. In an AP: (i) given , find and . (ii) given , find and . (iii) given , find and . (iv) given , find and . (v) given , find and . (vi) given , find and . (vii) given , find and . (viii) given , find and . (ix) given , find . (x) given , and there are total 9 terms. Find .

Understand the Question An Arithmetic Progression (AP) is a sequence of numbers where the difference between consecutive terms is constant. th term formula: Sum of first terms formulas: Where: = first term = common difference = number of terms (must be a positive integer, ) (or ) = th term (or last term) = sum of the first terms Part (i) Given , find and . Step 1 · Find the Number of Terms Using Step 2 · Calculate the Sum Using Answer > (i) Part (ii) Given , find and . Step 1 · Find the Common Difference Using with Step 2 · Calculate the Sum Using Answer > (ii) Part (iii) Given , find and . Step 1 · Find the First Term Using with Step 2 · Calculate the Sum Using Answer > (iii) Part (iv) Given , find and . Step 1 · Form Linear Equations in and For : For : Step 2 · Solve for and From equation (1), . Substitute into equation (2): Substitute into equation (1): Step 3 · Find Using Answer > (iv) Part (v) Given , find and . Step 1 · Find the First Term Using with Step 2 · Find the 9th Term Using Answer > (v) Part (vi) Given , find and . Step 1 · Form and Solve the Quadratic Equation for Using Using the quadratic formula : Since must be a positive integer, . Step 2 · Find Using with Answer > (vi) Part (vii) Given , find and . Step 1 · Find the Number of Terms Using Step 2 · Find the Common Difference Using with Answer > (vii) Part (viii) Given , find and . Step 1 · Express in terms of Using Step 2 · Form and Solve the Quadratic Equation for Using : Since must be a positive integer, . Step 3 · Find the First Term Substitute into equation (1): Answer > (viii) Part (ix) Given , find . Step 1 · Find the Common Difference Using with Answer > (ix) Part (x) Given , and there are total 9 terms. Find . Step 1 · Find the First Term Here, last term and . Using : Answer > (x) Common Mistakes Discarding Negative or Fractional : The number of terms represents a count and must be a positive integer (). Always reject negative or fractional roots when solving quadratic equations in . Using the Wrong Sum Formula: When the last term is known, using is much quicker than . Sign Errors with Negative Common Difference: When , remember that is subtracted, not added.

Arithmetic Progressions | Exercise 5.3

Question 3

In an AP:

(i) given $a = 5, d = 3, a_n = 50$ , find $n$ and $S_n$ .

(ii) given $a = 7, a_{13} = 35$ , find $d$ and $S_{13}$ .

(iii) given $a_{12} = 37, d = 3$ , find $a$ and $S_{12}$ .

(iv) given $a_3 = 15, S_{10} = 125$ , find $d$ and $a_{10}$ .

(v) given $d = 5, S_9 = 75$ , find $a$ and $a_9$ .

(vi) given $a = 2, d = 8, S_n = 90$ , find $n$ and $a_n$ .

(vii) given $a = 8, a_n = 62, S_n = 210$ , find $n$ and $d$ .

(viii) given $a_n = 4, d = 2, S_n = -14$ , find $n$ and $a$ .

(ix) given $a = 3, n = 8, S = 192$ , find $d$ .

(x) given $l = 28, S = 144$ , and there are total 9 terms. Find $a$ .

Solution

Understand the Question

An Arithmetic Progression (AP) is a sequence of numbers where the difference between consecutive terms is constant.

$n$ th term formula: $a_n = a + (n - 1)d$
Sum of first $n$ terms formulas: $S_n = \dfrac{n}{2}\left[2a + (n - 1)d\right] \quad \text{or} \quad S_n = \dfrac{n}{2}(a + a_n)$

Where:

$a$ = first term
$d$ = common difference
$n$ = number of terms (must be a positive integer, $n \in \mathbb{N}$ )
$a_n$ (or $l$ ) = $n$ th term (or last term)
$S_n$ = sum of the first $n$ terms

(i) Given $a = 5, d = 3, a_n = 50$ , find $n$ and $S_n$ .

Step 1 · Find the Number of Terms $n$

Using $a_n = a + (n-1)d$

Diagram 1

\begin{aligned} 50 &= 5 + (n-1)3 \\[0.6em] 50 - 5 &= (n-1)3 \\[0.6em] 45 &= (n-1)3 \\[0.6em] \dfrac{45}{3} &= n-1 \\[0.6em] 15 &= n-1 \\[0.6em] n &= 16 \end{aligned}

Step 2 · Calculate the Sum $S_n$

Using $S_n = \dfrac{n}{2}(a + a_n)$

\begin{aligned} S_{16} &= \dfrac{16}{2}(5 + 50) \\[0.6em] &= 8(55) \\[0.6em] &= 440 \end{aligned}

Answer

(i) $n = 16, \quad S_{16} = 440$

(ii) Given $a = 7, a_{13} = 35$ , find $d$ and $S_{13}$ .

Step 1 · Find the Common Difference $d$

Using $a_n = a + (n-1)d$ with $n = 13$

\begin{aligned} a_{13} &= a + (13-1)d \\[0.6em] 35 &= 7 + 12d \\[0.6em] 35 - 7 &= 12d \\[0.6em] 28 &= 12d \\[0.6em] d &= \dfrac{28}{12} = \dfrac{7}{3} \end{aligned}

Step 2 · Calculate the Sum $S_{13}$

Using $S_n = \dfrac{n}{2}(a + a_n)$

\begin{aligned} S_{13} &= \dfrac{13}{2}(7 + 35) \\[0.6em] &= \dfrac{13}{2}(42) \\[0.6em] &= 13 \times 21 \\[0.6em] &= 273 \end{aligned}

Answer

(ii) $d = \dfrac{7}{3}, \quad S_{13} = 273$

(iii) Given $a_{12} = 37, d = 3$ , find $a$ and $S_{12}$ .

Step 1 · Find the First Term $a$

Using $a_n = a + (n-1)d$ with $n = 12$

\begin{aligned} a_{12} &= a + (12-1)d \\[0.6em] 37 &= a + 11(3) \\[0.6em] 37 &= a + 33 \\[0.6em] a &= 37 - 33 \\[0.6em] a &= 4 \end{aligned}

Step 2 · Calculate the Sum $S_{12}$

Using $S_n = \dfrac{n}{2}(a + a_n)$

\begin{aligned} S_{12} &= \dfrac{12}{2}(4 + 37) \\[0.6em] &= 6(41) \\[0.6em] &= 246 \end{aligned}

Answer

(iii) $a = 4, \quad S_{12} = 246$

(iv) Given $a_3 = 15, S_{10} = 125$ , find $d$ and $a_{10}$ .

Step 1 · Form Linear Equations in $a$ and $d$

For $a_3 = 15$ :

\begin{aligned} a + (3-1)d &= 15 \\[0.6em] a + 2d &= 15 \quad \dots (1) \end{aligned}

For $S_{10} = 125$ :

\begin{aligned} S_{10} &= \dfrac{10}{2}[2a + (10-1)d] \\[0.6em] 125 &= 5(2a + 9d) \\[0.6em] \dfrac{125}{5} &= 2a + 9d \\[0.6em] 2a + 9d &= 25 \quad \dots (2) \end{aligned}

Step 2 · Solve for $a$ and $d$

From equation (1), $a = 15 - 2d$ . Substitute into equation (2):

\begin{aligned} 2(15 - 2d) + 9d &= 25 \\[0.6em] 30 - 4d + 9d &= 25 \\[0.6em] 30 + 5d &= 25 \\[0.6em] 5d &= 25 - 30 \\[0.6em] 5d &= -5 \\[0.6em] d &= -1 \end{aligned}

Substitute $d = -1$ into equation (1):

\begin{aligned} a &= 15 - 2(-1) \\[0.6em] &= 15 + 2 = 17 \end{aligned}

Step 3 · Find $a_{10}$

Using $a_n = a + (n-1)d$

\begin{aligned} a_{10} &= 17 + (10-1)(-1) \\[0.6em] &= 17 + 9(-1) \\[0.6em] &= 17 - 9 = 8 \end{aligned}

Answer

(iv) $d = -1, \quad a_{10} = 8$

(v) Given $d = 5, S_9 = 75$ , find $a$ and $a_9$ .

Step 1 · Find the First Term $a$

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$ with $n = 9$

\begin{aligned} S_9 &= \dfrac{9}{2}[2a + (9-1)5] \\[0.6em] 75 &= \dfrac{9}{2}(2a + 40) \\[0.6em] 75 \times 2 &= 9(2a + 40) \\[0.6em] 150 &= 18a + 360 \\[0.6em] 18a &= 150 - 360 \\[0.6em] 18a &= -210 \\[0.6em] a &= \dfrac{-210}{18} = -\dfrac{35}{3} \end{aligned}

Step 2 · Find the 9th Term $a_9$

Using $a_n = a + (n-1)d$

\begin{aligned} a_9 &= -\dfrac{35}{3} + (9-1)5 \\[0.6em] &= -\dfrac{35}{3} + 40 \\[0.6em] &= -\dfrac{35}{3} + \dfrac{120}{3} \\[0.6em] &= \dfrac{85}{3} \end{aligned}

Answer

(v) $a = -\dfrac{35}{3}, \quad a_9 = \dfrac{85}{3}$

(vi) Given $a = 2, d = 8, S_n = 90$ , find $n$ and $a_n$ .

Step 1 · Form and Solve the Quadratic Equation for $n$

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$

\begin{aligned} 90 &= \dfrac{n}{2}[2(2) + (n-1)8] \\[0.6em] 90 &= \dfrac{n}{2}(4 + 8n - 8) \\[0.6em] 90 &= \dfrac{n}{2}(8n - 4) \\[0.6em] 90 &= n(4n - 2) \\[0.6em] 4n^2 - 2n - 90 &= 0 \\[0.6em] 2n^2 - n - 45 &= 0 \end{aligned}

Using the quadratic formula $n = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ :

\begin{aligned} n &= \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-45)}}{2(2)} \\[0.6em] &= \dfrac{1 \pm \sqrt{1 + 360}}{4} \\[0.6em] &= \dfrac{1 \pm \sqrt{361}}{4} \\[0.6em] &= \dfrac{1 \pm 19}{4} \end{aligned}

$n = \dfrac{1 + 19}{4} = 5 \quad \text{or} \quad n = \dfrac{1 - 19}{4} = -\dfrac{18}{4} = -4.5$

Since $n$ must be a positive integer, $n = 5$ .

Step 2 · Find $a_n$

Using $a_n = a + (n-1)d$ with $n = 5$

\begin{aligned} a_5 &= 2 + (5-1)8 \\[0.6em] &= 2 + 4 \times 8 \\[0.6em] &= 2 + 32 = 34 \end{aligned}

Answer

(vi) $n = 5, \quad a_n = 34$

(vii) Given $a = 8, a_n = 62, S_n = 210$ , find $n$ and $d$ .

Step 1 · Find the Number of Terms $n$

Using $S_n = \dfrac{n}{2}(a + a_n)$

\begin{aligned} 210 &= \dfrac{n}{2}(8 + 62) \\[0.6em] 210 &= \dfrac{n}{2}(70) \\[0.6em] 210 &= 35n \\[0.6em] n &= \dfrac{210}{35} = 6 \end{aligned}

Step 2 · Find the Common Difference $d$

Using $a_n = a + (n-1)d$ with $n = 6$

\begin{aligned} a_6 &= a + (6-1)d \\[0.6em] 62 &= 8 + 5d \\[0.6em] 62 - 8 &= 5d \\[0.6em] 54 &= 5d \\[0.6em] d &= \dfrac{54}{5} \end{aligned}

Answer

(vii) $n = 6, \quad d = \dfrac{54}{5}$

(viii) Given $a_n = 4, d = 2, S_n = -14$ , find $n$ and $a$ .

Step 1 · Express $a$ in terms of $n$

Using $a_n = a + (n-1)d$

\begin{aligned} 4 &= a + (n-1)2 \\[0.6em] 4 &= a + 2n - 2 \\[0.6em] a &= 6 - 2n \quad \dots (1) \end{aligned}

Step 2 · Form and Solve the Quadratic Equation for $n$

Using $S_n = \dfrac{n}{2}(a + a_n)$ :

\begin{aligned} -14 &= \dfrac{n}{2}(a + 4) \\[0.6em] -14 &= \dfrac{n}{2}[(6 - 2n) + 4] \\[0.6em] -14 &= \dfrac{n}{2}(10 - 2n) \\[0.6em] -14 &= n(5 - n) \\[0.6em] -14 &= 5n - n^2 \\[0.6em] n^2 - 5n - 14 &= 0 \\[0.6em] (n - 7)(n + 2) &= 0 \end{aligned}

$n = 7 \quad \text{or} \quad n = -2$

Since $n$ must be a positive integer, $n = 7$ .

Step 3 · Find the First Term $a$

Substitute $n = 7$ into equation (1):

\begin{aligned} a &= 6 - 2(7) \\[0.6em] &= 6 - 14 = -8 \end{aligned}

Answer

(viii) $n = 7, \quad a = -8$

(ix) Given $a = 3, n = 8, S = 192$ , find $d$ .

Step 1 · Find the Common Difference $d$

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$ with $n = 8$

\begin{aligned} 192 &= \dfrac{8}{2}[2(3) + (8-1)d] \\[0.6em] 192 &= 4(6 + 7d) \\[0.6em] \dfrac{192}{4} &= 6 + 7d \\[0.6em] 48 &= 6 + 7d \\[0.6em] 48 - 6 &= 7d \\[0.6em] 42 &= 7d \\[0.6em] d &= \dfrac{42}{7} = 6 \end{aligned}

Answer

(ix) $d = 6$

(x) Given $l = 28, S = 144$ , and there are total 9 terms. Find $a$ .

Step 1 · Find the First Term $a$

Here, last term $l = a_n = 28$ and $n = 9$ .

Using $S_n = \dfrac{n}{2}(a + l)$ :

\begin{aligned} 144 &= \dfrac{9}{2}(a + 28) \\[0.6em] 144 \times 2 &= 9(a + 28) \\[0.6em] 288 &= 9(a + 28) \\[0.6em] \dfrac{288}{9} &= a + 28 \\[0.6em] 32 &= a + 28 \\[0.6em] a &= 32 - 28 = 4 \end{aligned}

Answer

(x) $a = 4$

Common Mistakes

Discarding Negative or Fractional $n$ : The number of terms $n$ represents a count and must be a positive integer ( $n \in \{1, 2, 3, \dots\}$ ). Always reject negative or fractional roots when solving quadratic equations in $n$ .
Using the Wrong Sum Formula: When the last term $a_n$ is known, using $S_n = \dfrac{n}{2}(a + a_n)$ is much quicker than $S_n = \dfrac{n}{2}[2a + (n-1)d]$ .
Sign Errors with Negative Common Difference: When $d < 0$ , remember that $(n-1)d$ is subtracted, not added.

Question 3

More questions in Exercise 5.3