Arithmetic Progressions

(i) Yes, it is an AP. The fare after each km is ₹15, ₹23, ₹31, …, because each time the fare increases by ₹8. The common difference is 8.

(ii) No, it is not an AP. If a fixed fraction of air is removed each time, the remaining amount becomes $\frac34$ of the previous amount, so the terms decrease multiplicatively, not by a fixed difference.

(iii) Yes, it is an AP. The cost is ₹150, ₹200, ₹250, … since each metre adds ₹50. The common difference is 50.

(iv) No, it is not an AP. With compound interest, the amount increases by multiplying by a fixed factor each year, not by adding a fixed number.

Using the AP rule: first term + common difference repeatedly.

(i) $a=10, d=10$: 10, 20, 30, 40

(ii) $a=-2, d=0$: -2, -2, -2, -2

(iii) $a=4, d=-3$: 4, 1, -2, -5

(iv) $a=-1, d=\frac12$: -1, -\frac12, 0, \frac12

(v) $a=-1.25, d=-0.25$: -1.25, -1.50, -1.75, -2.00

For an AP, the first term is the first number and the common difference is the difference of consecutive terms.

(i) 3, 1, -1, -3, …
- First term $a=3$
- Common difference $d=1-3=-2$

(ii) -5, -1, 3, 7, …
- $a=-5$
- $d=-1-(-5)=4$

(iii) $\frac13, \frac53, \frac93, \frac{13}{3}, \dots$
- $a=\frac13$
- $d=\frac53-\frac13=\frac43$

(iv) 0.6, 1.7, 2.8, 3.9, …
- $a=0.6$
- $d=1.7-0.6=1.1$

Check the difference between consecutive terms.

(i) 2, 4, 8, 16, …
- Differences: 2, 4, 8
- Not an AP

(ii) 2, $\frac52$, 3, $\frac72$, …
- Differences: $\frac12, \frac12, \frac12$
- AP, with $d=\frac12$
- Next three terms: 4, $\frac92$, 5

(iii) -1.2, -3.2, -5.2, -7.2, …
- Differences: -2, -2, -2
- AP, with $d=-2$
- Next three terms: -9.2, -11.2, -13.2

(iv) -10, -6, -2, 2, …
- Differences: 4, 4, 4
- AP, with $d=4$
- Next three terms: 6, 10, 14

(v) 3, $3+\sqrt2$, $3+2\sqrt2$, $3+3\sqrt2$, …
- Differences: $\sqrt2, \sqrt2, \sqrt2$
- AP, with $d=\sqrt2$
- Next three terms: $3+4\sqrt2, 3+5\sqrt2, 3+6\sqrt2$

(vi) 0.2, 0.22, 0.222, 0.2222, …
- Differences are not equal
- Not an AP

(vii) 0, -4, -8, -12, …
- Differences: -4, -4, -4
- AP, with $d=-4$
- Next three terms: -16, -20, -24

(viii) $-\frac12, -\frac12, -\frac12, -\frac12, \dots$
- Differences are 0
- AP, with $d=0$
- Next three terms: $-\frac12, -\frac12, -\frac12$

(ix) 1, 3, 9, 27, …
- Differences are not equal
- Not an AP

(x) a, 2a, 3a, 4a, …
- Differences: a, a, a
- AP, with $d=a$
- Next three terms: 5a, 6a, 7a

(xi) a, $a^2$, $a^3$, $a^4$, …
- Differences are not generally equal
- Not an AP

(xii) $\sqrt2, \sqrt8, \sqrt{18}, \sqrt{32}, \dots$
- These are $\sqrt2, 2\sqrt2, 3\sqrt2, 4\sqrt2$
- Differences: $\sqrt2, \sqrt2, \sqrt2$
- AP, with $d=\sqrt2$
- Next three terms: $5\sqrt2, 6\sqrt2, 7\sqrt2$

(xiii) $\sqrt3, \sqrt6, \sqrt9, \sqrt{12}, \dots$
- Differences are not equal
- Not an AP

(xiv) $1^2, 3^2, 5^2, 7^2, \dots$
- Terms are 1, 9, 25, 49
- Differences are not equal
- Not an AP

(xv) $1^2, 5^2, 7^2, 73, \dots$
- As printed, the list does not have constant difference
- Not an AP

Use $a_n=a+(n-1)d$.

(i) $a=7, d=3, n=8$
$$a_8=7+(8-1)\cdot3=7+21=28$$

(ii) $a=-18, n=10, a_n=0$
$$0=-18+9d \Rightarrow 9d=18 \Rightarrow d=2$$

(iii) $d=-3, n=18, a_n=-5$
$$-5=a+17(-3)=a-51 \Rightarrow a=46$$

(iv) $a=-18.9, d=2.5, a_n=3.6$
$$3.6=-18.9+(n-1)2.5$$
$$22.5=(n-1)2.5 \Rightarrow n-1=9 \Rightarrow n=10$$

(v) $a=3.5, d=0, n=105$
$$a_{105}=3.5+(105-1)\cdot0=3.5$$

Use $a_n=a+(n-1)d$.

For 10th term of AP $10,7,4,\dots$:
$$a=10,\ d=-3$$
$$a_{30}=10+29(-3)=10-87=-77$$
So the correct option is (C) -77.

For 11th term of AP $-3,-\frac12,2,\dots$:
$$d=-\frac12-(-3)=\frac52$$
$$a_{11}=-3+10\cdot\frac52=-3+25=22$$
So the correct option is (B) 22.

Find the missing terms by using equal differences.

(i) 2, [ ], 26
- There are 3 terms, so middle term is the arithmetic mean:
$$\frac{2+26}{2}=14$$
So the missing term is 14.

(ii) [ ], 13, [ ], 3
Let the terms be $a, 13, b, 3$.
The common difference is
$$d=\frac{3-13}{2}=-5$$
So the terms are 18, 13, 8, 3.
Missing terms: 18, 8.

(iii) 5, [ ], [ ], 9$\frac12$
Let the common difference be $d$.
$$5+3d=9.5 \Rightarrow 3d=4.5 \Rightarrow d=1.5$$
So the missing terms are:
$$5+1.5=6.5=6\frac12, \quad 5+2(1.5)=8$$
Missing terms: **6$\frac12$, 8**.

(iv) -4, [ ], [ ], [ ], [ ], 6
There are 6 terms, so
$$d=\frac{6-(-4)}{5}=2$$
Terms: -4, -2, 0, 2, 4, 6
Missing terms: -2, 0, 2, 4

(v) [ ], 38, [ ], [ ], [ ], -22
There are 6 terms, so
$$d=\frac{-22-38}{4}=-15$$
Terms: 53, 38, 23, 8, -7, -22
Missing terms: 53, 23, 8, -7

For AP $3,8,13,18,\dots$,
$$a=3,\ d=5$$
Let the term be 78:
$$78=3+(n-1)5$$
$$75=5(n-1)$$
$$n-1=15 \Rightarrow n=16$$
So 78 is the 16th term.

Use $a_n=a+(n-1)d$ or the last term formula.

(i) 7, 13, 19, ..., 205
$$a=7,\ d=6$$
$$205=7+(n-1)6 \Rightarrow 198=6(n-1) \Rightarrow n-1=33 \Rightarrow n=34$$

(ii) 18, 15$\frac12$, 13, ..., -47
$$a=18,\ d=-2.5$$
$$-47=18+(n-1)(-2.5) \Rightarrow -65=-2.5(n-1) \Rightarrow n-1=26 \Rightarrow n=27$$

For AP $11,8,5,2,\dots$,
$$a=11,\ d=-3$$
Let the term be $-150$:
$$-150=11+(n-1)(-3)$$
$$-161=-3(n-1)$$
$$n-1=\frac{161}{3}$$
Since $n$ is not an integer, -150 is not a term of the AP.

For an AP, $a_n=a+(n-1)d$.

Given:
$$a_{11}=38,\quad a_{16}=73$$
So,
$$a+10d=38 \quad (1)$$
$$a+15d=73 \quad (2)$$
Subtract (1) from (2):
$$5d=35 \Rightarrow d=7$$
From (1):
$$a+70=38 \Rightarrow a=-32$$
Now,
$$a_{31}=a+30d=-32+30\cdot7=-32+210=178$$
So the 31st term is 178.

Given 50 terms, 3rd term = 12 and last term = 106.

Let first term be $a$ and common difference be $d$.
$$a+2d=12 \quad (1)$$
$$a+49d=106 \quad (2)$$
Subtract:
$$47d=94 \Rightarrow d=2$$
From (1):
$$a+4=12 \Rightarrow a=8$$
29th term:
$$a_{29}=a+28d=8+28\cdot2=8+56=64$$
So the 29th term is 64.

Let the AP have first term $a$ and common difference $d$.

Given:
$$a_3=a+2d=4 \quad (1)$$
$$a_9=a+8d=-8 \quad (2)$$
Subtract (1) from (2):
$$6d=-12 \Rightarrow d=-2$$
From (1):
$$a+2(-2)=4 \Rightarrow a=8$$
To find which term is zero:
$$a_n=a+(n-1)d=0$$
$$8+(n-1)(-2)=0$$
$$8-2n+2=0 \Rightarrow 10-2n=0 \Rightarrow n=5$$
So, the 5th term is zero.

For an AP, the difference between the 17th and 10th terms is
$$a_{17}-a_{10}=(17-10)d=7d$$
Given this equals 7:
$$7d=7 \Rightarrow d=1$$
So the common difference is 1.

For AP $3,15,27,39,\dots$,
$$a=3,\ d=12$$
54th term:
$$a_{54}=3+(54-1)12=3+636=639$$
We need the term that is 132 more than this:
$$639+132=771$$
Now solve:
$$a_n=3+(n-1)12=771$$
$$12(n-1)=768 \Rightarrow n-1=64 \Rightarrow n=65$$
So, it is the 65th term.

If two APs have the same common difference, then the difference between their corresponding terms is constant.

Difference between 100th terms = 100.

Difference between 1000th terms is also the same because both terms increase by the same amount for each step.

More explicitly, for the same common difference,
$$(a_{100}^{(1)}-a_{100}^{(2)})=(a_1^{(1)}-a_1^{(2)})+99(d-d)=a_1^{(1)}-a_1^{(2)}$$
So the difference remains 100.

Three-digit numbers divisible by 7 form the AP:
$$105, 112, 119, \dots, 994$$
Here,
$$a=105,\ d=7,\ l=994$$
Use
$$l=a+(n-1)d$$
$$994=105+(n-1)7$$
$$889=7(n-1)$$
$$n-1=127 \Rightarrow n=128$$
So, there are 128 such numbers.

Multiples of 4 between 10 and 250 are:
$$12,16,20,\dots,248$$
This is an AP with
$$a=12,\ d=4,\ l=248$$
Use
$$248=12+(n-1)4$$
$$236=4(n-1)$$
$$n-1=59 \Rightarrow n=60$$
So, there are 60 multiples of 4.

First AP: 63, 65, 67, …
$$a=63,\ d=2$$
Its nth term:
$$63+(n-1)2=2n+61$$
Second AP: 3, 10, 17, …
$$a=3,\ d=7$$
Its nth term:
$$3+(n-1)7=7n-4$$
Set them equal:
$$2n+61=7n-4$$
$$65=5n \Rightarrow n=13$$
So, the nth terms are equal for n = 13.

Let the AP have first term $a$ and common difference $d$.

Given:
$$a_3=a+2d=16 \quad (1)$$
The 7th term exceeds the 5th term by 12:
$$a_7-a_5=12$$
But
$$(a+6d)-(a+4d)=2d=12$$
So,
$$d=6$$
From (1):
$$a+12=16 \Rightarrow a=4$$
Hence the AP is
$$4, 10, 16, 22, 28, \dots$$

Use $S_n=\frac n2[2a+(n-1)d]$.

(i) 2, 7, 12, … to 10 terms
$$a=2, d=5, n=10$$
$$S_{10}=\frac{10}{2}[2\cdot2+9\cdot5]=5[4+45]=245$$

(ii) -37, -33, -29, … to 12 terms
$$a=-37, d=4, n=12$$
$$S_{12}=\frac{12}{2}[2(-37)+11\cdot4]=6[-74+44]=6(-30)=-180$$

(iii) 0.6, 1.7, 2.8, … to 100 terms
$$a=0.6, d=1.1, n=100$$
$$S_{100}=\frac{100}{2}[2(0.6)+99(1.1)] =50[1.2+108.9]=50\times110.1=5505$$

(iv) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$, to 11 terms
Here the common difference is
$$\frac{1}{12}-\frac{1}{15}=\frac{1}{60}$$
So $a=\frac{1}{15}, d=\frac{1}{60}, n=11$.
$$S_{11}=\frac{11}{2}\left[2\cdot\frac{1}{15}+10\cdot\frac{1}{60}\right] =\frac{11}{2}\left[\frac{2}{15}+\frac{1}{6}\right] =\frac{11}{2}\left[\frac{4}{30}+\frac{5}{30}\right] =\frac{11}{2}\cdot\frac{9}{30} =\frac{33}{20}$$

(i) $7+10\frac12+14+\dots+84$

This is an AP with
$$a=7,\ d=3.5=\frac72,\ l=84$$
Find number of terms:
$$84=7+(n-1)\frac72$$
$$77=(n-1)\frac72$$
$$n-1=22 \Rightarrow n=23$$
Now
$$S=\frac{23}{2}(7+84)=\frac{23}{2}\cdot91=\frac{2093}{2}=1046.5$$

(ii) $34+32+30+\dots+10$
$$a=34, d=-2, l=10$$
Find terms:
$$10=34+(n-1)(-2)$$
$$-24=-2(n-1) \Rightarrow n-1=12 \Rightarrow n=13$$
$$S=\frac{13}{2}(34+10)=\frac{13}{2}\cdot44=286$$

(iii) $-5+(-8)+(-11)+\dots+(-230)$
$$a=-5, d=-3, l=-230$$
$$-230=-5+(n-1)(-3)$$
$$-225=-3(n-1) \Rightarrow n-1=75 \Rightarrow n=76$$
$$S=\frac{76}{2}(-5-230)=38(-235)=-8930$$

(i) $a=5, d=3, a_n=50$
$$50=5+(n-1)3 \Rightarrow 45=3(n-1) \Rightarrow n=16$$
$$S_{16}=\frac{16}{2}[2\cdot5+15\cdot3]=8[10+45]=440$$

(ii) $a=7, a_{13}=35$
$$35=7+12d \Rightarrow 12d=28 \Rightarrow d=\frac73$$
$$S_{13}=\frac{13}{2}[2\cdot7+12\cdot\frac73] =\frac{13}{2}[14+28]=\frac{13}{2}\cdot42=273$$

(iii) $a_{12}=37, d=3$
$$37=a+11\cdot3=a+33 \Rightarrow a=4$$
$$S_{12}=\frac{12}{2}[2\cdot4+11\cdot3]=6[8+33]=246$$

(iv) $a_3=15, S_{10}=125$
$$a+2d=15 \quad (1)$$
$$125=\frac{10}{2}[2a+9d]=5(2a+9d) \Rightarrow 2a+9d=25 \quad (2)$$
From (1), $a=15-2d$. Substitute in (2):
$$2(15-2d)+9d=25 \Rightarrow 30-4d+9d=25 \Rightarrow 5d=-5 \Rightarrow d=-1$$
$$a=15-2(-1)=17$$
Then
$$a_{10}=a+9d=17-9=8$$

(v) $d=5, S_9=75$
$$75=\frac92[2a+8\cdot5] \Rightarrow 75=\frac92(2a+40)$$
$$150=9(2a+40) \Rightarrow 2a+40=\frac{150}{9}=\frac{50}{3}$$
$$2a=\frac{50}{3}-40=\frac{50-120}{3}=-\frac{70}{3} \Rightarrow a=-\frac{35}{3}$$
$$a_9=a+8d=-\frac{35}{3}+40=\frac{85}{3}$$

(vi) $a=2, d=8, S_n=90$
$$90=\frac{n}{2}[4+8(n-1)] =\frac{n}{2}(8n-4) =n(4n-2)$$
$$4n^2-2n-90=0 \Rightarrow 2n^2-n-45=0$$
$$(2n+9)(n-5)=0$$
So $n=5$.
$$a_n=2+4\cdot8=34$$

(vii) $a=8, a_n=62, S_n=210$
$$62=8+(n-1)d \Rightarrow (n-1)d=54 \quad (1)$$
$$210=\frac{n}{2}(8+62)=35n \Rightarrow n=6$$
From (1):
$$5d=54 \Rightarrow d=\frac{54}{5}$$

(viii) $a_n=4, d=2, S_n=-14$
$$4=a+2(n-1) \Rightarrow a=6-2n$$
$$-14=\frac{n}{2}(a+4)$$
Substitute $a=6-2n$:
$$-14=\frac{n}{2}(10-2n)=n(5-n)$$
$$-n^2+5n+14=0 \Rightarrow n^2-5n-14=0$$
$$(n-7)(n+2)=0 \Rightarrow n=7$$
$$a=6-14=-8$$

(ix) $a=3, n=8, S=192$
$$192=\frac{8}{2}[2\cdot3+7d]=4(6+7d)$$
$$48=6+7d \Rightarrow 7d=42 \Rightarrow d=6$$

(x) $l=28, S=144$, total 9 terms
$$144=\frac92(a+28) \Rightarrow 288=9(a+28) \Rightarrow a+28=32 \Rightarrow a=4$$