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NCERT Solutions

Arithmetic Progressions

Himachal Pradesh Board · Class 10 · Mathematics

NCERT Solutions for Arithmetic Progressions — Himachal Pradesh Board Class 10 Mathematics.

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Illustrates the structure of an Arithmetic Progression (A.P.), showing the first term 'a' and common difference 'd', along with the general term formula.
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49 Questions Solved · 4 Sections

Exercise 5.1

1In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum.Show solution
(i) Taxi fare situation:

Given: Fare for 1st km = ₹15, each additional km = ₹8.

Fare after 1st km = ₹15
Fare after 2nd km = ₹15 + ₹8 = ₹23
Fare after 3rd km = ₹23 + ₹8 = ₹31
Fare after 4th km = ₹31 + ₹8 = ₹39

The list is: 15, 23, 31, 39, …

Differences: 23−15=823 - 15 = 8, 31−23=831 - 23 = 8, 39−31=839 - 31 = 8

Since each successive difference is the same (= 8), this list forms an AP with a=15a = 15 and d=8d = 8.

---

(ii) Air in cylinder:

Let initial amount of air = VV.

After 1st pump: V−V4=3V4V - \dfrac{V}{4} = \dfrac{3V}{4}

After 2nd pump: 3V4−14⋅3V4=3V4×34=9V16\dfrac{3V}{4} - \dfrac{1}{4}\cdot\dfrac{3V}{4} = \dfrac{3V}{4}\times\dfrac{3}{4} = \dfrac{9V}{16}

After 3rd pump: 9V16×34=27V64\dfrac{9V}{16}\times\dfrac{3}{4} = \dfrac{27V}{64}

The list is: V, 3V4, 9V16, 27V64,…V,\ \dfrac{3V}{4},\ \dfrac{9V}{16},\ \dfrac{27V}{64},\ldots

Difference: 3V4−V=−V4\dfrac{3V}{4} - V = -\dfrac{V}{4}

9V16−3V4=9V16−12V16=−3V16\dfrac{9V}{16} - \dfrac{3V}{4} = \dfrac{9V}{16} - \dfrac{12V}{16} = -\dfrac{3V}{16}

Since −V4≠−3V16-\dfrac{V}{4} \neq -\dfrac{3V}{16}, the differences are not equal.

This list does NOT form an AP.

---

(iii) Cost of digging a well:

Cost for 1st metre = ₹150
Cost for 2nd metre = ₹150 + ₹50 = ₹200
Cost for 3rd metre = ₹200 + ₹50 = ₹250
Cost for 4th metre = ₹250 + ₹50 = ₹300

The list is: 150, 200, 250, 300, …

Differences: 200−150=50200-150=50, 250−200=50250-200=50, 300−250=50300-250=50

Since each successive difference is the same (= 50), this list forms an AP with a=150a = 150 and d=50d = 50.

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(iv) Compound interest:

Amount after 1st year = 10000(1+8100)1=1080010000\left(1+\dfrac{8}{100}\right)^1 = 10800

Amount after 2nd year = 10000(1+8100)2=1166410000\left(1+\dfrac{8}{100}\right)^2 = 11664

Amount after 3rd year = 10000(1+8100)3=12597.1210000\left(1+\dfrac{8}{100}\right)^3 = 12597.12

Differences: 11664−10800=86411664 - 10800 = 864; 12597.12−11664=933.1212597.12 - 11664 = 933.12

Since the differences are not equal, this list does NOT form an AP.
2Write first four terms of the AP, when the first term aa and the common difference dd are given as follows:
(i) a=10a = 10, d=10d = 10
(ii) a=4a = 4, d=−3d = -3
(iii) a=−2a = -2, d=0d = 0
(iv) a=−1a = -1, d=12d = \frac{1}{2}
(v) a=−1.25a = -1.25, d=−0.25d = -0.25Show solution
The first four terms of an AP are: a, a+d, a+2d, a+3da,\ a+d,\ a+2d,\ a+3d.

(i) a=10a = 10, d=10d = 10

a1=10,a2=10+10=20,a3=20+10=30,a4=30+10=40a_1 = 10,\quad a_2 = 10+10 = 20,\quad a_3 = 20+10 = 30,\quad a_4 = 30+10 = 40

First four terms: 10, 20, 30, 40

---

(ii) a=4a = 4, d=−3d = -3

a1=4,a2=4+(−3)=1,a3=1+(−3)=−2,a4=−2+(−3)=−5a_1 = 4,\quad a_2 = 4+(-3) = 1,\quad a_3 = 1+(-3) = -2,\quad a_4 = -2+(-3) = -5

First four terms: 4, 1, −2, −5

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(iii) a=−2a = -2, d=0d = 0

a1=−2,a2=−2+0=−2,a3=−2,a4=−2a_1 = -2,\quad a_2 = -2+0 = -2,\quad a_3 = -2,\quad a_4 = -2

First four terms: −2, −2, −2, −2

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(iv) a=−1a = -1, d=12d = \dfrac{1}{2}

a1=−1,a2=−1+12=−12,a3=−12+12=0,a4=0+12=12a_1 = -1,\quad a_2 = -1+\frac{1}{2} = -\frac{1}{2},\quad a_3 = -\frac{1}{2}+\frac{1}{2} = 0,\quad a_4 = 0+\frac{1}{2} = \frac{1}{2}

First four terms: −1, −12, 0, 12-1,\ -\dfrac{1}{2},\ 0,\ \dfrac{1}{2}

---

(v) a=−1.25a = -1.25, d=−0.25d = -0.25

a1=−1.25,a2=−1.25+(−0.25)=−1.50,a3=−1.75,a4=−2.00a_1 = -1.25,\quad a_2 = -1.25+(-0.25) = -1.50,\quad a_3 = -1.75,\quad a_4 = -2.00

First four terms: −1.25, −1.50, −1.75, −2.00
3For the following APs, write the first term and the common difference:
(i) 3,1,−1,−3,…3, 1, -1, -3, \ldots
(ii) −5,−1,3,7,…-5, -1, 3, 7, \ldots
(iii) 13,53,93,133,…\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots
(iv) 0.6,1.7,2.8,3.9,…0.6, 1.7, 2.8, 3.9, \ldotsShow solution
(i) 3,1,−1,−3,…3, 1, -1, -3, \ldots

First term a=3a = 3

Common difference d=1−3=−2d = 1 - 3 = -2

---

(ii) −5,−1,3,7,…-5, -1, 3, 7, \ldots

First term a=−5a = -5

Common difference d=−1−(−5)=4d = -1 - (-5) = 4

---

(iii) 13,53,93,133,…\dfrac{1}{3}, \dfrac{5}{3}, \dfrac{9}{3}, \dfrac{13}{3}, \ldots

First term a=13a = \dfrac{1}{3}

Common difference d=53−13=43d = \dfrac{5}{3} - \dfrac{1}{3} = \dfrac{4}{3}

---

(iv) 0.6,1.7,2.8,3.9,…0.6, 1.7, 2.8, 3.9, \ldots

First term a=0.6a = 0.6

Common difference d=1.7−0.6=1.1d = 1.7 - 0.6 = 1.1
4Which of the following are APs? If they form an AP, find the common difference dd and write three more terms.
(i) 2,4,8,16,…2, 4, 8, 16, \ldots
(ii) 2,52,3,72,…2, \frac{5}{2}, 3, \frac{7}{2}, \ldots
(iii) −1.2,−3.2,−5.2,−7.2,…-1.2, -3.2, -5.2, -7.2, \ldots
(iv) −10,−6,−2,2,…-10, -6, -2, 2, \ldots
(v) 3,3+2,3+22,3+32,…3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \ldots
(vi) 0.2,0.22,0.222,0.2222,…0.2, 0.22, 0.222, 0.2222, \ldots
(vii) 0,−4,−8,−12,…0, -4, -8, -12, \ldots
(viii) −12,−12,−12,−12,…-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \ldots
(ix) 1,3,9,27,…1, 3, 9, 27, \ldots
(x) a,2a,3a,4a,…a, 2a, 3a, 4a, \ldots
(xii) 2,8,18,32,…\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots
(xiii) 3,6,9,12,…\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots
(xiv) 12,52,72,73,…1^2, 5^2, 7^2, 73, \ldotsShow solution
(i) 2,4,8,16,…2, 4, 8, 16, \ldots

a2−a1=4−2=2a_2 - a_1 = 4 - 2 = 2
a3−a2=8−4=4a_3 - a_2 = 8 - 4 = 4

Since 2≠42 \neq 4, this is NOT an AP.

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(ii) 2,52,3,72,…2, \dfrac{5}{2}, 3, \dfrac{7}{2}, \ldots

a2−a1=52−2=12a_2 - a_1 = \dfrac{5}{2} - 2 = \dfrac{1}{2}
a3−a2=3−52=12a_3 - a_2 = 3 - \dfrac{5}{2} = \dfrac{1}{2}
a4−a3=72−3=12a_4 - a_3 = \dfrac{7}{2} - 3 = \dfrac{1}{2}

This is an AP with d=12d = \dfrac{1}{2}.

Next three terms:
a5=72+12=4a_5 = \dfrac{7}{2} + \dfrac{1}{2} = 4
a6=4+12=92a_6 = 4 + \dfrac{1}{2} = \dfrac{9}{2}
a7=92+12=5a_7 = \dfrac{9}{2} + \dfrac{1}{2} = 5

Three more terms: 4, 92, 54,\ \dfrac{9}{2},\ 5

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(iii) −1.2,−3.2,−5.2,−7.2,…-1.2, -3.2, -5.2, -7.2, \ldots

a2−a1=−3.2−(−1.2)=−2a_2 - a_1 = -3.2 - (-1.2) = -2
a3−a2=−5.2−(−3.2)=−2a_3 - a_2 = -5.2 - (-3.2) = -2
a4−a3=−7.2−(−5.2)=−2a_4 - a_3 = -7.2 - (-5.2) = -2

This is an AP with d=−2d = -2.

Next three terms:
a5=−7.2+(−2)=−9.2a_5 = -7.2 + (-2) = -9.2
a6=−11.2a_6 = -11.2
a7=−13.2a_7 = -13.2

Three more terms: −9.2, −11.2, −13.2-9.2,\ -11.2,\ -13.2

---

(iv) −10,−6,−2,2,…-10, -6, -2, 2, \ldots

a2−a1=−6−(−10)=4a_2 - a_1 = -6 - (-10) = 4
a3−a2=−2−(−6)=4a_3 - a_2 = -2 - (-6) = 4
a4−a3=2−(−2)=4a_4 - a_3 = 2 - (-2) = 4

This is an AP with d=4d = 4.

Next three terms:
a5=2+4=6a_5 = 2 + 4 = 6
a6=10a_6 = 10
a7=14a_7 = 14

Three more terms: 6, 10, 146,\ 10,\ 14

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(v) 3,3+2,3+22,3+32,…3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \ldots

a2−a1=(3+2)−3=2a_2 - a_1 = (3+\sqrt{2}) - 3 = \sqrt{2}
a3−a2=(3+22)−(3+2)=2a_3 - a_2 = (3+2\sqrt{2}) - (3+\sqrt{2}) = \sqrt{2}
a4−a3=2a_4 - a_3 = \sqrt{2}

This is an AP with d=2d = \sqrt{2}.

Next three terms:
a5=3+42a_5 = 3+4\sqrt{2}
a6=3+52a_6 = 3+5\sqrt{2}
a7=3+62a_7 = 3+6\sqrt{2}

Three more terms: 3+42, 3+52, 3+623+4\sqrt{2},\ 3+5\sqrt{2},\ 3+6\sqrt{2}

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(vi) 0.2,0.22,0.222,0.2222,…0.2, 0.22, 0.222, 0.2222, \ldots

a2−a1=0.22−0.2=0.02a_2 - a_1 = 0.22 - 0.2 = 0.02
a3−a2=0.222−0.22=0.002a_3 - a_2 = 0.222 - 0.22 = 0.002

Since 0.02≠0.0020.02 \neq 0.002, this is NOT an AP.

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(vii) 0,−4,−8,−12,…0, -4, -8, -12, \ldots

a2−a1=−4−0=−4a_2 - a_1 = -4 - 0 = -4
a3−a2=−8−(−4)=−4a_3 - a_2 = -8 - (-4) = -4
a4−a3=−12−(−8)=−4a_4 - a_3 = -12 - (-8) = -4

This is an AP with d=−4d = -4.

Next three terms:
a5=−12+(−4)=−16a_5 = -12 + (-4) = -16
a6=−20a_6 = -20
a7=−24a_7 = -24

Three more terms: −16, −20, −24-16,\ -20,\ -24

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(viii) −12,−12,−12,−12,…-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, \ldots

a2−a1=0a_2 - a_1 = 0, a3−a2=0a_3 - a_2 = 0, a4−a3=0a_4 - a_3 = 0

This is an AP with d=0d = 0.

Next three terms: −12, −12, −12-\dfrac{1}{2},\ -\dfrac{1}{2},\ -\dfrac{1}{2}

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(ix) 1,3,9,27,…1, 3, 9, 27, \ldots

a2−a1=3−1=2a_2 - a_1 = 3 - 1 = 2
a3−a2=9−3=6a_3 - a_2 = 9 - 3 = 6

Since 2≠62 \neq 6, this is NOT an AP.

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(x) a,2a,3a,4a,…a, 2a, 3a, 4a, \ldots

a2−a1=2a−a=aa_2 - a_1 = 2a - a = a
a3−a2=3a−2a=aa_3 - a_2 = 3a - 2a = a
a4−a3=4a−3a=aa_4 - a_3 = 4a - 3a = a

This is an AP with d=ad = a.

Next three terms:
a5=5a,a6=6a,a7=7aa_5 = 5a,\quad a_6 = 6a,\quad a_7 = 7a

Three more terms: 5a, 6a, 7a5a,\ 6a,\ 7a

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(xii) 2,8,18,32,…\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots

Simplify: 2, 22, 32, 42,…\sqrt{2},\ 2\sqrt{2},\ 3\sqrt{2},\ 4\sqrt{2}, \ldots

a2−a1=22−2=2a_2 - a_1 = 2\sqrt{2} - \sqrt{2} = \sqrt{2}
a3−a2=32−22=2a_3 - a_2 = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}
a4−a3=2a_4 - a_3 = \sqrt{2}

This is an AP with d=2d = \sqrt{2}.

Next three terms:
a5=52=50a_5 = 5\sqrt{2} = \sqrt{50}
a6=62=72a_6 = 6\sqrt{2} = \sqrt{72}
a7=72=98a_7 = 7\sqrt{2} = \sqrt{98}

Three more terms: 50, 72, 98\sqrt{50},\ \sqrt{72},\ \sqrt{98}

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(xiii) 3,6,9,12,…\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots

a2−a1=6−3=3(2−1)a_2 - a_1 = \sqrt{6} - \sqrt{3} = \sqrt{3}(\sqrt{2}-1)
a3−a2=9−6=3−6a_3 - a_2 = \sqrt{9} - \sqrt{6} = 3 - \sqrt{6}

Since 3(2−1)≈0.732\sqrt{3}(\sqrt{2}-1) \approx 0.732 and 3−6≈0.5513 - \sqrt{6} \approx 0.551, these are not equal.

This is NOT an AP.

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(xiv) 12,52,72,73,…1^2, 5^2, 7^2, 73, \ldots i.e., 1,25,49,73,…1, 25, 49, 73, \ldots

a2−a1=25−1=24a_2 - a_1 = 25 - 1 = 24
a3−a2=49−25=24a_3 - a_2 = 49 - 25 = 24
a4−a3=73−49=24a_4 - a_3 = 73 - 49 = 24

This is an AP with d=24d = 24.

Next three terms:
a5=73+24=97a_5 = 73 + 24 = 97
a6=121a_6 = 121
a7=145a_7 = 145

Three more terms: 97, 121, 14597,\ 121,\ 145

Exercise 5.2

1Fill in the blanks in the following table, given that aa is the first term, dd the common difference and ana_n the nnth term of the AP:
(i) a=7a=7, d=3d=3, n=8n=8, an=?a_n=?
(ii) a=−18a=-18, d=?d=?, n=10n=10, an=0a_n=0
(iii) a=?a=?, d=−3d=-3, n=18n=18, an=−5a_n=-5
(iv) a=−18.9a=-18.9, d=2.5d=2.5, n=?n=?, an=3.6a_n=3.6
(v) a=3.5a=3.5, d=0d=0, n=105n=105, an=?a_n=?Show solution
Using the formula an=a+(n−1)da_n = a + (n-1)d:

(i) a=7a=7, d=3d=3, n=8n=8

a8=7+(8−1)×3=7+21=28a_8 = 7 + (8-1)\times 3 = 7 + 21 = 28

an=28\boxed{a_n = 28}

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(ii) a=−18a=-18, n=10n=10, a10=0a_{10}=0

0=−18+(10−1)d0 = -18 + (10-1)d
9d=189d = 18
d=2d = 2

d=2\boxed{d = 2}

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(iii) d=−3d=-3, n=18n=18, a18=−5a_{18}=-5

−5=a+(18−1)×(−3)-5 = a + (18-1)\times(-3)
−5=a−51-5 = a - 51
a=46a = 46

a=46\boxed{a = 46}

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(iv) a=−18.9a=-18.9, d=2.5d=2.5, an=3.6a_n=3.6

3.6=−18.9+(n−1)×2.53.6 = -18.9 + (n-1)\times 2.5
(n−1)×2.5=3.6+18.9=22.5(n-1)\times 2.5 = 3.6 + 18.9 = 22.5
n−1=9n-1 = 9
n=10n = 10

n=10\boxed{n = 10}

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(v) a=3.5a=3.5, d=0d=0, n=105n=105

a105=3.5+(105−1)×0=3.5a_{105} = 3.5 + (105-1)\times 0 = 3.5

an=3.5\boxed{a_n = 3.5}
2Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, ..., is
(A) 97 (B) 77 (C) -77 (D) -87
(ii) 11th term of the AP: -3, -1/2, 2, ..., is
(A) 28 (B) 22 (C) -38 (D) -48½Show solution
(i) AP: 10, 7, 4, …

a=10a = 10, d=7−10=−3d = 7 - 10 = -3

a30=a+(30−1)d=10+29×(−3)=10−87=−77a_{30} = a + (30-1)d = 10 + 29\times(-3) = 10 - 87 = -77

Correct option: (C) −77

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(ii) AP: −3,−12,2,…-3, -\dfrac{1}{2}, 2, \ldots

a=−3a = -3, d=−12−(−3)=−12+3=52d = -\dfrac{1}{2} - (-3) = -\dfrac{1}{2} + 3 = \dfrac{5}{2}

a11=−3+(11−1)×52=−3+10×52=−3+25=22a_{11} = -3 + (11-1)\times\frac{5}{2} = -3 + 10\times\frac{5}{2} = -3 + 25 = 22

Correct option: (B) 22
3In the following APs, find the missing terms in the boxes:
(i) 2, □, 26
(ii) □, 13, □, 3
(iii) 5, □, □, 9½
(iv) -4, □, □, □, □, 6
(v) □, 38, □, □, □, -22Show solution
(i) 2, □, 26

Here a1=2a_1 = 2, a3=26a_3 = 26.

a3=a+2d⇒26=2+2d⇒d=12a_3 = a + 2d \Rightarrow 26 = 2 + 2d \Rightarrow d = 12

a2=2+12=14a_2 = 2 + 12 = 14

Missing term: 14

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(ii) □, 13, □, 3

a2=13a_2 = 13, a4=3a_4 = 3

a4=a2+2d⇒3=13+2d⇒d=−5a_4 = a_2 + 2d \Rightarrow 3 = 13 + 2d \Rightarrow d = -5

a1=13−(−5)=18a_1 = 13 - (-5) = 18

a3=13+(−5)=8a_3 = 13 + (-5) = 8

Missing terms: 18, 8

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(iii) 5, □, □, 9129\dfrac{1}{2}

a1=5a_1 = 5, a4=192a_4 = \dfrac{19}{2}

a4=a+3d⇒192=5+3d⇒3d=92⇒d=32a_4 = a + 3d \Rightarrow \dfrac{19}{2} = 5 + 3d \Rightarrow 3d = \dfrac{9}{2} \Rightarrow d = \dfrac{3}{2}

a2=5+32=132=6.5a_2 = 5 + \dfrac{3}{2} = \dfrac{13}{2} = 6.5

a3=5+2×32=5+3=8a_3 = 5 + 2\times\dfrac{3}{2} = 5 + 3 = 8

Missing terms: 6126\dfrac{1}{2}, 88

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(iv) −4-4, □, □, □, □, 66

a1=−4a_1 = -4, a6=6a_6 = 6

a6=a+5d⇒6=−4+5d⇒d=2a_6 = a + 5d \Rightarrow 6 = -4 + 5d \Rightarrow d = 2

a2=−4+2=−2a_2 = -4+2 = -2
a3=−2+2=0a_3 = -2+2 = 0
a4=0+2=2a_4 = 0+2 = 2
a5=2+2=4a_5 = 2+2 = 4

Missing terms: −2, 0, 2, 4

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(v) □, 38, □, □, □, −22-22

a2=38a_2 = 38, a6=−22a_6 = -22

a6=a2+4d⇒−22=38+4d⇒4d=−60⇒d=−15a_6 = a_2 + 4d \Rightarrow -22 = 38 + 4d \Rightarrow 4d = -60 \Rightarrow d = -15

a1=38−(−15)=53a_1 = 38 - (-15) = 53
a3=38+(−15)=23a_3 = 38 + (-15) = 23
a4=23+(−15)=8a_4 = 23 + (-15) = 8
a5=8+(−15)=−7a_5 = 8 + (-15) = -7

Missing terms: 53, 23, 8, −7
4Which term of the AP: 3, 8, 13, 18, ..., is 78?Show solution
Given AP: 3, 8, 13, 18, …

a=3a = 3, d=8−3=5d = 8 - 3 = 5

Let an=78a_n = 78.

Using an=a+(n−1)da_n = a + (n-1)d:

78=3+(n−1)×578 = 3 + (n-1)\times 5
75=(n−1)×575 = (n-1)\times 5
n−1=15n - 1 = 15
n=16n = 16

78 is the 16th term of the AP.
5Find the number of terms in each of the following APs:
(i) 7, 13, 19, ..., 205
(ii) 18, 15½, 13, ..., -47Show solution
(i) 7, 13, 19, …, 205

a=7a = 7, d=13−7=6d = 13 - 7 = 6, an=205a_n = 205

205=7+(n−1)×6205 = 7 + (n-1)\times 6
198=(n−1)×6198 = (n-1)\times 6
n−1=33n - 1 = 33
n=34n = 34

There are 34 terms.

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(ii) 18,1512,13,…,−4718, 15\dfrac{1}{2}, 13, \ldots, -47

a=18a = 18, d=15.5−18=−2.5d = 15.5 - 18 = -2.5, an=−47a_n = -47

−47=18+(n−1)×(−2.5)-47 = 18 + (n-1)\times(-2.5)
−65=(n−1)×(−2.5)-65 = (n-1)\times(-2.5)
n−1=26n - 1 = 26
n=27n = 27

There are 27 terms.
6Check whether -150 is a term of the AP: 11, 8, 5, 2, ...Show solution
Given AP: 11, 8, 5, 2, …

a=11a = 11, d=8−11=−3d = 8 - 11 = -3

Assume an=−150a_n = -150.

−150=11+(n−1)×(−3)-150 = 11 + (n-1)\times(-3)
−161=(n−1)×(−3)-161 = (n-1)\times(-3)
n−1=1613=53.6‾n - 1 = \frac{161}{3} = 53.\overline{6}

Since nn is not a natural number, −150 is NOT a term of this AP.
7Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.Show solution
Given: a11=38a_{11} = 38 and a16=73a_{16} = 73.

Using an=a+(n−1)da_n = a + (n-1)d:

a+10d=38⋯(1)a + 10d = 38 \quad \cdots (1)
a+15d=73⋯(2)a + 15d = 73 \quad \cdots (2)

Subtracting (1) from (2):
5d=35⇒d=75d = 35 \Rightarrow d = 7

From (1): a=38−70=−32a = 38 - 70 = -32

a31=a+30d=−32+30×7=−32+210=178a_{31} = a + 30d = -32 + 30\times 7 = -32 + 210 = 178

The 31st term is 178.
8An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.Show solution
Given: n=50n = 50, a3=12a_3 = 12, a50=106a_{50} = 106.

a+2d=12⋯(1)a + 2d = 12 \quad \cdots (1)
a+49d=106⋯(2)a + 49d = 106 \quad \cdots (2)

Subtracting (1) from (2):
47d=94⇒d=247d = 94 \Rightarrow d = 2

From (1): a=12−4=8a = 12 - 4 = 8

a29=a+28d=8+28×2=8+56=64a_{29} = a + 28d = 8 + 28\times 2 = 8 + 56 = 64

The 29th term is 64.
9If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?Show solution
Given: a3=4a_3 = 4 and a9=−8a_9 = -8.

a+2d=4⋯(1)a + 2d = 4 \quad \cdots (1)
a+8d=−8⋯(2)a + 8d = -8 \quad \cdots (2)

Subtracting (1) from (2):
6d=−12⇒d=−26d = -12 \Rightarrow d = -2

From (1): a=4−2(−2)=4+4=8a = 4 - 2(-2) = 4 + 4 = 8

Let an=0a_n = 0:
0=8+(n−1)×(−2)0 = 8 + (n-1)\times(-2)
(n−1)×2=8(n-1)\times 2 = 8
n−1=4⇒n=5n - 1 = 4 \Rightarrow n = 5

The 5th term of this AP is zero.
10The 17th term of an AP exceeds its 10th term by 7. Find the common difference.Show solution
Given: a17−a10=7a_{17} - a_{10} = 7

[a+16d]−[a+9d]=7[a + 16d] - [a + 9d] = 7
7d=77d = 7
d=1d = 1

The common difference is 1.
11Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?Show solution
Given AP: 3, 15, 27, 39, …

a=3a = 3, d=15−3=12d = 15 - 3 = 12

First, find the 54th term:
a54=3+53×12=3+636=639a_{54} = 3 + 53\times 12 = 3 + 636 = 639

Let the required term be ana_n:
an=a54+132=639+132=771a_n = a_{54} + 132 = 639 + 132 = 771

771=3+(n−1)×12771 = 3 + (n-1)\times 12
768=(n−1)×12768 = (n-1)\times 12
n−1=64⇒n=65n - 1 = 64 \Rightarrow n = 65

The 65th term is 132 more than the 54th term.
12Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?Show solution
Let the two APs have first terms aa and bb respectively, with the same common difference dd.

100th term of 1st AP: a+99da + 99d
100th term of 2nd AP: b+99db + 99d

Given: (a+99d)−(b+99d)=100(a + 99d) - (b + 99d) = 100
⇒a−b=100\Rightarrow a - b = 100

1000th term of 1st AP: a+999da + 999d
1000th term of 2nd AP: b+999db + 999d

Difference =(a+999d)−(b+999d)=a−b=100= (a + 999d) - (b + 999d) = a - b = 100

The difference between their 1000th terms is also 100.
13How many three-digit numbers are divisible by 7?Show solution
The smallest three-digit number divisible by 7:
100÷7=14100 \div 7 = 14 remainder 22, so smallest = 100+(7−2)=105100 + (7-2) = 105

The largest three-digit number divisible by 7:
999÷7=142999 \div 7 = 142 remainder 55, so largest = 999−5=994999 - 5 = 994

AP: 105, 112, 119, …, 994 with a=105a = 105, d=7d = 7, an=994a_n = 994

994=105+(n−1)×7994 = 105 + (n-1)\times 7
889=(n−1)×7889 = (n-1)\times 7
n−1=127⇒n=128n - 1 = 127 \Rightarrow n = 128

There are 128 three-digit numbers divisible by 7.
14How many multiples of 4 lie between 10 and 250?Show solution
The smallest multiple of 4 greater than 10 is 12.
The largest multiple of 4 less than 250 is 248.

AP: 12, 16, 20, …, 248 with a=12a = 12, d=4d = 4, an=248a_n = 248

248=12+(n−1)×4248 = 12 + (n-1)\times 4
236=(n−1)×4236 = (n-1)\times 4
n−1=59⇒n=60n - 1 = 59 \Rightarrow n = 60

There are 60 multiples of 4 between 10 and 250.
15For what value of nn, are the nnth terms of two APs: 63, 65, 67, ... and 3, 10, 17, ... equal?Show solution
AP 1: 63, 65, 67, … → a1=63a_1 = 63, d1=2d_1 = 2

nnth term: 63+(n−1)×2=61+2n63 + (n-1)\times 2 = 61 + 2n

AP 2: 3, 10, 17, … → a2=3a_2 = 3, d2=7d_2 = 7

nnth term: 3+(n−1)×7=−4+7n3 + (n-1)\times 7 = -4 + 7n

Setting them equal:
61+2n=−4+7n61 + 2n = -4 + 7n
65=5n65 = 5n
n=13n = 13

The 13th terms of the two APs are equal.
16Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.Show solution
Given: a3=16a_3 = 16 and a7−a5=12a_7 - a_5 = 12.

From a7−a5=12a_7 - a_5 = 12:
[a+6d]−[a+4d]=12[a + 6d] - [a + 4d] = 12
2d=12⇒d=62d = 12 \Rightarrow d = 6

From a3=16a_3 = 16:
a+2d=16a + 2d = 16
a+12=16⇒a=4a + 12 = 16 \Rightarrow a = 4

The AP is: 4,10,16,22,28,…4, 10, 16, 22, 28, \ldots

The AP is 4, 10, 16, 22, 28, …
17Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253.Show solution
Given AP: 3, 8, 13, …, 253

a=3a = 3, d=5d = 5, last term l=253l = 253

The 20th term from the last term is the same as the 20th term of the AP written in reverse, which has first term 253 and common difference −5-5.

a20 (from last)=253+(20−1)×(−5)=253−95=158a_{20}\text{ (from last)} = 253 + (20-1)\times(-5) = 253 - 95 = 158

The 20th term from the last is 158.
18The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.Show solution
Given: a4+a8=24a_4 + a_8 = 24 and a6+a10=44a_6 + a_{10} = 44.

a4+a8=(a+3d)+(a+7d)=2a+10d=24a_4 + a_8 = (a+3d) + (a+7d) = 2a + 10d = 24
⇒a+5d=12⋯(1)\Rightarrow a + 5d = 12 \quad \cdots (1)

a6+a10=(a+5d)+(a+9d)=2a+14d=44a_6 + a_{10} = (a+5d) + (a+9d) = 2a + 14d = 44
⇒a+7d=22⋯(2)\Rightarrow a + 7d = 22 \quad \cdots (2)

Subtracting (1) from (2):
2d=10⇒d=52d = 10 \Rightarrow d = 5

From (1): a=12−25=−13a = 12 - 25 = -13

First three terms:
a1=−13a_1 = -13
a2=−13+5=−8a_2 = -13 + 5 = -8
a3=−8+5=−3a_3 = -8 + 5 = -3

The first three terms are −13, −8, −3.
19Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?Show solution
Given: a=5000a = 5000, d=200d = 200, an=7000a_n = 7000.

7000=5000+(n−1)×2007000 = 5000 + (n-1)\times 200
2000=(n−1)×2002000 = (n-1)\times 200
n−1=10⇒n=11n - 1 = 10 \Rightarrow n = 11

The 11th year from 1995 is 1995+10=20051995 + 10 = 2005.

Subba Rao's income reached ₹ 7000 in the year 2005.
20Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nnth week, her weekly savings become ₹ 20.75, find nn.Show solution
Given: a=5a = 5, d=1.75d = 1.75, an=20.75a_n = 20.75.

20.75=5+(n−1)×1.7520.75 = 5 + (n-1)\times 1.75
15.75=(n−1)×1.7515.75 = (n-1)\times 1.75
n−1=15.751.75=9n - 1 = \frac{15.75}{1.75} = 9
n=10n = 10

In the 10th week, her savings become ₹ 20.75.

Exercise 5.3

1Find the sum of the following APs:
(i) 2, 7, 12, …, to 10 terms.
(ii) -37, -33, -29, …, to 12 terms.
(iii) 0.6, 1.7, 2.8, …, to 100 terms.
(iv) 1/15, 1/12, 1/10, …, to 11 terms.Show solution
Using Sn=n2[2a+(n−1)d]S_n = \dfrac{n}{2}[2a + (n-1)d]:

(i) a=2a = 2, d=5d = 5, n=10n = 10

S10=102[2×2+9×5]=5[4+45]=5×49=245S_{10} = \frac{10}{2}[2\times2 + 9\times5] = 5[4 + 45] = 5\times49 = 245

Sum = 245

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(ii) a=−37a = -37, d=4d = 4, n=12n = 12

S12=122[2×(−37)+11×4]=6[−74+44]=6×(−30)=−180S_{12} = \frac{12}{2}[2\times(-37) + 11\times4] = 6[-74 + 44] = 6\times(-30) = -180

Sum = −180

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(iii) a=0.6a = 0.6, d=1.1d = 1.1, n=100n = 100

S100=1002[2×0.6+99×1.1]=50[1.2+108.9]=50×110.1=5505S_{100} = \frac{100}{2}[2\times0.6 + 99\times1.1] = 50[1.2 + 108.9] = 50\times110.1 = 5505

Sum = 5505

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(iv) a=115a = \dfrac{1}{15}, d=112−115=5−460=160d = \dfrac{1}{12} - \dfrac{1}{15} = \dfrac{5-4}{60} = \dfrac{1}{60}, n=11n = 11

S11=112[2×115+10×160]=112[215+16]S_{11} = \frac{11}{2}\left[2\times\frac{1}{15} + 10\times\frac{1}{60}\right] = \frac{11}{2}\left[\frac{2}{15} + \frac{1}{6}\right]

=112[430+530]=112×930=112×310=3320= \frac{11}{2}\left[\frac{4}{30} + \frac{5}{30}\right] = \frac{11}{2}\times\frac{9}{30} = \frac{11}{2}\times\frac{3}{10} = \frac{33}{20}

Sum = 3320\dfrac{33}{20}
2Find the sums given below:
(i) 7+1012+14+…+847 + 10\frac{1}{2} + 14 + \ldots + 84
(ii) 34+32+30+…+1034 + 32 + 30 + \ldots + 10
(iii) −5+(−8)+(−11)+…+(−230)-5 + (-8) + (-11) + \ldots + (-230)Show solution
(i) a=7a = 7, d=3.5d = 3.5, l=84l = 84

First find nn: 84=7+(n−1)×3.5⇒77=(n−1)×3.5⇒n−1=22⇒n=2384 = 7 + (n-1)\times3.5 \Rightarrow 77 = (n-1)\times3.5 \Rightarrow n-1 = 22 \Rightarrow n = 23

S23=232(7+84)=232×91=20932=1046.5S_{23} = \frac{23}{2}(7 + 84) = \frac{23}{2}\times91 = \frac{2093}{2} = 1046.5

Sum = 1046121046\dfrac{1}{2}

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(ii) a=34a = 34, d=−2d = -2, l=10l = 10

10=34+(n−1)×(−2)⇒−24=(n−1)×(−2)⇒n−1=12⇒n=1310 = 34 + (n-1)\times(-2) \Rightarrow -24 = (n-1)\times(-2) \Rightarrow n-1 = 12 \Rightarrow n = 13

S13=132(34+10)=132×44=13×22=286S_{13} = \frac{13}{2}(34 + 10) = \frac{13}{2}\times44 = 13\times22 = 286

Sum = 286

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(iii) a=−5a = -5, d=−3d = -3, l=−230l = -230

−230=−5+(n−1)×(−3)⇒−225=(n−1)×(−3)⇒n−1=75⇒n=76-230 = -5 + (n-1)\times(-3) \Rightarrow -225 = (n-1)\times(-3) \Rightarrow n-1 = 75 \Rightarrow n = 76

S76=762(−5+(−230))=38×(−235)=−8930S_{76} = \frac{76}{2}(-5 + (-230)) = 38\times(-235) = -8930

Sum = −8930
3In an AP:
(i) given a=5a = 5, d=3d = 3, an=50a_n = 50, find nn and SnS_n.
(ii) given a=7a = 7, a13=35a_{13} = 35, find dd and S13S_{13}.
(iii) given a12=37a_{12} = 37, d=3d = 3, find aa and S12S_{12}.
(iv) given a3=15a_3 = 15, S10=125S_{10} = 125, find dd and a10a_{10}.
(v) given d=5d = 5, S9=75S_9 = 75, find aa and a9a_9.
(vi) given a=2a = 2, d=8d = 8, Sn=90S_n = 90, find nn and ana_n.
(vii) given a=8a = 8, an=62a_n = 62, Sn=210S_n = 210, find nn and dd.
(viii) given an=4a_n = 4, d=2d = 2, Sn=−14S_n = -14, find nn and aa.
(ix) given a=3a = 3, n=8n = 8, S=192S = 192, find dd.
(x) given l=28l = 28, S=144S = 144, and there are total 9 terms. Find aa.Show solution
(i) a=5a = 5, d=3d = 3, an=50a_n = 50

50=5+(n−1)×3⇒45=3(n−1)⇒n=1650 = 5 + (n-1)\times3 \Rightarrow 45 = 3(n-1) \Rightarrow n = 16

S16=162(5+50)=8×55=440S_{16} = \frac{16}{2}(5 + 50) = 8\times55 = 440

n=16n = 16, Sn=440S_n = 440

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(ii) a=7a = 7, a13=35a_{13} = 35

35=7+12d⇒12d=28⇒d=7335 = 7 + 12d \Rightarrow 12d = 28 \Rightarrow d = \frac{7}{3}

S13=132(7+35)=132×42=13×21=273S_{13} = \frac{13}{2}(7 + 35) = \frac{13}{2}\times42 = 13\times21 = 273

d=73d = \dfrac{7}{3}, S13=273S_{13} = 273

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(iii) a12=37a_{12} = 37, d=3d = 3

37=a+11×3⇒a=37−33=437 = a + 11\times3 \Rightarrow a = 37 - 33 = 4

S12=122[2×4+11×3]=6[8+33]=6×41=246S_{12} = \frac{12}{2}[2\times4 + 11\times3] = 6[8 + 33] = 6\times41 = 246

a=4a = 4, S12=246S_{12} = 246

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(iv) a3=15a_3 = 15, S10=125S_{10} = 125

a+2d=15⋯(1)a + 2d = 15 \quad \cdots (1)

S10=102[2a+9d]=5(2a+9d)=125⇒2a+9d=25⋯(2)S_{10} = \dfrac{10}{2}[2a + 9d] = 5(2a + 9d) = 125 \Rightarrow 2a + 9d = 25 \quad \cdots (2)

From (1): a=15−2da = 15 - 2d. Substituting in (2):
2(15−2d)+9d=25⇒30−4d+9d=25⇒5d=−5⇒d=−12(15-2d) + 9d = 25 \Rightarrow 30 - 4d + 9d = 25 \Rightarrow 5d = -5 \Rightarrow d = -1

a=15−2(−1)=17a = 15 - 2(-1) = 17

a10=17+9×(−1)=17−9=8a_{10} = 17 + 9\times(-1) = 17 - 9 = 8

d=−1d = -1, a10=8a_{10} = 8

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(v) d=5d = 5, S9=75S_9 = 75

S9=92[2a+8×5]=92[2a+40]=75S_9 = \frac{9}{2}[2a + 8\times5] = \frac{9}{2}[2a + 40] = 75
2a+40=1509=5032a + 40 = \frac{150}{9} = \frac{50}{3}

Wait, let me redo:
92(2a+40)=75⇒2a+40=1509\frac{9}{2}(2a + 40) = 75 \Rightarrow 2a + 40 = \frac{150}{9}

That gives a non-integer. Let me recheck:
92[2a+(9−1)×5]=75\frac{9}{2}[2a + (9-1)\times5] = 75
92[2a+40]=75\frac{9}{2}[2a + 40] = 75
9(2a+40)=1509(2a + 40) = 150
2a+40=15092a + 40 = \frac{150}{9}

This seems odd. Let me use S9=92(2a+8d)S_9 = \frac{9}{2}(2a+8d):
75=92(2a+8×5)=92(2a+40)75 = \frac{9}{2}(2a + 8\times5) = \frac{9}{2}(2a+40)
1509=2a+40\frac{150}{9} = 2a + 40
503=2a+40\frac{50}{3} = 2a + 40
2a=503−40=50−1203=−7032a = \frac{50}{3} - 40 = \frac{50-120}{3} = -\frac{70}{3}
a=−353a = -\frac{35}{3}

a9=a+8d=−353+40=−35+1203=853a_9 = a + 8d = -\frac{35}{3} + 40 = \frac{-35+120}{3} = \frac{85}{3}

a=−353a = -\dfrac{35}{3}, a9=853a_9 = \dfrac{85}{3}

*(Note: Some textbook editions have S9=75S_9 = 75 giving a=−353a = -\frac{35}{3}; verify with your edition.)*

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(vi) a=2a = 2, d=8d = 8, Sn=90S_n = 90

90=n2[2×2+(n−1)×8]=n2[4+8n−8]=n2[8n−4]90 = \frac{n}{2}[2\times2 + (n-1)\times8] = \frac{n}{2}[4 + 8n - 8] = \frac{n}{2}[8n - 4]
90=n(4n−2)90 = n(4n - 2)
90=4n2−2n90 = 4n^2 - 2n
4n2−2n−90=04n^2 - 2n - 90 = 0
2n2−n−45=02n^2 - n - 45 = 0
2n2−10n+9n−45=02n^2 - 10n + 9n - 45 = 0
2n(n−5)+9(n−5)=02n(n-5) + 9(n-5) = 0
(2n+9)(n−5)=0(2n+9)(n-5) = 0
n = 5 \quad (\text{since } n > 0)

a5=2+4×8=2+32=34a_5 = 2 + 4\times8 = 2 + 32 = 34

n=5n = 5, an=34a_n = 34

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(vii) a=8a = 8, an=62a_n = 62, Sn=210S_n = 210

Sn=n2(a+an)⇒210=n2(8+62)=n2×70=35nS_n = \frac{n}{2}(a + a_n) \Rightarrow 210 = \frac{n}{2}(8 + 62) = \frac{n}{2}\times70 = 35n
n=6n = 6

a6=a+5d⇒62=8+5d⇒5d=54⇒d=545a_6 = a + 5d \Rightarrow 62 = 8 + 5d \Rightarrow 5d = 54 \Rightarrow d = \frac{54}{5}

Wait: 62=8+5d⇒5d=54⇒d=10.862 = 8 + 5d \Rightarrow 5d = 54 \Rightarrow d = 10.8

n=6n = 6, d=545d = \dfrac{54}{5}

*(Check: S6=62(8+62)=3×70=210S_6 = \frac{6}{2}(8+62) = 3\times70 = 210 ✓)*

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(viii) an=4a_n = 4, d=2d = 2, Sn=−14S_n = -14

Sn=n2(a+an)⇒−14=n2(a+4)⋯(1)S_n = \frac{n}{2}(a + a_n) \Rightarrow -14 = \frac{n}{2}(a + 4) \quad \cdots (1)

an=a+(n−1)d⇒4=a+(n−1)×2⋯(2)a_n = a + (n-1)d \Rightarrow 4 = a + (n-1)\times2 \quad \cdots (2)

From (2): a=4−2(n−1)=6−2na = 4 - 2(n-1) = 6 - 2n

Substitute in (1):
−14=n2(6−2n+4)=n2(10−2n)=n(5−n)-14 = \frac{n}{2}(6 - 2n + 4) = \frac{n}{2}(10 - 2n) = n(5 - n)
−14=5n−n2-14 = 5n - n^2
n2−5n−14=0n^2 - 5n - 14 = 0
(n−7)(n+2)=0(n-7)(n+2) = 0
n = 7 \quad (n > 0)

a=6−2×7=6−14=−8a = 6 - 2\times7 = 6 - 14 = -8

n=7n = 7, a=−8a = -8

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(ix) a=3a = 3, n=8n = 8, S=192S = 192

192=82[2×3+7d]=4[6+7d]192 = \frac{8}{2}[2\times3 + 7d] = 4[6 + 7d]
48=6+7d48 = 6 + 7d
7d=42⇒d=67d = 42 \Rightarrow d = 6

d=6d = 6

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(x) l=28l = 28, S=144S = 144, n=9n = 9

S=n2(a+l)⇒144=92(a+28)S = \frac{n}{2}(a + l) \Rightarrow 144 = \frac{9}{2}(a + 28)
2889=a+28\frac{288}{9} = a + 28
32=a+2832 = a + 28
a=4a = 4

a=4a = 4
4How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?Show solution
Given: a=9a = 9, d=8d = 8, Sn=636S_n = 636.

636=n2[2×9+(n−1)×8]=n2[18+8n−8]=n2[8n+10]636 = \frac{n}{2}[2\times9 + (n-1)\times8] = \frac{n}{2}[18 + 8n - 8] = \frac{n}{2}[8n + 10]
636=n(4n+5)636 = n(4n + 5)
4n2+5n−636=04n^2 + 5n - 636 = 0

Using the quadratic formula:
n=−5±25+4×4×6368=−5±25+101768=−5±102018=−5±1018n = \frac{-5 \pm \sqrt{25 + 4\times4\times636}}{8} = \frac{-5 \pm \sqrt{25 + 10176}}{8} = \frac{-5 \pm \sqrt{10201}}{8} = \frac{-5 \pm 101}{8}

Taking positive value: n=968=12n = \dfrac{96}{8} = 12

12 terms must be taken.
5The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.Show solution
Given: a=5a = 5, l=45l = 45, Sn=400S_n = 400.

Sn=n2(a+l)⇒400=n2(5+45)=25nS_n = \frac{n}{2}(a + l) \Rightarrow 400 = \frac{n}{2}(5 + 45) = 25n
n=16n = 16

Now, an=a+(n−1)da_n = a + (n-1)d:
45=5+15d⇒15d=40⇒d=8345 = 5 + 15d \Rightarrow 15d = 40 \Rightarrow d = \frac{8}{3}

Number of terms = 16, common difference = 83\dfrac{8}{3}
6The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?Show solution
Given: a=17a = 17, l=350l = 350, d=9d = 9.

350=17+(n−1)×9350 = 17 + (n-1)\times9
333=(n−1)×9333 = (n-1)\times9
n−1=37⇒n=38n - 1 = 37 \Rightarrow n = 38

S38=382(17+350)=19×367=6973S_{38} = \frac{38}{2}(17 + 350) = 19\times367 = 6973

There are 38 terms and their sum is 6973.
7Find the sum of first 22 terms of an AP in which d=7d = 7 and 22nd term is 149.Show solution
Given: d=7d = 7, a22=149a_{22} = 149, n=22n = 22.

S22=222(a1+a22)=11(a+149)S_{22} = \frac{22}{2}(a_1 + a_{22}) = 11(a + 149)

Find aa: 149=a+21×7=a+147⇒a=2149 = a + 21\times7 = a + 147 \Rightarrow a = 2

S22=11(2+149)=11×151=1661S_{22} = 11(2 + 149) = 11\times151 = 1661

Sum of first 22 terms = 1661
8Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.Show solution
Given: a2=14a_2 = 14, a3=18a_3 = 18.

d=a3−a2=18−14=4d = a_3 - a_2 = 18 - 14 = 4

a=a2−d=14−4=10a = a_2 - d = 14 - 4 = 10

S51=512[2×10+50×4]=512[20+200]=512×220=51×110=5610S_{51} = \frac{51}{2}[2\times10 + 50\times4] = \frac{51}{2}[20 + 200] = \frac{51}{2}\times220 = 51\times110 = 5610

Sum of first 51 terms = 5610
9If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first nn terms.Show solution
Given: S7=49S_7 = 49 and S17=289S_{17} = 289.

S7=72(2a+6d)=7(a+3d)=49⇒a+3d=7⋯(1)S_7 = \frac{7}{2}(2a + 6d) = 7(a + 3d) = 49 \Rightarrow a + 3d = 7 \quad \cdots (1)

S17=172(2a+16d)=17(a+8d)=289⇒a+8d=17⋯(2)S_{17} = \frac{17}{2}(2a + 16d) = 17(a + 8d) = 289 \Rightarrow a + 8d = 17 \quad \cdots (2)

Subtracting (1) from (2): 5d=10⇒d=25d = 10 \Rightarrow d = 2

From (1): a=7−6=1a = 7 - 6 = 1

Sn=n2[2×1+(n−1)×2]=n2×2n=n2S_n = \frac{n}{2}[2\times1 + (n-1)\times2] = \frac{n}{2}\times2n = n^2

Sn=n2S_n = n^2
10Show that a1,a2,…,an,…a_1, a_2, \ldots, a_n, \ldots form an AP where ana_n is defined as below:
(i) an=3+4na_n = 3 + 4n
(ii) an=9−5na_n = 9 - 5n
Also find the sum of the first 15 terms in each case.Show solution
(i) an=3+4na_n = 3 + 4n

a1=3+4=7a_1 = 3 + 4 = 7
a2=3+8=11a_2 = 3 + 8 = 11
a3=3+12=15a_3 = 3 + 12 = 15

ak+1−ak=[3+4(k+1)]−[3+4k]=4a_{k+1} - a_k = [3 + 4(k+1)] - [3 + 4k] = 4 (constant)

Since the difference between consecutive terms is constant (= 4), the sequence forms an AP with a=7a = 7 and d=4d = 4.

S15=152[2×7+14×4]=152[14+56]=152×70=525S_{15} = \frac{15}{2}[2\times7 + 14\times4] = \frac{15}{2}[14 + 56] = \frac{15}{2}\times70 = 525

Sum of first 15 terms = 525

---

(ii) an=9−5na_n = 9 - 5n

a1=9−5=4a_1 = 9 - 5 = 4
a2=9−10=−1a_2 = 9 - 10 = -1
a3=9−15=−6a_3 = 9 - 15 = -6

ak+1−ak=[9−5(k+1)]−[9−5k]=−5a_{k+1} - a_k = [9 - 5(k+1)] - [9 - 5k] = -5 (constant)

Since the difference is constant (= −5), the sequence forms an AP with a=4a = 4 and d=−5d = -5.

S15=152[2×4+14×(−5)]=152[8−70]=152×(−62)=−465S_{15} = \frac{15}{2}[2\times4 + 14\times(-5)] = \frac{15}{2}[8 - 70] = \frac{15}{2}\times(-62) = -465

Sum of first 15 terms = −465
11If the sum of the first nn terms of an AP is 4n−n24n - n^2, what is the first term (that is S1S_1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nnth terms.Show solution
Given: Sn=4n−n2S_n = 4n - n^2

First term =S1=4(1)−12=4−1=3= S_1 = 4(1) - 1^2 = 4 - 1 = 3

Sum of first two terms =S2=4(2)−22=8−4=4= S_2 = 4(2) - 2^2 = 8 - 4 = 4

Second term =S2−S1=4−3=1= S_2 - S_1 = 4 - 3 = 1

Third term =S3−S2=[4(3)−9]−4=3−4=−1= S_3 - S_2 = [4(3) - 9] - 4 = 3 - 4 = -1

10th term =S10−S9=[40−100]−[36−81]=−60−(−45)=−60+45=−15= S_{10} - S_9 = [40 - 100] - [36 - 81] = -60 - (-45) = -60 + 45 = -15

nnth term (for n≥2n \geq 2):
an=Sn−Sn−1=(4n−n2)−[4(n−1)−(n−1)2]a_n = S_n - S_{n-1} = (4n - n^2) - [4(n-1) - (n-1)^2]
=4n−n2−[4n−4−n2+2n−1]= 4n - n^2 - [4n - 4 - n^2 + 2n - 1]
=4n−n2−4n+4+n2−2n+1= 4n - n^2 - 4n + 4 + n^2 - 2n + 1
=5−2n= 5 - 2n

Check for n=1n=1: 5−2(1)=3=a15 - 2(1) = 3 = a_1 ✓

an=5−2na_n = 5 - 2n

Summary: a1=3a_1 = 3, a2=1a_2 = 1, a3=−1a_3 = -1, a10=−15a_{10} = -15, an=5−2na_n = 5 - 2n
12Find the sum of the first 40 positive integers divisible by 6.Show solution
The first 40 positive integers divisible by 6 are: 6, 12, 18, …, 240.

a=6a = 6, d=6d = 6, n=40n = 40

S40=402[2×6+39×6]=20[12+234]=20×246=4920S_{40} = \frac{40}{2}[2\times6 + 39\times6] = 20[12 + 234] = 20\times246 = 4920

Sum = 4920
13Find the sum of the first 15 multiples of 8.Show solution
First 15 multiples of 8: 8, 16, 24, …, 120.

a=8a = 8, d=8d = 8, n=15n = 15

S15=152[2×8+14×8]=152[16+112]=152×128=15×64=960S_{15} = \frac{15}{2}[2\times8 + 14\times8] = \frac{15}{2}[16 + 112] = \frac{15}{2}\times128 = 15\times64 = 960

Sum = 960
14Find the sum of the odd numbers between 0 and 50.Show solution
Odd numbers between 0 and 50: 1, 3, 5, …, 49.

a=1a = 1, d=2d = 2, l=49l = 49

Number of terms: 49=1+(n−1)×2⇒n=2549 = 1 + (n-1)\times2 \Rightarrow n = 25

S25=252(1+49)=252×50=625S_{25} = \frac{25}{2}(1 + 49) = \frac{25}{2}\times50 = 625

Sum = 625
15A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?Show solution
Penalty forms an AP: 200, 250, 300, …

a=200a = 200, d=50d = 50, n=30n = 30

S30=302[2×200+29×50]=15[400+1450]=15×1850=27750S_{30} = \frac{30}{2}[2\times200 + 29\times50] = 15[400 + 1450] = 15\times1850 = 27750

The contractor has to pay ₹ 27,750 as penalty.
16A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.Show solution
Let the prizes form an AP with first term aa and d=−20d = -20, n=7n = 7, S7=700S_7 = 700.

700=72[2a+6×(−20)]=72[2a−120]700 = \frac{7}{2}[2a + 6\times(-20)] = \frac{7}{2}[2a - 120]
200=2a−120200 = 2a - 120
2a=320⇒a=1602a = 320 \Rightarrow a = 160

The prizes are:
a1=160a_1 = 160, a2=140a_2 = 140, a3=120a_3 = 120, a4=100a_4 = 100, a5=80a_5 = 80, a6=60a_6 = 60, a7=40a_7 = 40

The values of the prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40.
17In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?Show solution
Each class has 3 sections.

Trees planted by Class kk = 3k3k (since each of 3 sections plants kk trees).

Total trees = 3×1+3×2+3×3+…+3×123\times1 + 3\times2 + 3\times3 + \ldots + 3\times12

=3(1+2+3+…+12)=3×12×132=3×78=234= 3(1 + 2 + 3 + \ldots + 12) = 3\times\frac{12\times13}{2} = 3\times78 = 234

A total of 234 trees will be planted.
18A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ... What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π=227\pi = \frac{22}{7})Show solution
Length of a semicircle with radius rr = πr\pi r.

The radii are: 0.5, 1.0, 1.5, 2.0, … (AP with a=0.5a = 0.5, d=0.5d = 0.5)

Lengths of semicircles: π×0.5, π×1.0, π×1.5,…\pi\times0.5,\ \pi\times1.0,\ \pi\times1.5,\ldots

This is an AP with first term l1=0.5πl_1 = 0.5\pi and common difference 0.5π0.5\pi, for n=13n = 13 terms.

S13=132[2×0.5π+12×0.5π]=132[π+6π]=132×7π=91π2S_{13} = \frac{13}{2}[2\times0.5\pi + 12\times0.5\pi] = \frac{13}{2}[\pi + 6\pi] = \frac{13}{2}\times7\pi = \frac{91\pi}{2}

=912×227=91×2214=200214=143 cm= \frac{91}{2}\times\frac{22}{7} = \frac{91\times22}{14} = \frac{2002}{14} = 143 \text{ cm}

The total length of the spiral is 143 cm.
19200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?Show solution
The number of logs in each row forms an AP: 20, 19, 18, …

a=20a = 20, d=−1d = -1, Sn=200S_n = 200

200=n2[2×20+(n−1)×(−1)]=n2[40−n+1]=n2[41−n]200 = \frac{n}{2}[2\times20 + (n-1)\times(-1)] = \frac{n}{2}[40 - n + 1] = \frac{n}{2}[41 - n]
400=n(41−n)=41n−n2400 = n(41 - n) = 41n - n^2
n2−41n+400=0n^2 - 41n + 400 = 0
(n−16)(n−25)=0(n - 16)(n - 25) = 0
n=16 or n=25n = 16 \text{ or } n = 25

For n=25n = 25: a25=20+24×(−1)=−4a_{25} = 20 + 24\times(-1) = -4 (not possible, logs can't be negative).

So n=16n = 16.

Top row (16th row): a16=20+15×(−1)=5a_{16} = 20 + 15\times(-1) = 5

The logs are placed in 16 rows and the top row has 5 logs.
20In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?Show solution
Distance to pick up the 1st potato and return = 2×5=102\times5 = 10 m

Distance to pick up the 2nd potato and return = 2×(5+3)=162\times(5+3) = 16 m

Distance to pick up the 3rd potato and return = 2×(5+6)=222\times(5+6) = 22 m

In general, distance for kkth potato = 2×[5+(k−1)×3]2\times[5 + (k-1)\times3]

The distances form an AP: 10, 16, 22, … with a=10a = 10, d=6d = 6, n=10n = 10.

S10=102[2×10+9×6]=5[20+54]=5×74=370 mS_{10} = \frac{10}{2}[2\times10 + 9\times6] = 5[20 + 54] = 5\times74 = 370 \text{ m}

The total distance the competitor has to run is 370 m.

Exercise 5.4 (Optional)

1Which term of the AP: 121, 117, 113, ..., is its first negative term?Show solution
Given AP: 121, 117, 113, …

a=121a = 121, d=117−121=−4d = 117 - 121 = -4

For the first negative term, we need a_n < 0:

a_n = 121 + (n-1)\times(-4) < 0
121 - 4(n-1) < 0
121 < 4(n-1)
n - 1 > \frac{121}{4} = 30.25
n > 31.25

The smallest integer value is n=32n = 32.

Verification: a_{32} = 121 + 31\times(-4) = 121 - 124 = -3 < 0 ✓

The 32nd term is the first negative term.
2The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.Show solution
Let aa be the first term and dd the common difference.

a3=a+2da_3 = a + 2d, a7=a+6da_7 = a + 6d

Given:
a3+a7=6⇒(a+2d)+(a+6d)=6⇒2a+8d=6⇒a+4d=3⋯(1)a_3 + a_7 = 6 \Rightarrow (a+2d) + (a+6d) = 6 \Rightarrow 2a + 8d = 6 \Rightarrow a + 4d = 3 \quad \cdots (1)

a3×a7=8⇒(a+2d)(a+6d)=8⋯(2)a_3 \times a_7 = 8 \Rightarrow (a+2d)(a+6d) = 8 \quad \cdots (2)

From (1): a=3−4da = 3 - 4d

Substitute in (2):
(3−4d+2d)(3−4d+6d)=8(3-4d+2d)(3-4d+6d) = 8
(3−2d)(3+2d)=8(3-2d)(3+2d) = 8
9−4d2=89 - 4d^2 = 8
4d2=1⇒d2=14⇒d=±124d^2 = 1 \Rightarrow d^2 = \frac{1}{4} \Rightarrow d = \pm\frac{1}{2}

Case 1: d=12d = \dfrac{1}{2}: a=3−4×12=3−2=1a = 3 - 4\times\dfrac{1}{2} = 3 - 2 = 1

Case 2: d=−12d = -\dfrac{1}{2}: a=3−4×(−12)=3+2=5a = 3 - 4\times(-\dfrac{1}{2}) = 3 + 2 = 5

S16=162[2a+15d]=8[2a+15d]S_{16} = \frac{16}{2}[2a + 15d] = 8[2a + 15d]

Case 1: S16=8[2+15×12]=8[2+7.5]=8×9.5=76S_{16} = 8[2 + 15\times\frac{1}{2}] = 8[2 + 7.5] = 8\times9.5 = 76

Case 2: S16=8[10+15×(−12)]=8[10−7.5]=8×2.5=20S_{16} = 8[10 + 15\times(-\frac{1}{2})] = 8[10 - 7.5] = 8\times2.5 = 20

The sum of first 16 terms is 76 or 20.
3A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2122\frac{1}{2} m apart, what is the length of the wood required for the rungs?Show solution
Distance between top and bottom rungs = 2122\dfrac{1}{2} m = 250 cm.

Number of rungs = 25025+1=10+1=11\dfrac{250}{25} + 1 = 10 + 1 = 11

The lengths of rungs form an AP: 45, …, 25 with n=11n = 11.

a=45a = 45, l=25l = 25, n=11n = 11

Total length of wood for rungs:
S11=112(45+25)=112×70=11×35=385 cmS_{11} = \frac{11}{2}(45 + 25) = \frac{11}{2}\times70 = 11\times35 = 385 \text{ cm}

The length of wood required for the rungs is 385 cm.
4The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of xx such that the sum of the numbers of the houses preceding the house numbered xx is equal to the sum of the numbers of the houses following it. Find this value of xx.Show solution
Sum of house numbers from 1 to (x−1)(x-1):
Sx−1=(x−1)x2S_{x-1} = \frac{(x-1)x}{2}

Sum of all house numbers from 1 to 49:
S49=49×502=1225S_{49} = \frac{49\times50}{2} = 1225

Sum of house numbers following house xx (i.e., from x+1x+1 to 49):
S49−Sx=1225−x(x+1)2S_{49} - S_x = 1225 - \frac{x(x+1)}{2}

Given condition: Sx−1=S49−SxS_{x-1} = S_{49} - S_x

(x−1)x2=1225−x(x+1)2\frac{(x-1)x}{2} = 1225 - \frac{x(x+1)}{2}

x(x−1)2+x(x+1)2=1225\frac{x(x-1)}{2} + \frac{x(x+1)}{2} = 1225

x[(x−1)+(x+1)]2=1225\frac{x[(x-1)+(x+1)]}{2} = 1225

x×2x2=1225\frac{x\times2x}{2} = 1225

x2=1225x^2 = 1225

x = 35 \quad (x > 0)

Since x=35x = 35 is a natural number between 1 and 49, such a value exists.

x=35x = 35
5A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 14\frac{1}{4} m and a tread of 12\frac{1}{2} m. Calculate the total volume of concrete required to build the terrace.Show solution
Volume of concrete for the 1st step:
V1=14×12×50=508=6.25 m3V_1 = \frac{1}{4}\times\frac{1}{2}\times50 = \frac{50}{8} = 6.25 \text{ m}^3

Volume of concrete for the 2nd step (height = 24\dfrac{2}{4} m):
V2=24×12×50=1008=12.5 m3V_2 = \frac{2}{4}\times\frac{1}{2}\times50 = \frac{100}{8} = 12.5 \text{ m}^3

In general, volume of kkth step:
Vk=k4×12×50=50k8=25k4V_k = \frac{k}{4}\times\frac{1}{2}\times50 = \frac{50k}{8} = \frac{25k}{4}

The volumes form an AP: 254,504,754,…\dfrac{25}{4}, \dfrac{50}{4}, \dfrac{75}{4}, \ldots with a=254a = \dfrac{25}{4}, d=254d = \dfrac{25}{4}, n=15n = 15.

S15=152[2×254+14×254]=152×254×16=152×100=750 m3S_{15} = \frac{15}{2}\left[2\times\frac{25}{4} + 14\times\frac{25}{4}\right] = \frac{15}{2}\times\frac{25}{4}\times16 = \frac{15}{2}\times100 = 750 \text{ m}^3

The total volume of concrete required is 750 m³.

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