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Arithmetic Progressions — CBSE Class 10 Maths Textbook Solutions

Step-by-step NCERT textbook solutions for every question in Arithmetic Progressions — all 4 exercises, 49 questions, solved in full.

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NCERT Class 10 Maths Chapter 5 (Arithmetic Progressions) solutions cover Exercises 5.1 to 5.4 in full: identifying an AP, finding its common difference, computing the nth term with an=a+(n−1)da_n=a+(n-1)da_n=a+(n-1)d, and finding the sum of n terms with Sn=n2[2a+(n−1)d]S_n=\frac{n}{2}[2a+(n-1)d]S_n=n/2[2a+(n-1)d]. All 49 questions across the four exercises are solved step by step here.

Syllabus reference (2026–27): NCERT (National Council of Educational Research and Training) · CBSE (Central Board of Secondary Education)
How these are built: Every solution is written from the official NCERT textbook and checked carefully, exercise by exercise.

About Arithmetic Progressions

Arithmetic Progressions is about number patterns that grow (or shrink) by a fixed amount at every step. These full NCERT textbook solutions work through every question in Exercises 5.1 to 5.4 — including the optional exercise — showing the formula used, the substitution, and the final answer at each step, matching the CBSE Class 10 Maths syllabus and the official NCERT answer key.

Definition of an AP and common differenceGeneral form of an AP: a, a+d, a+2d, ...nth term of an APSum of first n terms of an APFinding the number of terms in an APWord problems and real-life applications of AP

Where this fits in the exam

Arithmetic Progressions is part of the Algebra unit. Across the whole Algebra unit, CBSE Class 10 Maths board papers carry 20 marks in total — see the full Class 10 Maths marks weightage to see how every unit is scored.

Key concepts & formulas

Common difference

For a list to be an AP, ak+1−aka_{k+1}-a_ka_k+1-a_k must be the same value ddd for every kkk. ddd can be positive, negative, or zero.

nth term of an AP

an=a+(n−1)da_n=a+(n-1)da_n=a+(n-1)d, where aaa is the first term and ddd the common difference. Used to find any term, or to find nnn when the value of a term is given.

Sum of first n terms

Sn=n2[2a+(n−1)d]S_n=\frac{n}{2}[2a+(n-1)d]S_n=n/2[2a+(n-1)d], or equivalently Sn=n2(a+l)S_n=\frac{n}{2}(a+l)S_n=n/2(a+l) when the last term lll is known.

Relation between a_n and S_n

an=Sn−Sn−1a_n=S_n-S_{n-1}a_n=S_n-S_n-1. Useful when the sum of an AP is given as an expression in nnn.

Solving for an unknown

The two AP formulas link four quantities (a,d,n,ana,d,n,a_na,d,n,a_n or a,d,n,Sna,d,n,S_na,d,n,S_n). Given any relevant three (or two equations from two given terms), solve for the rest — often via simultaneous linear equations.

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Exercise-wise solutions

Every exercise in Arithmetic Progressions, in textbook order. Attempt each question first, then check your method against the solution.

ExerciseQuestions
Exercise 5.14
Exercise 5.220
Exercise 5.320
Exercise 5.4 (Optional)5

Exercise 5.1

Q1

In 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 14\frac{1}{4}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.

Solution

(i) Yes. The fares after 1, 2, 3, ... km are ₹15, ₹23, ₹31, ₹39, .... Each term is obtained by adding a fixed ₹8 to the preceding term, so this is an AP with a=15a=15a=15, d=8d=8d=8.

(ii) No. If VVV is the air initially present, the amounts left after each removal are V, 34V, (34)2V, (34)3V, …V,\ \frac{3}{4}V,\ \left(\frac{3}{4}\right)^2V,\ \left(\frac{3}{4}\right)^3V,\ \ldotsV, 3/4V, (3/4)^2V, (3/4)^3V,. Consecutive terms are obtained by multiplying by 34\frac{3}{4}3/4, not by adding a fixed number, so this is not an AP (it is a GP).

(iii) Yes. The costs after 1, 2, 3, ... metres are ₹150, ₹200, ₹250, .... Each term exceeds the previous one by a fixed ₹50, so this is an AP with a=150a=150a=150, d=50d=50d=50.

(iv) No. At compound interest, the amount after nnn years is 10000(1+8100)n10000\left(1+\frac{8}{100}\right)^n10000(1+8/100)^n. Consecutive amounts are obtained by multiplying by a fixed factor 1.081.081.08, not by adding a fixed number, so the list is not an AP.

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Q2

Write first four terms of the AP, when the first term aaa and the common difference ddd are given as follows:

(i) a=10, d=10a=10,\ d=10a=10, d=10

(ii) a=−2, d=0a=-2,\ d=0a=-2, d=0

(iii) a=4, d=−3a=4,\ d=-3a=4, d=-3

(iv) a=−1, d=12a=-1,\ d=\frac{1}{2}a=-1, d=1/2

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

Solution

Using the general form a, a+d, a+2d, a+3d, …a,\ a+d,\ a+2d,\ a+3d,\ \ldotsa, a+d, a+2d, a+3d,:

(i) 10, 20, 30, 4010,\ 20,\ 30,\ 4010, 20, 30, 40

(ii) −2, −2, −2, −2-2,\ -2,\ -2,\ -2-2, -2, -2, -2

(iii) 4, 1, −2, −54,\ 1,\ -2,\ -54, 1, -2, -5

(iv) −1, −12, 0, 12-1,\ -\frac{1}{2},\ 0,\ \frac{1}{2}-1, -1/2, 0, 1/2

(v) −1.25, −1.50, −1.75, −2.00-1.25,\ -1.50,\ -1.75,\ -2.00-1.25, -1.50, -1.75, -2.00

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Q3

For the following APs, write the first term and the common difference:

(i) 3, 1, −1, −3, …3,\ 1,\ -1,\ -3,\ \ldots3, 1, -1, -3,

(ii) −5, −1, 3, 7, …-5,\ -1,\ 3,\ 7,\ \ldots-5, -1, 3, 7,

(iii) 13, 53, 93, 133, …\frac{1}{3},\ \frac{5}{3},\ \frac{9}{3},\ \frac{13}{3},\ \ldots1/3, 5/3, 9/3, 13/3,

(iv) 0.6, 1.7, 2.8, 3.9, …0.6,\ 1.7,\ 2.8,\ 3.9,\ \ldots0.6, 1.7, 2.8, 3.9,

Solution

The common difference is d=a2−a1d=a_2-a_1d=a_2-a_1 (found using any two consecutive terms).

(i) a=3a=3a=3, d=1−3=−2d=1-3=-2d=1-3=-2

(ii) a=−5a=-5a=-5, d=−1−(−5)=4d=-1-(-5)=4d=-1-(-5)=4

(iii) a=13a=\frac{1}{3}a=1/3, d=53−13=43d=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}d=5/3-1/3=4/3

(iv) a=0.6a=0.6a=0.6, d=1.7−0.6=1.1d=1.7-0.6=1.1d=1.7-0.6=1.1

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Q4

Which of the following are APs? If they form an AP, find the common difference ddd and write three more terms.

(i) 2, 4, 8, 16, …2,\ 4,\ 8,\ 16,\ \ldots2, 4, 8, 16,

(ii) 2, 52, 3, 72, …2,\ \frac{5}{2},\ 3,\ \frac{7}{2},\ \ldots2, 5/2, 3, 7/2,

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

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

(v) 3, 3+2, 3+22, 3+32, …3,\ 3+\sqrt{2},\ 3+2\sqrt{2},\ 3+3\sqrt{2},\ \ldots3, 3+√2, 3+2√2, 3+3√2,

(vi) 0.2, 0.22, 0.222, 0.2222, …0.2,\ 0.22,\ 0.222,\ 0.2222,\ \ldots0.2, 0.22, 0.222, 0.2222,

(vii) 0, −4, −8, −12, …0,\ -4,\ -8,\ -12,\ \ldots0, -4, -8, -12,

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

(ix) 1, 3, 9, 27, …1,\ 3,\ 9,\ 27,\ \ldots1, 3, 9, 27,

(x) a, 2a, 3a, 4a, …a,\ 2a,\ 3a,\ 4a,\ \ldotsa, 2a, 3a, 4a,

(xi) a, a2, a3, a4, …a,\ a^2,\ a^3,\ a^4,\ \ldotsa, a^2, a^3, a^4,

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

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

(xiv) 12, 32, 52, 72, …1^2,\ 3^2,\ 5^2,\ 7^2,\ \ldots1^2, 3^2, 5^2, 7^2,

(xv) 12, 52, 72, 73, …1^2,\ 5^2,\ 7^2,\ 73,\ \ldots1^2, 5^2, 7^2, 73,

Solution

Check whether a2−a1=a3−a2=a4−a3a_2-a_1=a_3-a_2=a_4-a_3a_2-a_1=a_3-a_2=a_4-a_3 in each case.

(i) 4−2=24-2=24-2=2, 8−4=48-4=48-4=4 — not equal. Not an AP (it is a GP).

(ii) 52−2=12\frac{5}{2}-2=\frac{1}{2}5/2-2=1/2, 3−52=123-\frac{5}{2}=\frac{1}{2}3-5/2=1/2, 72−3=12\frac{7}{2}-3=\frac{1}{2}7/2-3=1/2 — equal. AP, d=12d=\frac{1}{2}d=1/2. Next three terms: 4, 92, 54,\ \frac{9}{2},\ 54, 9/2, 5.

(iii) −3.2−(−1.2)=−2-3.2-(-1.2)=-2-3.2-(-1.2)=-2, and so on, constant. AP, d=−2d=-2d=-2. Next three terms: −9.2, −11.2, −13.2-9.2,\ -11.2,\ -13.2-9.2, -11.2, -13.2.

(iv) −6−(−10)=4-6-(-10)=4-6-(-10)=4, −2−(−6)=4-2-(-6)=4-2-(-6)=4, 2−(−2)=42-(-2)=42-(-2)=4 — constant. AP, d=4d=4d=4. Next three terms: 6, 10, 146,\ 10,\ 146, 10, 14.

(v) Each term increases by 2\sqrt{2}√2. AP, d=2d=\sqrt{2}d=√2. Next three terms: 3+42, 3+52, 3+623+4\sqrt{2},\ 3+5\sqrt{2},\ 3+6\sqrt{2}3+4√2, 3+5√2, 3+6√2.

(vi) 0.22−0.2=0.020.22-0.2=0.020.22-0.2=0.02, 0.222−0.22=0.0020.222-0.22=0.0020.222-0.22=0.002 — not equal. Not an AP.

(vii) −4−0=−4-4-0=-4-4-0=-4, −8−(−4)=−4-8-(-4)=-4-8-(-4)=-4, −12−(−8)=−4-12-(-8)=-4-12-(-8)=-4 — constant. AP, d=−4d=-4d=-4. Next three terms: −16, −20, −24-16,\ -20,\ -24-16, -20, -24.

(viii) All terms equal −12-\frac{1}{2}-1/2, so the difference is always 000. AP, d=0d=0d=0. Next three terms: −12, −12, −12-\frac{1}{2},\ -\frac{1}{2},\ -\frac{1}{2}-1/2, -1/2, -1/2.

(ix) 3−1=23-1=23-1=2, 9−3=69-3=69-3=6 — not equal. Not an AP (it is a GP).

(x) 2a−a=a2a-a=a2a-a=a, 3a−2a=a3a-2a=a3a-2a=a, 4a−3a=a4a-3a=a4a-3a=a — constant. AP, d=ad=ad=a. Next three terms: 5a, 6a, 7a5a,\ 6a,\ 7a5a, 6a, 7a.

(xi) a2−a1=a2−aa_2-a_1=a^2-aa_2-a_1=a^2-a and a3−a2=a3−a2=a2(a−1)a_3-a_2=a^3-a^2=a^2(a-1)a_3-a_2=a^3-a^2=a^2(a-1); these are not equal in general (only for special values of aaa). Not an AP (it is a GP with common ratio aaa).

(xii) Simplify: 8=22\sqrt{8}=2\sqrt{2}√8=2√2, 18=32\sqrt{18}=3\sqrt{2}√18=3√2, 32=42\sqrt{32}=4\sqrt{2}√32=4√2, so the list is 2, 22, 32, 42\sqrt{2},\ 2\sqrt{2},\ 3\sqrt{2},\ 4\sqrt{2}√2, 2√2, 3√2, 4√2 — constant difference 2\sqrt{2}√2. AP, d=2d=\sqrt{2}d=√2. Next three terms: 50, 72, 98\sqrt{50},\ \sqrt{72},\ \sqrt{98}√50, √72, √98.

(xiii) 6−3≈0.72\sqrt{6}-\sqrt{3}\approx0.72√6-√30.72, 9−6≈0.55\sqrt{9}-\sqrt{6}\approx0.55√9-√60.55 — not equal. Not an AP.

(xiv) The list is 1, 9, 25, 491,\ 9,\ 25,\ 491, 9, 25, 49; differences are 8, 16, 248,\ 16,\ 248, 16, 24 — not equal. Not an AP.

(xv) The list is 1, 25, 49, 731,\ 25,\ 49,\ 731, 25, 49, 73; differences are 24, 24, 2424,\ 24,\ 2424, 24, 24 — constant. AP, d=24d=24d=24. Next three terms: 97, 121, 14597,\ 121,\ 14597, 121, 145.

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Exercise 5.2

Q1

Fill in the blanks in the following table, given that aaa is the first term, ddd the common difference and ana_na_n the nnnth term of the AP:

aaadddnnnana_na_n
(i)738?
(ii)−18-18-18?100
(iii)?−3-3-318−5-5-5
(iv)−18.9-18.9-18.92.5?3.6
(v)3.50105?
Solution

Use an=a+(n−1)da_n=a+(n-1)da_n=a+(n-1)d throughout.

(i) an=7+(8−1)(3)=7+21=28a_n=7+(8-1)(3)=7+21=28a_n=7+(8-1)(3)=7+21=28

(ii) 0=−18+(10−1)d⇒9d=18⇒d=20=-18+(10-1)d\Rightarrow9d=18\Rightarrow d=20=-18+(10-1)d9d=18 d=2

(iii) −5=a+(18−1)(−3)=a−51⇒a=46-5=a+(18-1)(-3)=a-51\Rightarrow a=46-5=a+(18-1)(-3)=a-51 a=46

(iv) 3.6=−18.9+(n−1)(2.5)⇒(n−1)(2.5)=22.5⇒n−1=9⇒n=103.6=-18.9+(n-1)(2.5)\Rightarrow(n-1)(2.5)=22.5\Rightarrow n-1=9\Rightarrow n=103.6=-18.9+(n-1)(2.5)(n-1)(2.5)=22.5 n-1=9 n=10

(v) an=3.5+(105−1)(0)=3.5a_n=3.5+(105-1)(0)=3.5a_n=3.5+(105-1)(0)=3.5

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Q2

Choose the correct choice in the following and justify:

(i) 30th term of the AP 10, 7, 4, …10,\ 7,\ 4,\ \ldots10, 7, 4, is
(A) 97 (B) 77 (C) −77-77-77 (D) −87-87-87

(ii) 11th term of the AP −3, −12, 2, …-3,\ -\frac{1}{2},\ 2,\ \ldots-3, -1/2, 2, is
(A) 28 (B) 22 (C) −38-38-38 (D) −4812-48\frac{1}{2}-481/2

Solution

(i) a=10a=10a=10, d=7−10=−3d=7-10=-3d=7-10=-3.
a30=a+29d=10+29(−3)=10−87=−77a_{30}=a+29d=10+29(-3)=10-87=-77a_30=a+29d=10+29(-3)=10-87=-77.
Correct option: (C) −77-77-77.

(ii) a=−3a=-3a=-3, d=−12−(−3)=52d=-\frac{1}{2}-(-3)=\frac{5}{2}d=-1/2-(-3)=5/2.
a11=a+10d=−3+10×52=−3+25=22a_{11}=a+10d=-3+10\times\frac{5}{2}=-3+25=22a_11=a+10d=-3+10×5/2=-3+25=22.
Correct option: (B) 22.

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Q3

In the following APs, find the missing terms in the boxes:

(i) 2, □, 262,\ \square,\ 262, , 26

(ii) □, 13, □, 3\square,\ 13,\ \square,\ 3, 13, , 3

(iii) 5, □, □, 9125,\ \square,\ \square,\ 9\frac{1}{2}5, , , 91/2

(iv) −4, □, □, □, □, 6-4,\ \square,\ \square,\ \square,\ \square,\ 6-4, , , , , 6

(v) □, 38, □, □, □, −22\square,\ 38,\ \square,\ \square,\ \square,\ -22, 38, , , , -22

Solution

(i) 3 terms: a=2a=2a=2, a3=26a_3=26a_3=26. 2d=26−2=24⇒d=122d=26-2=24\Rightarrow d=122d=26-2=24 d=12. Missing term =2+12=14=2+12=14=2+12=14.
AP: 2, 14, 262,\ 14,\ 262, 14, 26.

(ii) 4 terms: a2=13a_2=13a_2=13, a4=3a_4=3a_4=3. 2d=a4−a2=3−13=−10⇒d=−52d=a_4-a_2=3-13=-10\Rightarrow d=-52d=a_4-a_2=3-13=-10 d=-5. a1=a2−d=13−(−5)=18a_1=a_2-d=13-(-5)=18a_1=a_2-d=13-(-5)=18; a3=a2+d=13+(−5)=8a_3=a_2+d=13+(-5)=8a_3=a_2+d=13+(-5)=8.
AP: 18, 13, 8, 318,\ 13,\ 8,\ 318, 13, 8, 3.

(iii) 4 terms: a=5a=5a=5, a4=912=192a_4=9\frac{1}{2}=\frac{19}{2}a_4=91/2=19/2. 3d=192−5=92⇒d=323d=\frac{19}{2}-5=\frac{9}{2}\Rightarrow d=\frac{3}{2}3d=19/2-5=9/2 d=3/2. a2=5+32=612a_2=5+\frac{3}{2}=6\frac{1}{2}a_2=5+3/2=61/2; a3=612+32=8a_3=6\frac{1}{2}+\frac{3}{2}=8a_3=61/2+3/2=8.
AP: 5, 612, 8, 9125,\ 6\frac{1}{2},\ 8,\ 9\frac{1}{2}5, 61/2, 8, 91/2.

(iv) 6 terms: a=−4a=-4a=-4, a6=6a_6=6a_6=6. 5d=6−(−4)=10⇒d=25d=6-(-4)=10\Rightarrow d=25d=6-(-4)=10 d=2. a2=−2, a3=0, a4=2, a5=4a_2=-2,\ a_3=0,\ a_4=2,\ a_5=4a_2=-2, a_3=0, a_4=2, a_5=4.
AP: −4, −2, 0, 2, 4, 6-4,\ -2,\ 0,\ 2,\ 4,\ 6-4, -2, 0, 2, 4, 6.

(v) 6 terms: a2=38a_2=38a_2=38, a6=−22a_6=-22a_6=-22. 4d=−22−38=−60⇒d=−154d=-22-38=-60\Rightarrow d=-154d=-22-38=-60 d=-15. a1=38−d=38−(−15)=53a_1=38-d=38-(-15)=53a_1=38-d=38-(-15)=53; a3=38+d=23a_3=38+d=23a_3=38+d=23; a4=23+d=8a_4=23+d=8a_4=23+d=8; a5=8+d=−7a_5=8+d=-7a_5=8+d=-7.
AP: 53, 38, 23, 8, −7, −2253,\ 38,\ 23,\ 8,\ -7,\ -2253, 38, 23, 8, -7, -22.

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Q4

Which term of the AP 3, 8, 13, 18, …3,\ 8,\ 13,\ 18,\ \ldots3, 8, 13, 18, is 787878?

Solution

a=3a=3a=3, d=5d=5d=5. Let the nnnth term be 787878:
78=3+(n−1)(5)⇒75=5(n−1)⇒n−1=15⇒n=1678=3+(n-1)(5)\Rightarrow75=5(n-1)\Rightarrow n-1=15\Rightarrow n=1678=3+(n-1)(5)75=5(n-1) n-1=15 n=16.
So 787878 is the 16th term.

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Q5

Find the number of terms in each of the following APs:

(i) 7, 13, 19, …, 2057,\ 13,\ 19,\ \ldots,\ 2057, 13, 19, , 205

(ii) 18, 1512, 13, …, −4718,\ 15\frac{1}{2},\ 13,\ \ldots,\ -4718, 151/2, 13, , -47

Solution

(i) a=7a=7a=7, d=6d=6d=6, an=205a_n=205a_n=205.
205=7+(n−1)(6)⇒198=6(n−1)⇒n−1=33⇒n=34205=7+(n-1)(6)\Rightarrow198=6(n-1)\Rightarrow n-1=33\Rightarrow n=34205=7+(n-1)(6)198=6(n-1) n-1=33 n=34.
There are 34 terms.

(ii) a=18a=18a=18, d=1512−18=−52d=15\frac{1}{2}-18=-\frac{5}{2}d=151/2-18=-5/2, an=−47a_n=-47a_n=-47.
−47=18+(n−1)(−52)⇒−65=−52(n−1)⇒n−1=26⇒n=27-47=18+(n-1)\left(-\frac{5}{2}\right)\Rightarrow-65=-\frac{5}{2}(n-1)\Rightarrow n-1=26\Rightarrow n=27-47=18+(n-1)(-5/2)-65=-5/2(n-1) n-1=26 n=27.
There are 27 terms.

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Q6

Check whether −150-150-150 is a term of the AP 11, 8, 5, 2, …11,\ 8,\ 5,\ 2,\ \ldots11, 8, 5, 2,

Solution

a=11a=11a=11, d=−3d=-3d=-3. Suppose −150-150-150 is the nnnth term:
−150=11+(n−1)(−3)⇒−161=−3(n−1)⇒n−1=1613-150=11+(n-1)(-3)\Rightarrow-161=-3(n-1)\Rightarrow n-1=\frac{161}{3}-150=11+(n-1)(-3)-161=-3(n-1) n-1=161/3, which is not a whole number.
Since nnn must be a positive integer, −150-150-150 is not a term of this AP.

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Q7

Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution

a11=a+10d=38a_{11}=a+10d=38a_11=a+10d=38 and a16=a+15d=73a_{16}=a+15d=73a_16=a+15d=73.
Subtracting: 5d=35⇒d=75d=35\Rightarrow d=75d=35 d=7. Then a=38−10(7)=−32a=38-10(7)=-32a=38-10(7)=-32.
a31=a+30d=−32+30(7)=−32+210=178a_{31}=a+30d=-32+30(7)=-32+210=178a_31=a+30d=-32+30(7)=-32+210=178.
The 31st term is 178.

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Q8

An AP consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.

Solution

a3=a+2d=12a_3=a+2d=12a_3=a+2d=12 and a50=a+49d=106a_{50}=a+49d=106a_50=a+49d=106.
Subtracting: 47d=94⇒d=247d=94\Rightarrow d=247d=94 d=2. Then a=12−2(2)=8a=12-2(2)=8a=12-2(2)=8.
a29=a+28d=8+28(2)=8+56=64a_{29}=a+28d=8+28(2)=8+56=64a_29=a+28d=8+28(2)=8+56=64.
The 29th term is 64.

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Q9

If the 3rd and the 9th terms of an AP are 4 and −8-8-8 respectively, which term of this AP is zero?

Solution

a3=a+2d=4a_3=a+2d=4a_3=a+2d=4 and a9=a+8d=−8a_9=a+8d=-8a_9=a+8d=-8.
Subtracting: 6d=−12⇒d=−26d=-12\Rightarrow d=-26d=-12 d=-2. Then a=4−2(−2)=8a=4-2(-2)=8a=4-2(-2)=8.
Let an=0a_n=0a_n=0: 0=8+(n−1)(−2)⇒−8=−2(n−1)⇒n−1=4⇒n=50=8+(n-1)(-2)\Rightarrow-8=-2(n-1)\Rightarrow n-1=4\Rightarrow n=50=8+(n-1)(-2)-8=-2(n-1) n-1=4 n=5.
The 5th term is zero.

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Q10

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution

a17−a10=7a_{17}-a_{10}=7a_17-a_10=7. Now a17−a10=(a+16d)−(a+9d)=7da_{17}-a_{10}=(a+16d)-(a+9d)=7da_17-a_10=(a+16d)-(a+9d)=7d.
So 7d=7⇒d=17d=7\Rightarrow d=17d=7 d=1.
The common difference is 1.

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Q11

Which term of the AP 3, 15, 27, 39, …3,\ 15,\ 27,\ 39,\ \ldots3, 15, 27, 39, will be 132 more than its 54th term?

Solution

a=3a=3a=3, d=12d=12d=12.
a54=a+53d=3+53(12)=3+636=639a_{54}=a+53d=3+53(12)=3+636=639a_54=a+53d=3+53(12)=3+636=639.
Required term =639+132=771=639+132=771=639+132=771.
Let this be the nnnth term: 771=3+(n−1)(12)⇒768=12(n−1)⇒n−1=64⇒n=65771=3+(n-1)(12)\Rightarrow768=12(n-1)\Rightarrow n-1=64\Rightarrow n=65771=3+(n-1)(12)768=12(n-1) n-1=64 n=65.
The 65th term is 132 more than the 54th term.

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Q12

Two APs have the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000th terms?

Solution

Let the two APs have first terms aaa and a′a'a' and the same common difference ddd.
nnnth term of first AP: a+(n−1)da+(n-1)da+(n-1)d; of second: a′+(n−1)da'+(n-1)da'+(n-1)d.
Difference between corresponding terms =[a+(n−1)d]−[a′+(n−1)d]=a−a′=[a+(n-1)d]-[a'+(n-1)d]=a-a'=[a+(n-1)d]-[a'+(n-1)d]=a-a', which does not depend on nnn.
Since the difference between the 100th terms is 100100100, a−a′=100a-a'=100a-a'=100.
So the difference between the 1000th terms is also 100.

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Q13

How many three-digit numbers are divisible by 7?

Solution

The smallest 3-digit multiple of 7 is 105105105 (=15×7=15\times7=15×7); the largest is 994994994 (=142×7=142\times7=142×7).
This is an AP: a=105a=105a=105, d=7d=7d=7, an=994a_n=994a_n=994.
994=105+(n−1)(7)⇒889=7(n−1)⇒n−1=127⇒n=128994=105+(n-1)(7)\Rightarrow889=7(n-1)\Rightarrow n-1=127\Rightarrow n=128994=105+(n-1)(7)889=7(n-1) n-1=127 n=128.
There are 128 three-digit numbers divisible by 7.

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Q14

How many multiples of 4 lie between 10 and 250?

Solution

The multiples of 4 between 10 and 250 are 12, 16, 20, …, 24812,\ 16,\ 20,\ \ldots,\ 24812, 16, 20, , 248.
a=12a=12a=12, d=4d=4d=4, an=248a_n=248a_n=248.
248=12+(n−1)(4)⇒236=4(n−1)⇒n−1=59⇒n=60248=12+(n-1)(4)\Rightarrow236=4(n-1)\Rightarrow n-1=59\Rightarrow n=60248=12+(n-1)(4)236=4(n-1) n-1=59 n=60.
There are 60 multiples of 4 between 10 and 250.

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Q15

For what value of nnn, are the nnnth terms of two APs: 63, 65, 67, …63,\ 65,\ 67,\ \ldots63, 65, 67, and 3, 10, 17, …3,\ 10,\ 17,\ \ldots3, 10, 17, equal?

Solution

First AP: a=63a=63a=63, d=2d=2d=2, so an=63+(n−1)(2)=61+2na_n=63+(n-1)(2)=61+2na_n=63+(n-1)(2)=61+2n.
Second AP: a=3a=3a=3, d=7d=7d=7, so an=3+(n−1)(7)=7n−4a_n=3+(n-1)(7)=7n-4a_n=3+(n-1)(7)=7n-4.
Setting them equal: 61+2n=7n−4⇒65=5n⇒n=1361+2n=7n-4\Rightarrow65=5n\Rightarrow n=1361+2n=7n-465=5n n=13.
The nnnth terms are equal when n=13n=\textbf{13}n=13 (both terms then equal 61+26=8761+26=8761+26=87).

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Q16

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution

a7−a5=12⇒(a+6d)−(a+4d)=2d=12⇒d=6a_7-a_5=12\Rightarrow(a+6d)-(a+4d)=2d=12\Rightarrow d=6a_7-a_5=12(a+6d)-(a+4d)=2d=12 d=6.
a3=a+2d=16⇒a=16−12=4a_3=a+2d=16\Rightarrow a=16-12=4a_3=a+2d=16 a=16-12=4.
The AP is 4, 10, 16, 22, …4,\ 10,\ 16,\ 22,\ \ldots4, 10, 16, 22,

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Q17

Find the 20th term from the last term of the AP 3, 8, 13, …, 2533,\ 8,\ 13,\ \ldots,\ 2533, 8, 13, , 253.

Solution

a=3a=3a=3, d=5d=5d=5, l=253l=253l=253.
First find the total number of terms: 253=3+(n−1)(5)⇒250=5(n−1)⇒n=51253=3+(n-1)(5)\Rightarrow250=5(n-1)\Rightarrow n=51253=3+(n-1)(5)250=5(n-1) n=51.
The 20th term from the last term is the (51−20+1)=32(51-20+1)=32(51-20+1)=32nd term from the start.
a32=a+31d=3+31(5)=3+155=158a_{32}=a+31d=3+31(5)=3+155=158a_32=a+31d=3+31(5)=3+155=158.

(Alternative: reverse the AP so a=253a=253a=253, d=−5d=-5d=-5; then a20=253+19(−5)=253−95=158a_{20}=253+19(-5)=253-95=158a_20=253+19(-5)=253-95=158 — same answer.)

The 20th term from the last term is 158.

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Q18

The 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.

Solution

a4+a8=24⇒(a+3d)+(a+7d)=2a+10d=24⇒a+5d=12a_4+a_8=24\Rightarrow(a+3d)+(a+7d)=2a+10d=24\Rightarrow a+5d=12a_4+a_8=24(a+3d)+(a+7d)=2a+10d=24 a+5d=12 ... (1)
a6+a10=44⇒(a+5d)+(a+9d)=2a+14d=44⇒a+7d=22a_6+a_{10}=44\Rightarrow(a+5d)+(a+9d)=2a+14d=44\Rightarrow a+7d=22a_6+a_10=44(a+5d)+(a+9d)=2a+14d=44 a+7d=22 ... (2)

(2) −-- (1): 2d=10⇒d=52d=10\Rightarrow d=52d=10 d=5. From (1): a=12−25=−13a=12-25=-13a=12-25=-13.

First three terms: −13, −8, −3-13,\ -8,\ -3-13, -8, -3.

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Q19

Subba 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?

Solution

Salaries form an AP with a=5000a=5000a=5000 (year 1995), d=200d=200d=200.
Let the salary reach ₹7000 in the nnnth year:
7000=5000+(n−1)(200)⇒2000=200(n−1)⇒n−1=10⇒n=117000=5000+(n-1)(200)\Rightarrow2000=200(n-1)\Rightarrow n-1=10\Rightarrow n=117000=5000+(n-1)(200)2000=200(n-1) n-1=10 n=11.
The 11th year of his job is 1995+(11−1)=20051995+(11-1)=20051995+(11-1)=2005.
His income reached ₹7000 in the year 2005 (the 11th year of his employment).

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Q20

Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nnnth week, her weekly savings become ₹20.75, find nnn.

Solution

a=5a=5a=5, d=1.75d=1.75d=1.75, an=20.75a_n=20.75a_n=20.75.
20.75=5+(n−1)(1.75)⇒15.75=1.75(n−1)⇒n−1=9⇒n=1020.75=5+(n-1)(1.75)\Rightarrow15.75=1.75(n-1)\Rightarrow n-1=9\Rightarrow n=1020.75=5+(n-1)(1.75)15.75=1.75(n-1) n-1=9 n=10.
So n=10n=\textbf{10}n=10.

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Exercise 5.3

Q1

Find the sum of the following APs:

(i) 2, 7, 12, …2,\ 7,\ 12,\ \ldots2, 7, 12,, to 10 terms

(ii) −37, −33, −29, …-37,\ -33,\ -29,\ \ldots-37, -33, -29,, to 12 terms

(iii) 0.6, 1.7, 2.8, …0.6,\ 1.7,\ 2.8,\ \ldots0.6, 1.7, 2.8,, to 100 terms

(iv) 115, 112, 110, …\frac{1}{15},\ \frac{1}{12},\ \frac{1}{10},\ \ldots1/15, 1/12, 1/10,, to 11 terms

Solution

Use Sn=n2[2a+(n−1)d]S_n=\frac{n}{2}[2a+(n-1)d]S_n=n/2[2a+(n-1)d].

(i) a=2a=2a=2, d=5d=5d=5, n=10n=10n=10: S10=102[4+9(5)]=5(49)=245S_{10}=\frac{10}{2}[4+9(5)]=5(49)=245S_10=10/2[4+9(5)]=5(49)=245.

(ii) a=−37a=-37a=-37, d=4d=4d=4, n=12n=12n=12: S12=122[−74+11(4)]=6(−30)=−180S_{12}=\frac{12}{2}[-74+11(4)]=6(-30)=-180S_12=12/2[-74+11(4)]=6(-30)=-180.

(iii) a=0.6a=0.6a=0.6, d=1.1d=1.1d=1.1, n=100n=100n=100: S100=1002[1.2+99(1.1)]=50(110.1)=5505S_{100}=\frac{100}{2}[1.2+99(1.1)]=50(110.1)=5505S_100=100/2[1.2+99(1.1)]=50(110.1)=5505.

(iv) a=115a=\frac{1}{15}a=1/15, d=112−115=5−460=160d=\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}d=1/12-1/15=5-4/60=1/60, n=11n=11n=11:
S11=112[215+10×160]=112[430+530]=112×930=112×310=3320S_{11}=\frac{11}{2}\left[\frac{2}{15}+10\times\frac{1}{60}\right]=\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}S_11=11/2[2/15+10×1/60]=11/2[4/30+5/30]=11/2×9/30=11/2×3/10=33/20.

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Q2

Find the sums given below:

(i) 7+1012+14+…+847+10\frac{1}{2}+14+\ldots+847+101/2+14++84

(ii) 34+32+30+…+1034+32+30+\ldots+1034+32+30++10

(iii) −5+(−8)+(−11)+…+(−230)-5+(-8)+(-11)+\ldots+(-230)-5+(-8)+(-11)++(-230)

Solution

First find nnn using an=a+(n−1)da_n=a+(n-1)da_n=a+(n-1)d, then use Sn=n2(a+l)S_n=\frac{n}{2}(a+l)S_n=n/2(a+l).

(i) a=7a=7a=7, d=312=72d=3\frac{1}{2}=\frac{7}{2}d=31/2=7/2, l=84l=84l=84.
84=7+(n−1)72⇒77=72(n−1)⇒n−1=22⇒n=2384=7+(n-1)\frac{7}{2}\Rightarrow77=\frac{7}{2}(n-1)\Rightarrow n-1=22\Rightarrow n=2384=7+(n-1)7/277=7/2(n-1) n-1=22 n=23.
S23=232(7+84)=232(91)=20932=104612S_{23}=\frac{23}{2}(7+84)=\frac{23}{2}(91)=\frac{2093}{2}=1046\frac{1}{2}S_23=23/2(7+84)=23/2(91)=2093/2=10461/2.

(ii) a=34a=34a=34, d=−2d=-2d=-2, l=10l=10l=10.
10=34+(n−1)(−2)⇒n−1=12⇒n=1310=34+(n-1)(-2)\Rightarrow n-1=12\Rightarrow n=1310=34+(n-1)(-2) n-1=12 n=13.
S13=132(34+10)=132(44)=286S_{13}=\frac{13}{2}(34+10)=\frac{13}{2}(44)=286S_13=13/2(34+10)=13/2(44)=286.

(iii) a=−5a=-5a=-5, d=−3d=-3d=-3, l=−230l=-230l=-230.
−230=−5+(n−1)(−3)⇒−225=−3(n−1)⇒n−1=75⇒n=76-230=-5+(n-1)(-3)\Rightarrow-225=-3(n-1)\Rightarrow n-1=75\Rightarrow n=76-230=-5+(n-1)(-3)-225=-3(n-1) n-1=75 n=76.
S76=762(−5−230)=38(−235)=−8930S_{76}=\frac{76}{2}(-5-230)=38(-235)=-8930S_76=76/2(-5-230)=38(-235)=-8930.

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Q3

In an AP:

(i) given a=5, d=3, an=50a=5,\ d=3,\ a_n=50a=5, d=3, a_n=50, find nnn and SnS_nS_n.

(ii) given a=7, a13=35a=7,\ a_{13}=35a=7, a_13=35, find ddd and S13S_{13}S_13.

(iii) given a12=37, d=3a_{12}=37,\ d=3a_12=37, d=3, find aaa and S12S_{12}S_12.

(iv) given a3=15, S10=125a_3=15,\ S_{10}=125a_3=15, S_10=125, find ddd and a10a_{10}a_10.

(v) given d=5, S9=75d=5,\ S_9=75d=5, S_9=75, find aaa and a9a_9a_9.

(vi) given a=2, d=8, Sn=90a=2,\ d=8,\ S_n=90a=2, d=8, S_n=90, find nnn and ana_na_n.

(vii) given a=8, an=62, Sn=210a=8,\ a_n=62,\ S_n=210a=8, a_n=62, S_n=210, find nnn and ddd.

(viii) given an=4, d=2, Sn=−14a_n=4,\ d=2,\ S_n=-14a_n=4, d=2, S_n=-14, find nnn and aaa.

(ix) given a=3, n=8, Sn=192a=3,\ n=8,\ S_n=192a=3, n=8, S_n=192, find ddd.

(x) given l=28, Sn=144l=28,\ S_n=144l=28, S_n=144, and there are a total of 9 terms. Find aaa.

Solution

(i) 50=5+(n−1)(3)⇒45=3(n−1)⇒n=1650=5+(n-1)(3)\Rightarrow45=3(n-1)\Rightarrow n=1650=5+(n-1)(3)45=3(n-1) n=16.
S16=162(5+50)=8(55)=440S_{16}=\frac{16}{2}(5+50)=8(55)=440S_16=16/2(5+50)=8(55)=440.
n=16, Sn=440n=16,\ S_n=440n=16, S_n=440.

(ii) a13=a+12d=35⇒7+12d=35⇒d=2812=73a_{13}=a+12d=35\Rightarrow7+12d=35\Rightarrow d=\frac{28}{12}=\frac{7}{3}a_13=a+12d=357+12d=35 d=28/12=7/3.
S13=132(a+a13)=132(7+35)=132(42)=273S_{13}=\frac{13}{2}(a+a_{13})=\frac{13}{2}(7+35)=\frac{13}{2}(42)=273S_13=13/2(a+a_13)=13/2(7+35)=13/2(42)=273.
d=73, S13=273d=\frac{7}{3},\ S_{13}=273d=7/3, S_13=273.

(iii) a12=a+11d=37⇒a=37−33=4a_{12}=a+11d=37\Rightarrow a=37-33=4a_12=a+11d=37 a=37-33=4.
S12=122(a+a12)=6(4+37)=246S_{12}=\frac{12}{2}(a+a_{12})=6(4+37)=246S_12=12/2(a+a_12)=6(4+37)=246.
a=4, S12=246a=4,\ S_{12}=246a=4, S_12=246.

(iv) a3=a+2d=15⇒a=15−2da_3=a+2d=15\Rightarrow a=15-2da_3=a+2d=15 a=15-2d.
S10=102[2a+9d]=5(2a+9d)=125⇒2a+9d=25S_{10}=\frac{10}{2}[2a+9d]=5(2a+9d)=125\Rightarrow2a+9d=25S_10=10/2[2a+9d]=5(2a+9d)=1252a+9d=25.
Substitute: 2(15−2d)+9d=25⇒30+5d=25⇒d=−12(15-2d)+9d=25\Rightarrow30+5d=25\Rightarrow d=-12(15-2d)+9d=2530+5d=25 d=-1.
a=15−2(−1)=17a=15-2(-1)=17a=15-2(-1)=17; a10=a+9d=17−9=8a_{10}=a+9d=17-9=8a_10=a+9d=17-9=8.
d=−1, a10=8d=-1,\ a_{10}=8d=-1, a_10=8.

(v) S9=92[2a+8d]=9(a+4d)=75⇒a+4d=253S_9=\frac{9}{2}[2a+8d]=9(a+4d)=75\Rightarrow a+4d=\frac{25}{3}S_9=9/2[2a+8d]=9(a+4d)=75 a+4d=25/3.
a=253−4(5)=253−20=−353a=\frac{25}{3}-4(5)=\frac{25}{3}-20=-\frac{35}{3}a=25/3-4(5)=25/3-20=-35/3.
a9=a+8d=−353+40=853a_9=a+8d=-\frac{35}{3}+40=\frac{85}{3}a_9=a+8d=-35/3+40=85/3.
a=−353, a9=853a=-\frac{35}{3},\ a_9=\frac{85}{3}a=-35/3, a_9=85/3.

(vi) Sn=n2[4+8(n−1)]=n(4n−2)=90⇒4n2−2n−90=0⇒2n2−n−45=0S_n=\frac{n}{2}[4+8(n-1)]=n(4n-2)=90\Rightarrow4n^2-2n-90=0\Rightarrow2n^2-n-45=0S_n=n/2[4+8(n-1)]=n(4n-2)=904n^2-2n-90=02n^2-n-45=0.
Solving, n=1±1+3604=1±194n=\frac{1\pm\sqrt{1+360}}{4}=\frac{1\pm19}{4}n=1±√1+360/4=1±19/4, so n=5n=5n=5 (rejecting the negative root).
a5=a+4d=2+32=34a_5=a+4d=2+32=34a_5=a+4d=2+32=34.
n=5, an=34n=5,\ a_n=34n=5, a_n=34.

(vii) Sn=n2(a+an)=n2(8+62)=35n=210⇒n=6S_n=\frac{n}{2}(a+a_n)=\frac{n}{2}(8+62)=35n=210\Rightarrow n=6S_n=n/2(a+a_n)=n/2(8+62)=35n=210 n=6.
an=a+(n−1)d⇒62=8+5d⇒d=545a_n=a+(n-1)d\Rightarrow62=8+5d\Rightarrow d=\frac{54}{5}a_n=a+(n-1)d62=8+5d d=54/5.
n=6, d=545n=6,\ d=\frac{54}{5}n=6, d=54/5.

(viii) an=a+(n−1)d=4⇒a=4−2(n−1)=6−2na_n=a+(n-1)d=4\Rightarrow a=4-2(n-1)=6-2na_n=a+(n-1)d=4 a=4-2(n-1)=6-2n.
Sn=n2(a+an)=n2(6−2n+4)=n(5−n)=−14⇒n2−5n−14=0⇒(n−7)(n+2)=0⇒n=7S_n=\frac{n}{2}(a+a_n)=\frac{n}{2}(6-2n+4)=n(5-n)=-14\Rightarrow n^2-5n-14=0\Rightarrow(n-7)(n+2)=0\Rightarrow n=7S_n=n/2(a+a_n)=n/2(6-2n+4)=n(5-n)=-14 n^2-5n-14=0(n-7)(n+2)=0 n=7.
a=6−14=−8a=6-14=-8a=6-14=-8.
n=7, a=−8n=7,\ a=-8n=7, a=-8.

(ix) Sn=n2[2a+(n−1)d]=4[6+7d]=24+28d=192⇒d=6S_n=\frac{n}{2}[2a+(n-1)d]=4[6+7d]=24+28d=192\Rightarrow d=6S_n=n/2[2a+(n-1)d]=4[6+7d]=24+28d=192 d=6.

(x) Sn=n2(a+l)=92(a+28)=144⇒a+28=32⇒a=4S_n=\frac{n}{2}(a+l)=\frac{9}{2}(a+28)=144\Rightarrow a+28=32\Rightarrow a=4S_n=n/2(a+l)=9/2(a+28)=144 a+28=32 a=4.

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Q4

How many terms of the AP 9, 17, 25, …9,\ 17,\ 25,\ \ldots9, 17, 25, must be taken to give a sum of 636?

Solution

a=9a=9a=9, d=8d=8d=8.
Sn=n2[18+8(n−1)]=n(4n+5)=636⇒4n2+5n−636=0S_n=\frac{n}{2}[18+8(n-1)]=n(4n+5)=636\Rightarrow4n^2+5n-636=0S_n=n/2[18+8(n-1)]=n(4n+5)=6364n^2+5n-636=0.
By the quadratic formula, n=−5±25+101768=−5±1018n=\frac{-5\pm\sqrt{25+10176}}{8}=\frac{-5\pm101}{8}n=-5±√25+10176/8=-5±101/8, giving n=12n=12n=12 or n=−534n=-\frac{53}{4}n=-53/4.
Since nnn must be a positive integer, n=12n=\textbf{12}n=12.

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Q5

The 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.

Solution

Sn=n2(a+l)=n2(5+45)=25n=400⇒n=16S_n=\frac{n}{2}(a+l)=\frac{n}{2}(5+45)=25n=400\Rightarrow n=16S_n=n/2(a+l)=n/2(5+45)=25n=400 n=16.
l=a+(n−1)d⇒45=5+15d⇒d=4015=83l=a+(n-1)d\Rightarrow45=5+15d\Rightarrow d=\frac{40}{15}=\frac{8}{3}l=a+(n-1)d45=5+15d d=40/15=8/3.
Number of terms =16=16=16, common difference =83=\frac{8}{3}=8/3.

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Q6

The 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?

Solution

l=a+(n−1)d⇒350=17+(n−1)(9)⇒333=9(n−1)⇒n−1=37⇒n=38l=a+(n-1)d\Rightarrow350=17+(n-1)(9)\Rightarrow333=9(n-1)\Rightarrow n-1=37\Rightarrow n=38l=a+(n-1)d350=17+(n-1)(9)333=9(n-1) n-1=37 n=38.
S38=382(17+350)=19(367)=6973S_{38}=\frac{38}{2}(17+350)=19(367)=6973S_38=38/2(17+350)=19(367)=6973.
There are 38 terms, and their sum is 6973.

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Q7

Find the sum of first 22 terms of an AP in which d=7d=7d=7 and the 22nd term is 149.

Solution

a22=a+21d=149⇒a=149−147=2a_{22}=a+21d=149\Rightarrow a=149-147=2a_22=a+21d=149 a=149-147=2.
S22=222(a+a22)=11(2+149)=11(151)=1661S_{22}=\frac{22}{2}(a+a_{22})=11(2+149)=11(151)=1661S_22=22/2(a+a_22)=11(2+149)=11(151)=1661.

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Q8

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution

d=18−14=4d=18-14=4d=18-14=4; a=14−d=10a=14-d=10a=14-d=10.
S51=512[2(10)+50(4)]=512[20+200]=512(220)=51(110)=5610S_{51}=\frac{51}{2}[2(10)+50(4)]=\frac{51}{2}[20+200]=\frac{51}{2}(220)=51(110)=5610S_51=51/2[2(10)+50(4)]=51/2[20+200]=51/2(220)=51(110)=5610.

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Q9

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first nnn terms.

Solution

S7=72[2a+6d]=7(a+3d)=49⇒a+3d=7S_7=\frac{7}{2}[2a+6d]=7(a+3d)=49\Rightarrow a+3d=7S_7=7/2[2a+6d]=7(a+3d)=49 a+3d=7 ... (1)
S17=172[2a+16d]=17(a+8d)=289⇒a+8d=17S_{17}=\frac{17}{2}[2a+16d]=17(a+8d)=289\Rightarrow a+8d=17S_17=17/2[2a+16d]=17(a+8d)=289 a+8d=17 ... (2)

(2) −-- (1): 5d=10⇒d=25d=10\Rightarrow d=25d=10 d=2; from (1), a=7−6=1a=7-6=1a=7-6=1.
Sn=n2[2(1)+(n−1)(2)]=n2[2+2n−2]=n2(2n)=n2S_n=\frac{n}{2}[2(1)+(n-1)(2)]=\frac{n}{2}[2+2n-2]=\frac{n}{2}(2n)=n^2S_n=n/2[2(1)+(n-1)(2)]=n/2[2+2n-2]=n/2(2n)=n^2.
So Sn=n2S_n=n^2S_n=n^2.

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Q10

Show that a1, a2, …, an, …a_1,\ a_2,\ \ldots,\ a_n,\ \ldotsa_1, a_2, , a_n, form an AP where ana_na_n is defined as below. Also find the sum of the first 15 terms in each case.

(i) an=3+4na_n=3+4na_n=3+4n

(ii) an=9−5na_n=9-5na_n=9-5n

Solution

(i) a1=7, a2=11, a3=15,…a_1=7,\ a_2=11,\ a_3=15,\ldotsa_1=7, a_2=11, a_3=15,. Since an+1−an=[3+4(n+1)]−[3+4n]=4a_{n+1}-a_n=[3+4(n+1)]-[3+4n]=4a_n+1-a_n=[3+4(n+1)]-[3+4n]=4, a constant, the list is an AP with a=7a=7a=7, d=4d=4d=4.
S15=152[2(7)+14(4)]=152[14+56]=152(70)=525S_{15}=\frac{15}{2}[2(7)+14(4)]=\frac{15}{2}[14+56]=\frac{15}{2}(70)=525S_15=15/2[2(7)+14(4)]=15/2[14+56]=15/2(70)=525.

(ii) a1=4, a2=−1, a3=−6,…a_1=4,\ a_2=-1,\ a_3=-6,\ldotsa_1=4, a_2=-1, a_3=-6,. Since an+1−an=[9−5(n+1)]−[9−5n]=−5a_{n+1}-a_n=[9-5(n+1)]-[9-5n]=-5a_n+1-a_n=[9-5(n+1)]-[9-5n]=-5, a constant, the list is an AP with a=4a=4a=4, d=−5d=-5d=-5.
S15=152[2(4)+14(−5)]=152[8−70]=152(−62)=−465S_{15}=\frac{15}{2}[2(4)+14(-5)]=\frac{15}{2}[8-70]=\frac{15}{2}(-62)=-465S_15=15/2[2(4)+14(-5)]=15/2[8-70]=15/2(-62)=-465.

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Q11

If the sum of the first nnn terms of an AP is 4n−n24n-n^24n-n^2, what is the first term (that is S1S_1S_1)? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nnnth terms.

Solution

Sn=4n−n2S_n=4n-n^2S_n=4n-n^2.
S1=4(1)−12=3S_1=4(1)-1^2=3S_1=4(1)-1^2=3, so a1=3a_1=3a_1=3.
S2=4(2)−22=8−4=4S_2=4(2)-2^2=8-4=4S_2=4(2)-2^2=8-4=4; this is the sum of the first two terms.
a2=S2−S1=4−3=1a_2=S_2-S_1=4-3=1a_2=S_2-S_1=4-3=1.
S3=4(3)−32=12−9=3S_3=4(3)-3^2=12-9=3S_3=4(3)-3^2=12-9=3; a3=S3−S2=3−4=−1a_3=S_3-S_2=3-4=-1a_3=S_3-S_2=3-4=-1.
S9=4(9)−92=36−81=−45S_9=4(9)-9^2=36-81=-45S_9=4(9)-9^2=36-81=-45; S10=4(10)−102=40−100=−60S_{10}=4(10)-10^2=40-100=-60S_10=4(10)-10^2=40-100=-60; a10=S10−S9=−60−(−45)=−15a_{10}=S_{10}-S_9=-60-(-45)=-15a_10=S_10-S_9=-60-(-45)=-15.

In general, Sn−1=4(n−1)−(n−1)2=4n−4−(n2−2n+1)=6n−5−n2S_{n-1}=4(n-1)-(n-1)^2=4n-4-(n^2-2n+1)=6n-5-n^2S_n-1=4(n-1)-(n-1)^2=4n-4-(n^2-2n+1)=6n-5-n^2.
an=Sn−Sn−1=(4n−n2)−(6n−5−n2)=5−2na_n=S_n-S_{n-1}=(4n-n^2)-(6n-5-n^2)=5-2na_n=S_n-S_n-1=(4n-n^2)-(6n-5-n^2)=5-2n.

So: first term =3=3=3; sum of first two terms =4=4=4; second term =1=1=1; third term =−1=-1=-1; tenth term =−15=-15=-15; an=5−2na_n=5-2na_n=5-2n.

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Q12

Find the sum of the first 40 positive integers divisible by 6.

Solution

The numbers are 6, 12, 18, …6,\ 12,\ 18,\ \ldots6, 12, 18, (40 terms): a=6a=6a=6, d=6d=6d=6, n=40n=40n=40.
S40=402[2(6)+39(6)]=20[12+234]=20(246)=4920S_{40}=\frac{40}{2}[2(6)+39(6)]=20[12+234]=20(246)=4920S_40=40/2[2(6)+39(6)]=20[12+234]=20(246)=4920.

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Q13

Find the sum of the first 15 multiples of 8.

Solution

a=8a=8a=8, d=8d=8d=8, n=15n=15n=15.
S15=152[2(8)+14(8)]=152[16+112]=152(128)=960S_{15}=\frac{15}{2}[2(8)+14(8)]=\frac{15}{2}[16+112]=\frac{15}{2}(128)=960S_15=15/2[2(8)+14(8)]=15/2[16+112]=15/2(128)=960.

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Q14

Find the sum of the odd numbers between 0 and 50.

Solution

The odd numbers between 0 and 50 are 1, 3, 5, …, 491,\ 3,\ 5,\ \ldots,\ 491, 3, 5, , 49: a=1a=1a=1, d=2d=2d=2, l=49l=49l=49.
Number of terms: 49=1+(n−1)(2)⇒n=2549=1+(n-1)(2)\Rightarrow n=2549=1+(n-1)(2) n=25.
S25=252(1+49)=252(50)=625S_{25}=\frac{25}{2}(1+49)=\frac{25}{2}(50)=625S_25=25/2(1+49)=25/2(50)=625.

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Q15

A contract on a 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 does the contractor have to pay as penalty, if he has delayed the work by 30 days?

Solution

The penalties form an AP: a=200a=200a=200, d=50d=50d=50, n=30n=30n=30.
S30=302[2(200)+29(50)]=15[400+1450]=15(1850)=27750S_{30}=\frac{30}{2}[2(200)+29(50)]=15[400+1450]=15(1850)=27750S_30=30/2[2(200)+29(50)]=15[400+1450]=15(1850)=27750.
The contractor has to pay ₹27750 as penalty.

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Q16

A 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.

Solution

Let the largest prize be aaa; the prizes form an AP with n=7n=7n=7, d=−20d=-20d=-20, S7=700S_7=700S_7=700.
S7=72[2a+6(−20)]=7(a−60)=700⇒a−60=100⇒a=160S_7=\frac{7}{2}[2a+6(-20)]=7(a-60)=700\Rightarrow a-60=100\Rightarrow a=160S_7=7/2[2a+6(-20)]=7(a-60)=700 a-60=100 a=160.
The prizes are 160, 140, 120, 100, 80, 60, 40160,\ 140,\ 120,\ 100,\ 80,\ 60,\ 40160, 140, 120, 100, 80, 60, 40 (in ₹).

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Q17

In 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?

Solution

Trees planted by one section of each class form the AP 1, 2, 3, …, 121,\ 2,\ 3,\ \ldots,\ 121, 2, 3, , 12 (Class I to Class XII): a=1a=1a=1, d=1d=1d=1, n=12n=12n=12.
S12=122(1+12)=6(13)=78S_{12}=\frac{12}{2}(1+12)=6(13)=78S_12=12/2(1+12)=6(13)=78.
Since there are 3 sections in every class, total trees =3×78=234=3\times78=234=3×78=234.

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Q18

A spiral is made up of successive semicircles, with centres alternately at AAA and BBB, starting with centre at AAA, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …0.5\text{ cm},\ 1.0\text{ cm},\ 1.5\text{ cm},\ 2.0\text{ cm},\ \ldots0.5 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}π=22/7.)

CBSE Class 10 Maths — Arithmetic Progressions, Ex 5.3: A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5\t
Solution

The length of a semicircle of radius rrr is πr\pi rπ r. The radii are 0.5, 1.0, 1.5, …0.5,\ 1.0,\ 1.5,\ \ldots0.5, 1.0, 1.5, (13 terms), an AP with a=0.5a=0.5a=0.5, d=0.5d=0.5d=0.5, n=13n=13n=13.

Sum of the 13 radii: S13=132[2(0.5)+12(0.5)]=132[1+6]=132(7)=45.5 cmS_{13}=\frac{13}{2}[2(0.5)+12(0.5)]=\frac{13}{2}[1+6]=\frac{13}{2}(7)=45.5\text{ cm}S_13=13/2[2(0.5)+12(0.5)]=13/2[1+6]=13/2(7)=45.5 cm.

Total length of the spiral =π×(sum of radii)=227×45.5=22×6.5=143 cm=\pi\times(\text{sum of radii})=\frac{22}{7}\times45.5=22\times6.5=143\text{ cm}=π×(sum of radii)=22/7×45.5=22×6.5=143 cm.

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Q19

200 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?

Solution

a=20a=20a=20, d=−1d=-1d=-1, Sn=200S_n=200S_n=200.
Sn=n2[40−(n−1)]=n2(41−n)=200⇒41n−n2=400⇒n2−41n+400=0S_n=\frac{n}{2}[40-(n-1)]=\frac{n}{2}(41-n)=200\Rightarrow41n-n^2=400\Rightarrow n^2-41n+400=0S_n=n/2[40-(n-1)]=n/2(41-n)=20041n-n^2=400 n^2-41n+400=0.
By the quadratic formula, n=41±1681−16002=41±92n=\frac{41\pm\sqrt{1681-1600}}{2}=\frac{41\pm9}{2}n=41±√1681-1600/2=41±9/2, so n=25n=25n=25 or n=16n=16n=16.

Check n=25n=25n=25: a25=20+24(−1)=−4a_{25}=20+24(-1)=-4a_25=20+24(-1)=-4, which is not possible (the number of logs can't be negative), so this root is rejected.

So n=16n=16n=16: a16=20+15(−1)=5a_{16}=20+15(-1)=5a_16=20+15(-1)=5.
There are 16 rows, with 5 logs in the top row.

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Q20

In 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 continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Solution

Distance from the bucket to the nnnth potato is 5+3(n−1)5+3(n-1)5+3(n-1) metres, so the round trip for the nnnth potato is 2[5+3(n−1)]2[5+3(n-1)]2[5+3(n-1)] metres.

These round-trip distances form an AP with a=2(5)=10a=2(5)=10a=2(5)=10, d=2(3)=6d=2(3)=6d=2(3)=6, n=10n=10n=10.
S10=102[2(10)+9(6)]=5[20+54]=5(74)=370S_{10}=\frac{10}{2}[2(10)+9(6)]=5[20+54]=5(74)=370S_10=10/2[2(10)+9(6)]=5[20+54]=5(74)=370.
The competitor has to run a total distance of 370 m.

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Exercise 5.4 (Optional)

Q1

Which term of the AP 121, 117, 113, …121,\ 117,\ 113,\ \ldots121, 117, 113, is its first negative term?

Solution

a=121a=121a=121, d=−4d=-4d=-4. We need the smallest nnn for which an<0a_n<0a_n<0:
121+(n−1)(−4)<0⇒121−4n+4<0⇒125<4n⇒n>31.25121+(n-1)(-4)<0\Rightarrow121-4n+4<0\Rightarrow125<4n\Rightarrow n>31.25121+(n-1)(-4)<0121-4n+4<0125<4n n>31.25.
The smallest integer satisfying this is n=32n=32n=32.
Check: a31=121+30(−4)=1>0a_{31}=121+30(-4)=1>0a_31=121+30(-4)=1>0 and a32=121+31(−4)=−3<0a_{32}=121+31(-4)=-3<0a_32=121+31(-4)=-3<0.
So the 32nd term is the first negative term.

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Q2

The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.

Solution

a3+a7=6a_3+a_7=6a_3+a_7=6 gives (a+2d)+(a+6d)=2a+8d=6⇒a+4d=3(a+2d)+(a+6d)=2a+8d=6\Rightarrow a+4d=3(a+2d)+(a+6d)=2a+8d=6 a+4d=3 ... (1)
a3⋅a7=8a_3\cdot a_7=8a_3· a_7=8.

Let a3=xa_3=xa_3=x, so a7=6−xa_7=6-xa_7=6-x (from the sum). Then x(6−x)=8⇒x2−6x+8=0⇒(x−2)(x−4)=0⇒x=2x(6-x)=8\Rightarrow x^2-6x+8=0\Rightarrow(x-2)(x-4)=0\Rightarrow x=2x(6-x)=8 x^2-6x+8=0(x-2)(x-4)=0 x=2 or x=4x=4x=4.

Case 1: a3=2, a7=4a_3=2,\ a_7=4a_3=2, a_7=4. Then 4d=a7−a3=2⇒d=124d=a_7-a_3=2\Rightarrow d=\frac{1}{2}4d=a_7-a_3=2 d=1/2; from (1), a=3−4(12)=1a=3-4\left(\frac{1}{2}\right)=1a=3-4(1/2)=1.
S16=162[2(1)+15(12)]=8[2+7.5]=8(9.5)=76S_{16}=\frac{16}{2}\left[2(1)+15\left(\frac{1}{2}\right)\right]=8[2+7.5]=8(9.5)=76S_16=16/2[2(1)+15(1/2)]=8[2+7.5]=8(9.5)=76.

Case 2: a3=4, a7=2a_3=4,\ a_7=2a_3=4, a_7=2. Then 4d=2−4=−2⇒d=−124d=2-4=-2\Rightarrow d=-\frac{1}{2}4d=2-4=-2 d=-1/2; from (1), a=3−4(−12)=5a=3-4\left(-\frac{1}{2}\right)=5a=3-4(-1/2)=5.
S16=162[2(5)+15(−12)]=8[10−7.5]=8(2.5)=20S_{16}=\frac{16}{2}\left[2(5)+15\left(-\frac{1}{2}\right)\right]=8[10-7.5]=8(2.5)=20S_16=16/2[2(5)+15(-1/2)]=8[10-7.5]=8(2.5)=20.

Both cases are valid (both satisfy the given conditions), so S16=76S_{16}=76S_16=76 or S16=20S_{16}=20S_16=20.

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Q3

A 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}21/2 m apart, what is the length of the wood required for the rungs?

CBSE Class 10 Maths — Arithmetic Progressions, Ex 5.4: A 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 t
Solution

Distance between the top and bottom rungs =212 m=250 cm=2\frac{1}{2}\text{ m}=250\text{ cm}=21/2 m=250 cm; rungs are 25 cm apart, so the number of gaps is 25025=10\frac{250}{25}=10250/25=10, giving n=11n=11n=11 rungs.

The rung lengths form an AP with a=45a=45a=45 (bottom), l=25l=25l=25 (top), n=11n=11n=11.
Total wood required =S11=112(a+l)=112(45+25)=112(70)=385=S_{11}=\frac{11}{2}(a+l)=\frac{11}{2}(45+25)=\frac{11}{2}(70)=385=S_11=11/2(a+l)=11/2(45+25)=11/2(70)=385.
The length of wood required for the rungs is 385 cm.

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Q4

The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of xxx such that the sum of the numbers of the houses preceding the house numbered xxx is equal to the sum of the numbers of the houses following it. Find this value of xxx.

Solution

Sum of house numbers before house xxx: 1+2+⋯+(x−1)=(x−1)x21+2+\cdots+(x-1)=\frac{(x-1)x}{2}1+2+·s+(x-1)=(x-1)x/2.

Sum of house numbers after house xxx: (x+1)+⋯+49=49×502−x(x+1)2=1225−x(x+1)2(x+1)+\cdots+49=\frac{49\times50}{2}-\frac{x(x+1)}{2}=1225-\frac{x(x+1)}{2}(x+1)+·s+49=49×50/2-x(x+1)/2=1225-x(x+1)/2.

Setting these equal:
(x−1)x2=1225−x(x+1)2\frac{(x-1)x}{2}=1225-\frac{x(x+1)}{2}(x-1)x/2=1225-x(x+1)/2

(x−1)x+x(x+1)=2450(x-1)x+x(x+1)=2450(x-1)x+x(x+1)=2450

x2−x+x2+x=2450⇒2x2=2450⇒x2=1225⇒x=35x^2-x+x^2+x=2450\Rightarrow2x^2=2450\Rightarrow x^2=1225\Rightarrow x=35x^2-x+x^2+x=24502x^2=2450 x^2=1225 x=35 (taking the positive value, since 1≤x≤491\le x\le491≤ x≤49).

So x=35x=\textbf{35}x=35: the numbers of houses 1 to 34 sum to the same total as houses 36 to 49, confirming such a value of xxx exists.

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Q5

A small terrace at a football ground comprises 15 steps, each of which is 50 m long and built of solid concrete. Each step has a rise of 14\frac{1}{4}1/4 m and a tread of 12\frac{1}{2}1/2 m. Calculate the total volume of concrete required to build the terrace.

CBSE Class 10 Maths — Arithmetic Progressions, Ex 5.4: A small terrace at a football ground comprises 15 steps, each of which is 50 m long and built of solid concrete. Each step ha
Solution

The height of the kkkth step above the ground is k4\frac{k}{4}k/4 m (each step rises 14\frac1414 m over the one before). The volume of concrete in the kkkth step is
Vk=(height)×(tread)×(length)=k4×12×50=25k4 m3.V_k=(\text{height})\times(\text{tread})\times(\text{length})=\frac{k}{4}\times\frac{1}{2}\times50=\frac{25k}{4}\text{ m}^3.V_k=(height)×(tread)×(length)=k/4×1/2×50=25k/4 m^3.

The volumes V1, V2, …, V15V_1,\ V_2,\ \ldots,\ V_{15}V_1, V_2, , V_15 form an AP with first term and common difference both equal to 254\frac{25}{4}25/4 (since Vk=k×254V_k=k\times\frac{25}{4}V_k=k×25/4).

Total volume =∑k=11525k4=254(1+2+⋯+15)=254×15×162=254×120=750=\sum_{k=1}^{15}\frac{25k}{4}=\frac{25}{4}(1+2+\cdots+15)=\frac{25}{4}\times\frac{15\times16}{2}=\frac{25}{4}\times120=750=_k=1^1525k/4=25/4(1+2+·s+15)=25/4×15×16/2=25/4×120=750.

The total volume of concrete required is 750 m³.

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All CBSE Class 10 Maths Chapters

Real NumbersPolynomialsPair of Linear Equations in Two VariablesQuadratic EquationsArithmetic ProgressionsTrianglesCoordinate GeometryIntroduction to TrigonometrySome Applications of TrigonometryCirclesAreas Related to CirclesSurface Areas and VolumesStatisticsProbability
Arithmetic Progressions — Important QuestionsA shorter, exam-focused shortlist for CBSE Class 10 Maths — best closer to test day

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