NCERT Solutions Class 10 Maths Chapter 5

NCERT Solutions For Class 10 Mathematics Chapter 5 Arithmetic Progressions

Mathematics is one subject that most students find difficult. The only way to actually score good marks in Mathematics is to practice solving as many problems as one can. That is why students find NCERT Solutions For Class 10 Mathematics Chapter 5 by Extramarks really useful in their preparation. If students ever get stuck on any of the exercise questions, they can always refer to these solutions. These are also great resources for completing assignments and for last-minute revision. 

Access NCERT Solutions for Mathematics Chapter 5 – Arithmetic Progression

Chapter 5 of Class 10 Mathematics NCERT textbook introduces students to Arithmetic Progressions, which are nothing but lists of numbers where each term, except the first term, can be derived by adding or subtracting a fixed number to its preceding term. These are really useful to express patterns that we see in our daily lives and in nature. The chapter further discusses the different types of Arithmetic Progressions, the sum of Arithmetic Progressions, deriving a general formula for the nth term of an Arithmetic Progression, and much more. 

What is Arithmetic Progression?

An Arithmetic Progression is a series of numbers where each number can be derived from its predecessor by adding or subtracting a fixed number. For example, consider the series of numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Here, each term can be derived from its predecessor by adding 1 to it.

The fixed difference between the terms of an AP is called the common difference. In the above example, the common difference is 1.

NCERT Solutions for Class 10 Mathematics

Extramarks provides detailed NCERT Solutions for Class 10 Mathematics. Since the only way to get good at Mathematics is to solve problems, these solutions can be very useful for students. Extramarks provides detailed solutions to the questions given in all of the NCERT Mathematics textbooks chapters listed below:

  • Chapter 1 – Real Numbers
  • Chapter 2 – Polynomials
  • Chapter 3 – Pair of Linear Equations in Two Variables
  • Chapter 4 – Quadratic Equations
  • Chapter 5 – Arithmetic Progressions
  • Chapter 6 – Triangles
  • Chapter 7 – Coordinate Geometry
  • Chapter 8 – Introduction to Trigonometry
  • Chapter 9 – Some Applications of Trigonometry
  • Chapter 10 – Circles
  • Chapter 11 – Constructions
  • Chapter 12 – Areas Related to Circles
  • Chapter 13 – Surface Areas and Volumes
  • Chapter 14 – Statistics
  • Chapter 15–Probability

The General Form of an Arithmetic Progression

The general form of an AP can be expressed as follows:

A, A + d, A + 2d, A + 3d,….

Here, A is the first term and d is the common difference.

Position Of Terms Representation of terms Value of Terms
1 a1 A=a+(1-1) d
2 a2 A+d=a+(2-1) d
3 a3 A+2d=a+(3-1) d
4 a4 A+3d=a+(4-1) d
.
.
.
.
N An A+(n-1)d

The Formulas

There are some essential formulae that students will need to solve problems related to Arithmetic Progressions. These formulae have been listed below:

  • Arithmetic Progression’s nth Term (AP)

Students will face many problems related to this chapter where they will be expected to calculate the nth term of an Arithmetic Progression. The formula to calculate the nth term of an AP is shown below:

An = a + (n – 1) d

Here,

a= The initial term 

d= The difference value

n= the number of terms

An= the nth term

Consider the following problem. Calculate the 15th term in the AP:  1, 2, 3, 4, 5 … 

Here, we are supposed to calculate the 15th term. So, the value of n is 15. The first term, a, is 1 and the common difference is also 1. Using the formula to calculate the nth term of an AP, the 15th term can be calculated as follows:

A15 = 1 + (15 – 1) 1 = 1 + 14 = 15.

  • The Sum of the First n Terms

Another common type of problem that students will face related to Arithmetic Progressions is to calculate the sum of the first n terms of a given AP. The formula to do so is shown below:

S = (n / 2)*(2a+(n−1)d)

Here,

n is the number of terms

a is the first term in the AP

d is the common difference

Here, using the formula the nth term of an AP, we can further simplify this expression to the one shown below:

S = (n/2)*(a+An)/2

Here,

An is the nth term of the AP

Also, if the nth term is actually the last term of the AP, then the formula is further simplified to the following:

S = n / 2 (first term + last term)

We’ve also included all of the crucial formulas in this chapter in a table for easy reference.

General Form of AP A, a + d, a + 2d, a + 3d, a + 4d, …, a + nd
The nth term of AP An = a + (n – 1) x d
Sum of n terms in AP S= (n/2)*(2a + (n – 1)d)
Sum of all terms in a finite AP with the last term as I (n/2)*(a + I)

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The Benefits of Referring to NCERT Solutions Mathematics 

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  • All the answers are written following the CBSE guidelines. When students study from it, they will get an edge over their peers. 
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Q.1 Write first four terms of the AP, when the first terma and the common difference d are given as follows:(i) a=10, d=10 (ii) a=−2, d=0(iii) a=4, d=−3 (iv) a=−1, d=12(v) a=−1.25, d=−0.25

Ans.

(i) We have,First term=a=10 and Common difference=d=10Now,Second term=a+d=10+10=20,Third term=a+2d=10+2×10=30,Fourth term=a+3d=10+3×10=40,Therefore, first four terms of the AP are:10, 20, 30, 40(ii) We have,First term=a=−2and Common difference=d=0Now,Second term=a+d=−2+0=−2,Third term=a+2d=−2+2×0=−2,Fourth term=a+3d=−2+3×0=−2,Therefore, first four terms of the AP are:−2, −2, −2, −2(iii) We have,First term=a=4and Common difference=d=−3Now,Second term=a+d=4−3=1,Third term=a+2d=4+2×(−3)=−2,Fourth term=a+3d=4+3×(−3)=−5,Therefore, first four terms of the AP are:4, 1,  −2,  −5(iv) We have,First term=a=−1and Common difference=d=12Now,Second term=a+d=−1+12=−12Third term=a+2d=−1+2×12=0Fourth term=a+3d=−1+3×12=12Therefore, first four terms of the AP are:−1, −12, 0, 12(v)We have,First term=a=−1.25and Common difference=d=−0.25Now,Second term=a+d=−1.25−0.25=−1.50,Third term=a+2d=−1.25+2×(−0.25)=−1.75,Fourth term=a+3d=−1.25+3×(−0.25)=−2,Therefore, first four terms of the AP are:−1.25, −1.50,  −1.75,  −2

Q.2 For the following APs, write the first term and thecommon difference:(i) 3,  1, −1, −3, . . . (ii)−5,  −1, 3, 7, . . .(iii) 13,  53, 93,  133,. . . (iv) 0.6, 1.7, 2.8, 3.9, . . .

Ans.

(i) Following is the given AP. 3,  1, −1, −3, . . . Here, first term=3andcommon difference =difference between a term and                                               its preceding term                                            =1−3=−2(ii) Following is the given AP. −5,  −1,  3,  7, . . . Here, first term=−5andcommon difference =difference between a term and                                               its preceding term                                            =−1−5=−6(iii) Following is the given AP. 13,   53, 93, 139, . . . Here, first term=13andcommon difference =difference between a term and                                               its preceding term                                            =53−13=43(iv) Following is the given AP. 0.6, 1.7, 2.8, 3.9, . . . Here, first term=0.6andcommon difference =difference between a term and                                               its preceding term                                            =1.7−0.6= 1.1

Q.3 Which of the following are APs? If they form an AP,find the common difference d and write three moreterms.(i) 2, 4, 8, 16, . . .(ii) 2, 52, 3, 72, . . .(iii) −1.2, −3.2, −5.2, −7.2, . . .(iv) −10, −6,−2, 2, . . .(v) 3, 3+2,  3+22,  3+32,  . . .(vi) 0.2, 0.22, 0.222, 0.2222,  . . .(vii) 0, −4, −8, −12, . . .(viii) −12, −12, −12, −12, . . .(ix) 1, 3, 9, 27, ...(x) a, 2a, 3a, 4a, ...(xi) a, a2, a3, a4, ...(xii) 2, 8, 18, 32, ...(xiii) 3, 6, 9, 12, ...(xiv) 12, 32, 52, 72, ...(xv) 12, 52, 72, 73, ...

Ans.

(i) 2, 4, 8, 16, . . .Here, 4−2=2≠8−4 Therefore, the given list of numbers is not an AP.(ii) 2, 52, 3, 72, . . .We have, a2−a1=52−2=12 a3−a2=3−52=12 a4−a3=72−3=12i.e., ak+1−ak is the same every time.Therefore, the given list of numbers is an AP with the commondifference d=12.The next three terms are: 72+12=4,  4+12=92 and 92+12=5.(iii) −1.2, −3.2, −5.2, −7.2, . . .We have, a2−a1=−3.2−(−1.2)=−3.2+1.2=−2 a3−a2=−5.2−(−3.2)=−5.2+3.2=−2 a4−a3=−7.2−(−5.2)=−7.2+5.2=−2i.e., ak+1−ak is the same every time.Therefore, the given list of numbers is an AP with the commondifference d=−2.The next three terms are: −7.2−2=−9.2,  −9.2−2=−11.2 and −11.2−2=−13.2.(iv) −10, −6,−2, 2, . . .We have, a2−a1=−6−(−10)=−6+10=4 a3−a2=−2−(−6)=−2+6=4 a4−a3=2−(−2)=2+2=4i.e., ak+1−ak is the same every time.Therefore, the given list of numbers is an AP with the commondifference d=4.The next three terms are: 2+4=6,  6+4=10 and 10+4=14.(v) 3, 3+2,  3+22,  3+32,  . . .We have, a2−a1=3+2−3=2 a3−a2=3+22−(3+2)=2 a4−a3=3+32−(3+22)=2i.e., ak+1−ak is the same every time.Therefore, the given list of numbers is an AP with the commondifference d=2.The next three terms are: 3+32+2=3+42,  3+42+2=3+52 and 3+52+2=3+62.(vi) 0.2, 0.22, 0.222, 0.2222,  . . .We have, a2−a1=0.22−0.2=0.02 a3−a2=0.222−0.22=0.002So, a2−a1≠a3−a2Therefore, the given list of numbers is not an AP. (vii) 0, −4, −8, −12, . . .We have, a2−a1=−4−0=−4 a3−a2=−8−(−4)=−8+4=−4 a4−a3=−12−(−8)=−12+8=−4i.e., ak+1−ak is the same every time.Therefore, the given list of numbers is an AP with the commondifference d=−4.The next three terms are: −12−4=−16,  −16−4=−20and −20−4=−24.(viii) −12, −12, −12, −12,  . . .Obviously, ak+1−ak is the same every time.Therefore, the given list of numbers is an AP with the commondifference d=0.The next three terms are: −12, −12, −12.(ix) 1, 3, 9, 27, ...We have, a2−a1=3−1=2 a3−a2=9−3=6So, a2−a1≠a3−a2Therefore, the given list of numbers is not an AP. (x) a, 2a, 3a, 4a, ...We have, a2−a1=2a−a=a a3−a2=3a−2a=a a4−a3=4a−3a=ai.e., ak+1−ak is the same every time.Therefore, the given list of numbers is an AP with the commondifference d=a.The next three terms are: 4a+a=5a,  5a+a=6aand 6a+a=7a.(xi) a, a2, a3, a4, ...Obviously, ak+1−ak is not the same every time.Therefore, the given list of numbers is not an AP.(xii) 2, 8, 18, 32, ...We have, a2−a1=8−2=22−2=2 a3−a2=18−8=32−22=2 a4−a3=32−18=42−32=2i.e., ak+1−ak is the same every time.Therefore, the given list of numbers is an AP with the commondifference d=2.The next three terms are: 32+2=42+2=52=50,  50+2=52+2=62=72 and 72+2=62+2=72=98.(xiii) 3, 6, 9, 12, ...We have, a2−a1=6−3=2×3−3=32−1 a3−a2=9−6=3×3−2×3=33−2So, a2−a1≠a3−a2Therefore, the given list of numbers is not an AP. (xiv) 12, 32, 52, 72, ...We have, a2−a1=32−12=9−1=8 a3−a2=52−32=25−9=16So, a2−a1≠a3−a2Therefore, the given list of numbers is not an AP. (xv) 12, 52, 72, 73, ...We have, a2−a1=52−12=25−1=24 a3−a2=72−52=49−25=24          a4−a3=73−72=73−49=24i.e., ak+1−ak is the same every time.Therefore, the given list of numbers is an AP with the commondifference d=24.The next three terms are: 73+24=97,  97+24=121 and 121+24=145.

Q.4 Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:

Ans.

(i) We have a=7,    d=3,    n=8We know that      an=a+(n−1)d⇒an=7+(8−1)3=28(ii) We have a=−18,  n=10, an=0We know that      an=a+(n−1)d⇒0=−18+(10−1)d⇒d=189=2(iii) We have d=−3, n=18, an=−5We know that      an=a+(n−1)d⇒−5=a+(18−1)(−3)⇒a=51−5=46(iv) We have a=−18.9,  d=2.5, an=3.6We know that      an=a+(n−1)d⇒3.6=−18.9+(n−1)(2.5)⇒(n−1)(2.5)=3.6+18.9=22.5⇒n=22.52.5+1=10(v) We have a=3.5,    d=0,    n=105We know that      an=a+(n−1)d⇒an=3.5+(105−1)×0=3.5

Q.5 Choose 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, −12, 2, . . ., is (A) 28 (B) 22 (C) −38 (D) −4812

Ans.

(i)The given AP is: 10, 7, 4, . . . Here, a=10, d=7−10=−3, n=30We know that an=a+(n−1)d⇒ a30=10+(30−1)(−3)=10−87=−77Hence, the correct answer is (C).(ii)The given AP is: −3, −12, 2, . . .Here, a=−3, d=2−−12=52, n=11We know that an=a+(n−1)d⇒ a11=−3+(11−1)×52=−3+25=22Hence, the correct answer is (B).

Q.6 In the following APs, find the missing terms in the boxes:i 2, , 26ii , 13, , 3iii 5,  ,  , 912iv−4, ,  , , , 6v , 38, ,  ,  , −22

Ans.

iThe given AP is:2,, 26Let the missing term is ‘t‘. Then, we havet−2=26−t⇒2t=26+2=28⇒t=282=14iiThe given AP is:, 13,, 3Let the common difference is ‘d‘. Then, we havea4−a2=2d=3−13⇒2d=−10⇒d=−102=−5Therefore,first term=a1=a2−d=13−−5=13+5=18third term=a3=a4−d=3−−5=3+5=8Hence, the given APis:18,13,8,3iiiThe given AP is:5, , , 912Let the common difference is ‘d‘. Then, we havea2=a1+d,a3=a2+d=a1+d+d=a1+2d,a4=a3+d=a1+2d+d=a1+3d⇒192=5+3d⇒3d=192−5=92⇒d=92×13=32Therefore, missing terms of the given AP are:a2=a1+d=5+32=132,a3=a2+d=132+32=162=8Hence, the given AP is:5,132,8,912ivThe given AP is:−4, , , , , 6Let the common difference is ‘d‘. Then, we havea6=a1+5d⇒6=−4+5d⇒5d=6+4=10⇒d=105=2Therefore, missing terms of the given AP are:a2=a1+d=−4+2=−2a3=a1+2d=−4+2×2=0,a4=a1+3d=−4+3×2=2,a5=a1+4d=−4+4×2=4Hence, the given AP is:−4,−2,0,2,4,6vThe given AP is:, 38,, , , −22Let the common difference is ‘d‘. Then, we havea6=a1+5d⇒a6=a1+d+4d=a2+4d⇒−22=38+4d⇒4d=−22−38=−60⇒d=−604=−15Therefore, missing terms of the given AP are:a1=a2+d=38−−15=38+15=53,a3=a1+2d=53+2×−15=23,a4=a1+3d=53+3×−15=8a5=a1+4d=53+4×−15=−7Hence, the given AP is:53,38,23,8,−7,−22

Q.7 Which term of the AP: 3, 8, 13, 18, . . . , is 78?

Ans.

The given AP is: 3, 8, 13, 18, . . . Let nth term of the given AP is 78. We know that an=a+(n−1)d⇒78=3+(n−1)5⇒(n−1)5=78−3=75⇒n−1=755=15⇒n=15+1=16Therefore, 16th term of the given AP is 78.

Q.8 Find the number of terms in each of the following APs:(i) 7, 13, 19, . . . , 205 (ii) 18, 1512, 13, . . . , −47

Ans.

(i) 7, 13, 19, . . . , 205 Let the given AP has n terms. We know that an=a+(n−1)d⇒205=7+(n−1)6⇒(n−1)6=205−7=198⇒n−1=1986=33⇒n=33+1=34(ii) 18, 1512, 13, . . . , −47Let the given AP has n terms. We know that an=a+(n−1)d⇒−47=18+(n−1)(312−18)=18+(n−1)(−52)⇒(n−1)(−52)=−47−18=−65⇒n−1=−65×−25=26⇒n=26+1=27

Q.9 Check whether –150 is a term of the AP: 11, 8, 5, 2, . . .

Ans.

The given AP is  11, 8, 5, 2, . . .Let nth term of the given AP is −150.We know that an=a+(n−1)d⇒−150=11+(n−1)(8−11)=11+(n−1)(−3)⇒(n−1)(−3)=−150−11=−161⇒n−1=−161×−13=5323⇒n=5323+1=5423But 5423 is not a natural number. Therefore, −150 is nota term of the given AP.

Q.10 Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Ans.

We have a11=38 and a16=73Or a11=a1+10d=38                          ...(1) and     a16=a1+15d=73                          ...(2) Subtracting (1) from (2), we get          5d=35⇒        d=7On putting this value of d in (1), we get        a1+10d=38⇒ a1+10×7=38⇒ a1=38−70=−32Now,a31=a1+30d=−32+30×7=−32+210=178

Q.11 An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Ans.

We have a3=12 and a50=106Or a3=a1+2d=12                           ...(1) and     a50=a1+49d=106                          ...(2) Subtracting (1) from (2), we get          47d=94⇒        d=9447=2On putting this value of d in (1), we get        a1+2×2=12⇒ a1=12−4=8Now,a29=a1+28d=8+28×2=8+56=64

Q.12 If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero?

Ans.

We have a3=4 and a9=−8Or a3=a1+2d=4                           ...(1) and     a9=a1+8d=−8                                ...(2) Subtracting (1) from (2), we get          6d=−12⇒        d=−126=−2On putting this value of d in (1), we get        a1+2×(−2)=4⇒ a1=4+4=8Let nth term of the given AP is zero. Then, we have      an=a1+(n−1)d⇒0=8+(n−1)(−2)⇒n−1=−8−2=4⇒n=4+1=5Hence, 5th term of the given AP is zero.

Q.13 The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Ans.

We have a17=a10+7 ⇒ a17−a10=7⇒a1+16d−(a1+9d)=7 [an=a+(n−1)d]⇒7d=7⇒d=1Therefore, the common difference is 1.

Q.14 Which term of the AP: 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Ans.

Let nth term be 132 more than 54th term of the given AP.We have an=a54+132 ⇒ an−a54=132⇒a1+(n−1)d−(a1+53d)=132 [an=a+(n−1)d]⇒(n−54)d=132⇒n−54=132d=13212 [The given AP is: 3, 15, 27, 39, ...]⇒n=11+54=65Therefore, 65th term is 132 more than 54th term of the given AP.

Q.15 Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Ans.

It is given that two APs have the same common difference.Let an and bn represent the terms of the first and secondAP respectively for all natural numbers n.We have a100−b100=100⇒ a1+99d−b1−99d=100                               [an=a+(n−1)d]⇒a1−b1=100 Now,a1000−b1000=a1+999d−b1−999d=a1−b1=100 Therefore, the difference between 1000th terms of the twoAPs is 100.

Q.16 How many three-digit numbers are divisible by 7?

Ans.

The smallest three-digit number divisible by 7 is 105.The largest three-digit number divisible by 7 is 994.Three-digit numbers divisible by 7 are:105, 112, 119, ...,994These multiples of 7 form an AP with common difference 7,first term 105 and last term 994. Thus, we havea1=105, an=994 and d=7.We know that        an=a1+(n−1)d⇒994=105+(n−1)7⇒(n−1)7=994−105=889 ⇒n−1=8897=127⇒n=127+1=128 Therefore, 128 three-digit numbers are divisible by 7.

Q.17 How many multiples of 4 lie between 10 and 250?

Ans.

Multiples of 4 between 10 and 250 are:12, 16, 20, ...,248These multiples of 4 form an AP with common difference 4,first term 12 and last term 248. Thus, we havea1=12, an=248 and d=4.We know that        an=a1+(n−1)d⇒248=12+(n−1)4⇒(n−1)4=248−12=236⇒n−1=2364=59⇒n=59+1=60Therefore, 60 multiples of 4 lie between 10 and 250.

Q.18 For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

Ans.

Let nth term of the given APs are equal.Let an and bn represent the terms of the first and secondAP respectively for all natural numbers n. Also, let d1 and d2 arecommon diffrences of first and second AP respectively.The given two APs are:63, 65, 67, . . . and 3, 10, 17, . . .We have an=bn⇒ a1+(n−1)d1=b1+(n−1)d2                          [an=a+(n−1)d]⇒63+(n−1)2=3+(n−1)7⇒(n−1)2−(n−1)7=3−63⇒−5(n−1)=−60⇒n−1=−60−5=12⇒n=12+1=13Therefore, 13th terms of the given APs are equal.

Q.19 Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Ans.

We have     a3=16 and a7=a5+12⇒a1+2d=16 and a7−a5=12                        [an=a+(n−1)d]⇒a1+2d=16 and a1+6d−a1−4d=12    ⇒a1+2d=16 and 2d=12    ⇒a1+2d=16 and d=6    ⇒a1+2×6=16 and d=6 ⇒a1=4 and d=6       Therefore, the required AP is: 4, 10, 16, 22, ...

Q.20 Find the 20th term from the last term of the AP: 3, 8, 13, . . ., 253.

Ans.

Here, a=3, d=8−3=5, l=253 where l=a+(n−1)dTo find the 20th term from the last term, we will find thetotal number of terms in the AP.So, 253=3+(n−1)5i.e., 253−3=(n−1)5i.e., n−1=2505=50or n=50+1=51So, there are 51 terms in the given AP. The 20th term from the last term will be the 32nd term(51-20+1) from the beginning.So, a32=3+31×5=3+155=158i.e., the 20th term from the last term is 158.

Q.21 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.

Ans.

Here,    a4+a8=24⇒a+3d+a+7d=24⇒2a+10d=24⇒a+5d=12                                             ...(1)Also,      a6+a10=44⇒a+5d+a+9d=44⇒2a+14d=44⇒a+7d=22                                            ...(2)We subtract (2) from (1) and get     2d=10⇒d=5On putting this value of d in (1), we get      a+5×5=12⇒a=12−25=−13First three terms of the AP are a, a+d and a+2d i.e.,−13, −13+5 and −13+2×5 i.e., −13, −8 and −3.

Q.22 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?

Ans.

Subba Rao’s annual salary in 1995 was ₹ 5000.He received an increment of ₹ 200 each year.So, Subba Rao’s annual salaries forms an AP witha=5000 and d=200.Let 7000 is nth term of this AP.We know that         an=a+(n-1)d⇒7000=5000+(n-1)200⇒(n-1)200=7000-5000⇒n-1=2000200=10⇒n=10+1=11Therefore, in 11th year, Subba Rao’s salary reach ₹ 7000.

Q.23 Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.

Ans.

₹  ₹ ..So, ‘s forms an AP witha=5 and d=..It is given that in nth week her weekly savings become‘ ..∴an=.We know that         an=a+(n−1)d⇒.=5+(n−1)×.⇒(n−1)×.=.−5⇒n−1=15.75.=9⇒n=9+1=10

Q.24

Find 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) 115 , 112 , 110 , . . ., to 11 terms.

Ans.

(i)Here, a=2,  d=a2−a1=7−2=5,  n=10We know that sum of first  n terms of an AP is given by            S=n2[2a+(n−1)d]So,    S=102[2×2+(10−1)5]=5[4+45]or S=245So, the sum of the first 10 terms of the given AP is 245.(ii) Here, a=−37,  d=a2−a1=−33−(−37)=4,  n=12We know that sum of first  n terms of an AP is given by            S=n2[2a+(n−1)d]So,    S=122[2×(−37)+(12−1)4]=6[−74+44]or S=−180So, the sum of the first 12 terms of the given AP is −180.(iii)Here, a=0.6,  d=a2−a1=1.7−0.6=1.1,  n=100We know that sum of first  n terms of an AP is given by            S=n2[2a+(n−1)d]So,    S=1002[2×0.6+(100−1)×1.1]=50[1.2+108.9]or S=5505So, the sum of the first 100 terms of the given AP is 5505.(iv) Here, a= 115,  d=a2−a1=112−115=160,  n=11We know that sum of first  n terms of an AP is given by            S=n2[2a+(n−1)d]So,    S=112[2×115+(11−1)×160]=112[215+16]or S=3320So, the sum of the first 11 terms of the given AP is 3320.

Q.25

Find the sums given below :(i) 7+1012+14+ . . . +84 (ii) 34+32+30+ . . . +10(iii) −5+(−8)+(−11)+. . .+(−230)

Ans.

(i) 7+1012+14+ . . . +84 Here, a=7,  d=a2−a1=1012−7=312=72,  l=84We know that         l=a+(n−1)dSo,  84=7+(n−1)72or 84−7=(n−1)72or 77×27+1=nor n=23We know that sum of first  n terms of an AP is given by            Sn=n2(a+l)So,    S23=232(7+84)=232×91or S23=1046.5So, the requird sum is 1046.5.(ii) 34+32+30+ . . . +10Here, a=34,  d=a2−a1=32−34=−2,  l=10We know that         l=a+(n−1)dSo,  10=34+(n−1)(−2)or 10−34=(n−1)(−2)or −24−2=n−1or n=13We know that sum of first  n terms of an AP is given by            Sn=n2(a+l)So,    S13=132(34+10)=132×44or S13=286So, the requird sum is 286.(iii) −5+(−8)+(−11)+. . .+(−230)Here, a=−5,  d=a2−a1=−8−(−5)=−3,  l=−230We know that         l=a+(n−1)dSo,  −230=−5+(n−1)(−3)or −230+5=(n−1)(−3)or −225−3+1=nor n=76We know that sum of first  n terms of an AP is given by            Sn=n2(a+l)So,    S75=762(−5−230)=38×(−235)or S75=−8930So, the requird sum is −8930.

Q.26

In an AP:(i) given a=5, d=3, an=50, find n and Sn.(ii) given a=7, a13=35, find d and S13.(iii) given a12=37, d=3, find a and S12.(iv) given a3=15, S10=125, find d and a10.(v) given d=5, S9=75, find a and a9.(vi) given a=2, d=8, Sn=90, find n and an.(vii) given a=8, an=62, Sn=210, find n and d.(viii) given an=4, d=2, Sn=−14, find n and a.(ix) given a=3, n=8, S=192, find d.(x) given l=28, S=144, and there are total 9 terms. Find a.

Ans.

(i) Here, a=5,  d=3,  an=50We have to find n and Sn.We know that         an=a+(n−1)dSo,  50=5+(n−1)3or 50−5=(n−1)3or 453+1=nor n=16We know that sum of first  n terms of an AP is given by            Sn=n2(a+an)So,    S16=162(5+50)=8×55or S16=440(ii) Here, a=7,   a13=35We have to find d and S13.We know that         an=a+(n−1)dSo,  a13=7+(13−1)dor 35−7=12dor 2812=dor d=73We know that sum of first  n terms of an AP is given by            Sn=n2(a+an)So,    S13=132(7+35)=132×42or S13=273(iii) Here,  a12=37, d=3We have to find a and S12.We know that         an=a+(n−1)dSo,  a12=a+(12−1)3or 37=a+33or    a=37−33=4We know that sum of first  n terms of an AP is given by            Sn=n2(a+an)So,    S12=122(a+a12)=6(4+37)or S12=246(iv) Here,  a3=15, S10=125We have to find d and a10.We know that         an=a+(n−1)dSo,  a3=a+(3−1)dor    a+2d=15                                                                            ...(1)We know that sum of first  n terms of an AP is given by            Sn=n2[2a+(n−1)d]So,    S10=102[2a+(10−1)d]=5[2a+9d]or 125=5[2a+9d]or 2a+9d=25                                                                      or      2(15−2d)+9d=25 [From (1), a=15−2d]or         5d=25−30=−5or d=−1 putting this value of d in (1), we get a=17.Now,             a10=a+(10−1)d=17+9×(−1)=8(v) Here,  d=5, S9=75We have to find a and a9.We know that         an=a+(n−1)dSo,  a9=a+(9−1)5or    a9−a=40                                                                            ...(1)We know that sum of first  n terms of an AP is given by            Sn=n2[2a+(n−1)d]So,    S9=92[2a+(9−1)5]=92[2a+40]or 75=9[a+20]or a=759−20=75−1809=−1059=−353or      a=−353On putting this value of a in (1), we get          a9−−353=40⇒ a9=  40−353=120−353=853(vi) Here,  a=2, d=8, Sn=90We have to find n and an.We know that         an=a+(n−1)dSo,  an=2+(n−1)8                                                                        ...(1)We know that sum of first  n terms of an AP is given by            Sn=n2[2a+(n−1)d]So,    Sn=n2[2×2+(n−1)8]or 90=2n+4n(n−1)=2n+4n2−4nor 90=4n2−2nor       2n2−n−45=0or 2n2−10n+9n−45=0or 2n(n−5)+9(n−5)=0or (2n+9)(n−5)=0or n=5On putting this value of n in (1), we get          a5=2+(5−1)8=34 (vii)Here,  a=8, an=62, Sn=210We have to find n and d.We know that         an=a+(n−1)dSo,  62=8+(n−1)d                                                                       ...(1)We know that sum of first  n terms of an AP is given by            Sn=n2(a+an)So,    210=n2(8+62)or 210=35nor n=21035=6On putting this value of n in (1), we get          62=8+(n−1)d =8+(6−1)d=8+5dor d=62−85 =545(viii) Here,  an=4, d=2, Sn=−14We have to find n and a.We know that         an=a+(n−1)dSo,  4=a+2(n−1)=a+2n−2  or    a+2n=6or    a=6−2n                                                                         ...(1)We know that sum of first  n terms of an AP is given by            Sn=n2(a+an)So, −14=n2(a+4)or n(a+4)=−28or n(6−2n+4)=−28                                             [From (1)]or 10n−2n2+28=0or n2−5n−14=0or n2−7n+2n−14=0or n(n−7)+2(n−7)=0or (n−7)(n+2)=0or n−7=0or n=7 [n is a natural number]On putting this value of n in (1), we get          a=6−2n  =6−2×7=−8(ix) Here,   a=3, n=8, S=192We know that sum of first  n terms of an AP is given by            S=n2[2a+(n−1)d]So,    192=82[2×3+(8−1)d]or 192=4[6+7d]or 6+7d=1924=48or 7d=48−6=42or d=427=6(x) Here,  l=28, S=144, n=9We have to find a.We know that sum of first  n terms of an AP is given by            S=n2(a+l)So,   144=92(a+28)or a+28= 144×29=16×2=32or a=32−28=4  

Q.27

How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Ans.

Here,   a=9, d=a2−a1=17−9=8, Sn=636We know that sum of first  n terms of an AP is given by            S=n2[2a+(n−1)d]So,   636=n2[2×9+(n−1)8]or 636=n2×2[9+4n−4]=4n2+5nor 4n2+5n−636=0or 4n2+53n−48n−636=0or n(4n+53)−12(4n+53)=0or (4n+53)(n−12)=0or n=12 [n is a natural number]Therefore, 12 terms of the given AP must be taken to give a sum of 636.

Q.28 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.

Ans.

Here,  a=5, an=l=45, Sn=400We have to find n and d.We know that         an=a+(n−1)dSo,  45=5+(n−1)d                                                                       ...(1)We know that sum of first  n terms of an AP is given by            Sn=n2(a+an)So,    400=n2(5+45)or 400=25nor n=40025=16On putting this value of n in (1), we get          45=5+(n−1)d =5+(16−1)d=5+15dor d=45−515 =4015=83

Q.29 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?

Ans.

Here,  a=17, an=l=350, d=9We have to find n and Sn.We know that         an=a+(n−1)dSo,  350=17+(n−1)9 or n−1=350−179=3339=37or n=37+1=38We know that sum of first  n terms of an AP is given by            Sn=n2(a+an)So,    S38=382(17+350)=19×367or S38=6973

Q.30 Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Ans.

Here,   a22=149, d=7We have to find S22.We know that         an=a+(n−1)dSo,  a22=a+(22−1)dor 149=a+21×7 =a+147or a=149−147=2We know that sum of first n terms of an AP is given by            Sn=n2(a+an)So,    S22=222(a+a22)=11(2+149)=11×151or S22=1661

Q.31 Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Ans.

Here,   a2=14, a3=18Therefore, d=a3−a2=18−14=4We know that         an=a+(n−1)dSo,  a2=a+(2−1)dor 14=a+1×4 =a+4or a=14−4=10We know that sum of first n terms of an AP is given by            Sn=n2[2a+(n−1)d]So,    S51=512[2×10+(51−1)4]=512(20+200)=51×110or S51=5610

Q.32 If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Ans.

Here,   S7=49, S17=289We know that sum of first n terms of an AP is given by            Sn=n2[2a+(n−1)d]So,    S7=72[2a+(7−1)d]or 49=72(2a+6d)or 497=22(a+3d)or a+3d=7 ...(1)Also,            S17=172[2a+(17−1)d]or 289=172(2a+16d)=172×2(a+8d)or a+8d=17or 7−3d+8d=17 [From (1), a=7−3d]or 5d=17−7=10or d=105=2On putting this value of d in (1), we get             a=7−3d=7−3×2=1Therefore, sum of first n terms of the AP with a=1 and d=2 isgiven by Sn=n2[2×1+(n−1)2]=n(1+n−1)=n2

Q.33 Show that a1, a2, . . ., an, . . . form an AP where an is defined as below :
(i) an = 3 + 4n (ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.

Ans.

(i)Here,          an=3+4n⇒an+1=3+4(n+1)=3+4n+4=7+4nNow,an+1−an=7+4n−3−4n=4i.e., difference between a term and its preceding term is always 4.Therefore, a1, a2, . . ., an, . . . form an AP where an=3+4n.Now, we haved=an+1−an=4anda=a1=3+4×1=7 We know that sum of first n terms of an AP is given by            Sn=n2[2a+(n−1)d]So,    S15=152[2×7+(15−1)4]or S15=152×2[7+14×2]=15×35=525(ii)Here,         an=9−5n⇒an+1=9−5(n+1)=9−5n−5=4−5nNow,an+1−an=4−5n−(9−5n)=4−5n−9+5n=−5i.e., difference between a term and its preceding term isalways −5.Therefore, a1, a2, . . ., an, . . . form an AP where an=9−5n.Now, we haved=an+1−an=−5anda=a1=9−5×1=4We know that sum of first n terms of an AP is given by            Sn=n2[2a+(n−1)d]So,    S15=152[2×4+(15−1)(−5)]or S15=152×2[4−5×7]=−31×15=−465

Q.34 If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Ans.

Here,          S=−n2⇒   S1=×1−12=3i.e., a=S1=3S2=×2−22=4Now,      a2= S2−S1=×2−22−3=8−4−3=1Similarly,a3= S3−S2=×3−32−×2−22     =12−9−8−4          =3−4=−1a10= S10−S9=×10−102−×9−92             =40−100−36−81=−60−−45                        =−60+45=−15an= Sn−Sn−1=×n−n2−×(n−1)−(n−1)2                            =n(4−n)−n−14−n+1                            =n(4−n)−n−15−n                            =4n−n2−5n+n2+5−n                           =−2n+5       =5−2n

Q.35 Find the sum of the first 40 positive integers divisible by 6.

Ans.

First 40 positive integers divisible by 6 forms an AP with a=6 and d=6.We know that sum of first n terms of an AP is given by         Sn=n2[2a+(n−1)d]So, S40=402[2×6+(40−1)6]=20[12+39×6]or S40=20[12+234]=20×246=4920

Q.36 Find the sum of the first 15 multiples of 8.

Ans.

First 15 multiples of 8 forms an AP with a=8 and d=8.We know that sum of first n terms of an AP is given by         Sn=n2[2a+(n−1)d]So, S15=152[2×8+(15−1)8]=152×8[2+14]or S15=60×16=960

Q.37 Find the sum of the odd numbers between 0 and 50.

Ans.

Odd numbers between 0 and 50 forms an AP with a=1 and d=2.Also, last term of this AP is 49.Therefore,    l=an=49⇒a+(n−1)d=49⇒1+(n−1)2=49⇒n−1=49−12=24⇒n=24+1=25We know that sum of first n terms of an AP is given by         Sn=n2[2a+(n−1)d]So, S25=252[2×1+(25−1)2]=252×2[1+24]or S25=25×25=625

Q.38 A 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?

Ans.

The given 30 penalties forms an AP with first term as 200and common difference as 50.So, we have a=200, d=50 and n=30.We know that sum of first n terms of an AP is given by         Sn=n2[2a+(n−1)d]So, S30=302[2×200+(30−1)50]=15[400+29×50]or S30=15[400+1450]=1850×15=27,750Therefore, the contractor has to pay ₹ 27,750 as penalty.

Q.39 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.

Ans.

The given 7 cash prizes forms an AP with common differenceas −20 and sum of all the 7 prizes as 700.So, we have S7=700, d=−20 and n=7.We know that sum of first n terms of an AP is given by         Sn=n2[2a+(n−1)d]So, S7=72[2a+(7−1)(−20)]=72×2[a−60]or 700=7[a−60]or a−60=7007=100or a=100+60=160Therefore, the values of all the 7 prizes are ₹ 160, ₹ 140, ₹ 120,₹ 100, ₹ 80, ₹ 60 and ₹ 40.

Q.40 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?

Ans.

It is given that a section of Class I will plant 1 tree, a sectionof Class II will plant 2 trees and so on till Class XII. The number of trees planted by a section of each classforms an AP with a=1 and d=1.So, we have a=1, d=1 and n=12.We know that sum of first n terms of an AP is given by         Sn=n2[2a+(n−1)d]So, S12=122[2×1+(12−1)1]=6[2+11]or S12=78Therefore,       Sum of trees planted by a section of each class=78⇒ Sum of trees planted by 3 sections of each class=78×3                                                                                                            =234

Q.41 A 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, . . . as shown in the following figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π =22/7)

[Hint: Length of successive semicircles is l1, l2, l3, l4, . . . with centres at A, B, A, B, . . ., respectively.]

Ans.

Here, successive semicircles are of raddi 0.5 cm, 1.0 cm,1.5 cm, 2.0 cm, .… Let l1, l2, l3, l4, . . . are the lengths of successive semicirclesin the spiral.We know that length of semicircle is πr.So, lengths of successive semicircles are 0.5π cm, π cm,1.5π cm, 2π cm, .… Sum of lengths of 13 semicircles is π(0.5+1.0+1.5+2.0+ ...to 13th terms) cm.Now,List of numbers involved in sum 0.5+1.0+1.5+2.0+ ...to 13th terms forms an AP with a=0.5 and d=0.5We know that sum of first n terms of an AP is given by         Sn=n2[2a+(n−1)d]So, S13=132[2×0.5+(13−1)(0.5)]=132×0.5[2+12]or S13=132×0.5×14=45.5Therefore, Sum of lengths of 13 semicircles=π(0.5+1.0+1.5+2.0+...to 13th terms) cm=45.5π cm=45.5×227 cm=143 cmTherefore, total length of the spiral is 143 cm.

Q.42 200 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 (see the following figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Ans.

It is given that 200 logs are stacked in such a way that20 logs are in the bottom row, 19 logs are in the next row,18 logs are in the row next to it and so on.List of numbers 20, 19, 18, ... involved in stacking of logsform an AP with a=20, d=−1 and Sn=200.We know that sum of first n terms of an AP is given by         Sn=n2[2a+(n−1)d]So, 200=n2[2×20+(n−1)(−1)]=n2[40+−n+1]or 200=n2[41−n]or 400=41n−n2or n2−41n+400=0or n2−16n−25n+400=0or n(n−16)−25(n−16)=0or (n−16)(n−25)=0or n=16, n=25Now,a25=a+24d=20+24×(−1)=−4But, number of logs can not be negative.Therefore, number of rows of logs can not be 25 i.e., n≠25.Also,a16=a+15d=20+15×(−1)=5Therefore, 200 logs are placed in 16 rows and 5 logs areplaced in the top row.

Q.43 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(see the following figure)

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?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]

Ans.

To pick up the first potato and the second potato, the totaldistance in metres run by a competitor is 2×5+2×5 + 3.Similarly, the total distance in metres run by a competitorto pick up 10 potatoes is 2×5+2×5+3+2×5+3+3+...up to10th terms.Now,Total distance = 2×5+2×5+3+2×5+3+3+...up to  10th terms.      = 25+8+11+...up to 10th terms                               = 2×1022×5+(10-1)3               Sn=n2[2a+(n-1)d]                               = 1010+27                               = 370 m

Q.44 Which term of the AP: 121, 117, 113, . . ., is its first negative term?
[Hint: Find n for an < 0]

Ans.

The given AP is: 121, 117, 113, . . .Here, a=121, d=117−121=−4Now,an=a+(n−1) d =121+(n−1)(−4)      = 121−4n+4      = 125−4nWe have to find the first negative term of this A.P.So,     an<0⇒125−4n<0⇒4n>125⇒n>1254⇒n>31.25∴n=32Hence, 32nd term of the given AP is first negative term.

Q.45 The 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.

Ans.

We have a3+a7=6, a3×a7=8We know that nth terms of an AP is given by         an=a+(n−1)dSo,          a3=a+2d and a7=a+6dNow,         a3+a7=6⇒    a+2d+a+6d=6⇒ 2a+8d=6⇒    a+4d=3⇒    a=3−4d ...(1)Again,         a3×a7=8⇒ (a+2d)×(a+6d)=8⇒   (3−4d+2d)(3−4d+6d)=8 [From (1)]⇒     (3−2d)(3+2d)=8⇒      9−4d2=8⇒      4d2=1⇒       d=12 or −12On putting this value of d in (1), we geta=1, when d=12anda=5, when d=−12When a=1 and d=12 then,S16=162[2×1+15×12]=76When a=5 and d=−12 then,S16=162[2×5+15×(−12)]=20

Q.46

A ladder has rungs 25 cm apart. see the following figure. The rungs decrease uniformly in length from 45 cm at thebottom to 25 cm at the top. If the top and the bottomrungs are 212 m apart, what is the length of the woodrequired for the rungs?Hint:Number of rungs=25025

Ans.

It is given that the rungs are 25 cm apart and the top andbottom rungs are 12m apart.∴Total number of rungs=212×10025+1=25025+1=11 Lengths of the rungs form an AP with a=45 and a11=l=25.Sn=n2a+l=11245+25=112×70=385 Therefore, the length of the wood required for the rungs is 385 cm.

Q.47 The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint: Sx–1 = S49 – Sx]

Ans.

The houses are numbered consecutively as 1, 2, 3, ..., 49.It can be observed that the number of houses are in an APhaving a as 1 and d also as 1.Let there is a number x in this AP such that the sum of thenumbers of the houses preceding the house numbered x isequal to the sum of the numbers of the houses following it.We know that,Sum of first n terms of an A.P is given by     Sn=n2[2a+(n−1)d]∴ Sum of number of houses preceding xth house           =Sx−1          =x−12[2×1+(x−1−1)1]         =x−12[2+(x−2)]         =x(x−1)2Sum of number of houses following xth house =S49−Sx          =492[2+(49−1)1]−x2[2+(x−1)1]          =49×25−x2(1+x)It is given that               Sx−1=S49−Sx⇒      x(x−1)2=49×25−x2(1+x)⇒     x2−x=2450−x−x2⇒     2x2=2450⇒      x2=1225⇒      x=±35 But, number of houses can not be negative.∴x=35

Q.48

A small terrace at a football ground comprises of 15 steps each of which is 50 m long and  built of solidconcrete. Each step has a rise of 14m and a tread of12m. see the following figure. Calculate the totalvolume of concrete required to build the terrace.[Hint: Volume of concrete required to build the first step= 14×12×50 m3]

Ans.

According to question, widths of successive steps in meterfrom top are 12, 1, 32, 2,  ... which form an AP with a=d=12.Each step is 50 m long and  has a rise of 14m. Now,Volume of concrete required to build the 1st step =14×12×50 m3=254 m3Volume of concrete required to build the 2nd step =14×1×50 m3=252 m3Volume of concrete required to build the 3rd step =14×32×50 m3=754 m3List of numbers 254, 252, 754,  ... form an AP with a=254 andd=252−254=254.We know that       Sn=n2[2a+(n−1)d]So,   S15=152[2×254+(15−1)254]or S15=152[252+14×254]=152[252+1752]=152[100]=750Therefore,Volume of concrete required to build the terrace is 750 m3.

Q.49

In which of the following situations, does the list ofnumbers 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 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.

Ans.

(i) Fare for the first km       =  ₹15Fare for the second km =  ₹15 +  ₹8 =  ₹23Fare for the third km =  ₹15 + ₹8 +  ₹8 =  ₹31and so on.Therefore, fare for each successive kilometre in rupeesare:15, 23, 31, 39, ...Difference between fares for a particular kilometre and itspreceding kilometre is ‘8.Therefore, 15, 23, 31, 39, ... forms an AP.(ii) Let the amount of air present in the cylinder initially is V.The vaccum pump removes 14 of air remaining in thecylinder each time.Therefore, amounts of air in the cylinder after eachsuccessive removal by the vacuum pump will beV, 3V4, 3V4−3V4×14, 3V4−3V4×14−3V4−3V4×14×14, ...i.e., V, 3V4, 342V, 343V, ...Here, the differences between a term and its preceding termare not equal. Therefore, list of numbers involved doesnot make an A.P.(iii)According to question, cost in rupees of digging firstmetre and thereafter each subsequent metre arerespectively:150, 150+50, 150+50+50, 150+50+50+50, ...150, 200, 250, 300, ...Differences between a term and its preceding term in the above list of numbers are equal.Therefore, 150, 200, 250, 300, ...forms an AP.(iv)Amount at the end of first year=100001+81001= ₹10800Amount at the end of second year=108001+81001= ₹11664Amount at the end of third year=116641+81001= ₹12597.12Obviously, differences between amounts in any two successiveyears are not equal. Therefore, list of numbers involved heredoes not make an AP.

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NCERT Solutions for Class 10 Maths Related Chapters

Real Numbers

Polynomials

Pair of Linear Equations in Two Variables

Quadratic Equations

Triangles

Coordinate Geometry

Introduction to Trigonometry

Some Applications of Trigonometry

Circles

Constructions

Areas Related to Circles

Surface Areas and Volumes

Statistics

Probability

FAQs (Frequently Asked Questions)

1. What is the best way to find the Sum of an arithmetic progression?

 To find the Sum of an arithmetic progression, you’ll need to know the first term’s value, the number of terms, and the common difference between each term. To arrive at the final solution, use the formula given below.

S = (n / 2)*(2a+(n−1)d)

How many types of progressions are there in Mathematics?

Ans: In Mathematics, there are three types of progressions. They are: s:

  • Arithmetic progression (A.P.)
  • Geometric progression (G.P.)
  • Harmonic progression (H.P.)