# Arithmetic Progressions Chapter 5 NCERT Solutions For Class 10 CBSE Mathematics (Exercise 3 and 4)

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## Solutions:

(i) Given, 2, 7, 12 ,…, to 10 terms For this A.P.,

first term, a = 2 And common difference, d = a2 − a1 = 7−2 = 5

n = 10 We know that, the formula for sum of nth term in AP series is, Sn = n/2 [2a +(n-1)d]

S10 = 10/2 [2(2)+(10 -1)×5] = 5[4+(9)×(5)] = 5 × 49 = 245

(ii) Given, −37, −33, −29 ,…, to 12 terms

For this A.P., first term, a = −37 And common difference, d = a2− a1 d= (−33)−(−37) = − 33 + 37 = 4 n = 12

We know that, the formula for sum of nth term in AP series is, Sn = n/2 [2a+(n-1)d]

S12 = 12/2 [2(-37)+(12-1)×4] = 6[-74+11×4] = 6[-74+44] = 6(-30) = -180

(iii) Given, 0.6, 1.7, 2.8 ,…, to 100 terms

For this A.P., first term, a = 0.6 Common difference, d = a2 − a1 = 1.7 − 0.6 = 1.1 n = 100

We know that, the formula for sum of nth term in AP series is, Sn = n/2[2a +(n-1)d]

S12 = 50/2 [1.2+(99)×1.1] = 50[1.2+108.9] = 50[110.1] = 5505

(iv) Given, 1/15, 1/12, 1/10, …… , to 11 terms

For this A.P., First term, a = 1/5 Common difference, d = a2 –a1 = (1/12)-(1/5) = 1/60 And number of terms n = 11 We know that, the formula for sum of nth term in AP series is,

Sn = n/2 [2a + (n – 1) d]

= 11/2(2/15 + 10/60)

= 11/2 (9/30) = 33/20

## Solutions:

(i) First term, a = 7 n th term, an = 84

Let 84 be the n th term of this A.P.,

then as per the nth term formula,

an = a(n-1)d

84 = 7+(n – 1)×7/2

77 = (n-1)×7/2

22 = n−1 n = 23

We know that, sum of n term is; Sn = n/2 (a + l) , l = 84

Sn = 23/2 (7+84)

Sn = (23×91/2) = 2093/2

(ii) Given, 34 + 32 + 30 + ……….. + 10

For this A.P., first term, a = 34 common difference, d = a2−a1 = 32−34 = −2 n th term, an= 10

Let 10 be the n th term of this A.P., therefore,

an= a +(n−1)d

10 = 34+(n−1)(−2)

−24 = (n −1)(−2)

12 = n −1

n = 13

We know that, sum of n terms is;

Sn = n/2 (a +l) , l = 10 = 13/2 (34 + 10) = (13×44/2) = 13 × 22 = 286

(iii) Given, (−5) + (−8) + (−11) + ………… + (−230)

For this A.P., First term, a = −5 nth term, an= −230 Common difference, d = a2−a1 = (−8)−(−5) ⇒d = − 8+5 = −3

Let −230 be the n th term of this A.P.,

and by the nth term formula we know,

### YOU ARE READING: Arithmetic Progressions Chapter 5 NCERT Solutions For Class 10 CBSE Mathematics

an= a+(n−1)d −230 = − 5+(n−1)(−3)

−225 = (n−1)(−3)

(n−1) = 75

n = 76

And, Sum of n term,

Sn = n/2 (a + l) = 76/2 [(-5) + (-230)] = 38(-235) = -8930

## Solutions:

(i) Given that, a = 5, d = 3, an = 50

As we know, from the formula of the nth term in an AP, an = a +(n −1)d,

Therefore, putting the given values, we get,

⇒ 50 = 5+(n -1)×3

⇒ 3(n -1) = 45

⇒ n -1 = 15

⇒ n = 16

Now, sum of n terms, Sn = n/2 (a +an)

Sn = 16/2 (5 + 50) = 440

(ii) Given that, a = 7, a13 = 35

As we know, from the formula of the nth term in an AP, an = a+(n−1)d,

Therefore, putting the given values, we get,

⇒ 35 = 7+(13-1)d

⇒ 12d = 28

⇒ d = 28/12 = 2.33

Now, Sn = n/2 (a+an)

S13 = 13/2 (7+35) = 273

(iii) Given that, a12 = 37, d = 3

As we know, from the formula of the nth term in an AP, an = a+(n −1)d,

Therefore, putting the given values, we get,

⇒ a12 = a+(12−1)3

⇒ 37 = a+33

⇒ a = 4

Now, sum of n terms,

Sn = n/2 (a+an)

Sn = 12/2 (4+37) = 246

### YOU ARE READING: Arithmetic Progressions Chapter 5 NCERT Solutions For Class 10 CBSE Mathematics

(iv) Given that, a3 = 15, S10 = 125

As we know, from the formula of the nth term in an AP, an = a +(n−1)d,

Therefore, putting the given values, we get,

a3 = a+(3−1)d 15 = a+2d ………………………….. (i)

Sum of the nth term, Sn = n/2 [2a+(n-1)d]

S10 = 10/2 [2a+(10-1)d]

125 = 5(2a+9d)

25 = 2a+9d ……………………….. (ii)

On multiplying equation (i) by (ii), we will get;

30 = 2a+4d ………………………………. (iii)

By subtracting equation (iii) from (ii), we get,

−5 = 5d

d = −1

From equation (i),

15 = a+2(−1)

15 = a−2

a = 17 = First term a10 = a+(10−1)d

a10 = 17+(9)(−1)

a10 = 17−9 = 8

(v) Given that, d = 5, S9 = 75

As, sum of n terms in AP is, Sn = n/2 [2a +(n -1)d]

Therefore, the sum of first nine terms are;

S9 = 9/2 [2a +(9-1)5]

25 = 3(a+20)

25 = 3a+60

3a = 25−60

a = -35/3

As we know, the nth term can be written as;

an = a+(n−1)d

a9 = a+(9−1)(5)

= -35/3+8(5)

= -35/3+40

= (35+120/3)

= 85/3

### YOU ARE READING: Arithmetic Progressions Chapter 5 NCERT Solutions For Class 10 CBSE Mathematics

(vii) Given that, a = 8, an = 62, Sn = 210

As, sum of n terms in an AP is, Sn = n/2 (a + an)

210 = n/2 (8 +62)

⇒ 35n = 210

⇒ n = 210/35 = 6

Now, 62 = 8+5d

⇒ 5d = 62-8 = 54

⇒ d = 54/5 = 10.8

(viii) Given that, n th term, an = 4, common difference, d = 2, sum of n terms, Sn = −14.

As we know, from the formula of the nth term in an AP, an = a+(n −1)d,

Therefore, putting the given values, we get,

4 = a+(n −1)2

4 = a+2n−2 a+2n = 6

a = 6 − 2n …………………………………………. (i)

As we know, the sum of n terms is; Sn = n/2 (a+an)

-14 = n/2 (a+4)

−28 = n (a+4)

−28 = n (6 −2n +4)

{From equation (i)}

−28 = n (− 2n +10)

n (n−7)+2(n −7) = 0

(n −7)(n +2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

However, n can neither be negative nor fractional.

Therefore, n = 7 From equation (i), we get

a = 6−2n

a = 6−2(7)

= 6−14 = −8

(ix) Given that, first term, a = 3,

Number of terms, n = 8 And sum of n terms, S = 192

As we know, Sn = n/2 [2a+(n -1)d]

192 = 8/2 [2×3+(8 -1)d]

192 = 4[6 +7d]

48 = 6+7d

42 = 7d

d = 6

### YOU ARE READING: Arithmetic Progressions Chapter 5 NCERT Solutions For Class 10 CBSE Mathematics

(x) Given that, l = 28,S = 144 and there are total of 9 terms.

Sum of n terms formula, Sn = n/2 (a + l)

144 = 9/2(a+28)

(16)×(2) = a+28

32 = a+28

a = 4

## Solutions:

Let there be n terms of the AP. 9, 17, 25 …

For this A.P., First term, a = 9 Common difference, d = a2−a1 = 17−9 = 8

As, the sum of n terms, is;

Sn = n/2 [2a+(n -1)d]

636 = n/2 [2×a+(8-1)×8]

636 = n/2 [18+(n-1)×8]

636 = n [9 +4n −4]

636 = n (4n +5)

4n2 +5n −636 = 0

4n2+53n −48n −636 = 0

n (4n + 53)−12 (4n + 53) = 0

(4n +53)(n −12) = 0

Either 4n+53 = 0 or n−12 = 0

n = (-53/4) or n = 12

n cannot be negative or fraction,

therefore, n = 12 only.

## Solution:

Given that, first term, a = 5 last term, l = 45 Sum of the AP, Sn = 400

As we know, the sum of AP formula is;

Sn = n/2 (a+l)

400 = n/2(5+45)

400 = n/2(50)

Number of terms, n =16

### YOU ARE READING: Arithmetic Progressions Chapter 5 NCERT Solutions For Class 10 CBSE Mathematics

As we know, the last term of AP series can be written as;

l = a+(n −1)d

45 = 5 +(16 −1)d

40 = 15d

Common difference, d = 40/15 = 8/3

## Solution:

Given that, First term, a = 17 Last term, l = 350 Common difference, d = 9

Let there be n terms in the A.P., thus the formula for last term can be written as;

l = a+(n −1)d

350 = 17+(n −1)9

333 = (n−1)9

(n−1) = 37

n = 38

Sn = n/2 (a+l)

S38= 13/2 (17+350) = 19×367 = 6973

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

## Solution:

Given, Common difference, d = 7

22nd term,

a22= 149

Sum of first 22 term, S22= ?

By the formula of nth term, an = a+(n−1)d

a22 = a+(22−1)d

149 = a+21×7

149 = a+147

a = 2 = First term

Sum of n terms, Sn = n/2(a+an)

S22 = 22/2 (2+149) = 11×151 = 1661

## Solution:

Given that, Second term, a2 = 14 Third term, a3 = 18 Common difference, d = a3−a2 = 18−14 = 4

a2 = a+d

14 = a+4

a = 10 = First term

Sum of n terms;

Sn = n/2 [2a + (n – 1)d]

S51= 51/2 [2×10 (51-1) 4]

= 51/2 [2+(20)×4]

= 51 × 220/2

= 51 × 110

= 5610

## Solution:

Given that, S7= 49

S17= 289

We know, Sum of nth term;

Sn = n/2 [2a + (n – 1)d]

Therefore,

### YOU ARE READING: Arithmetic Progressions Chapter 5 NCERT Solutions For Class 10 CBSE Mathematics

S7 = 7/2 [2a +(n -1)d]

S7 = 7/2 [2a + (7 -1)d]

49 = 7/2 [2a +16d]

7 = (a+3d)

a + 3d = 7 …………………………………. (i)

In the same way,

S17 = 17/2 [2a+(17-1)d]

289 = 17/2 (2a +16d)

17 = (a+8d)

a +8d = 17 ………………………………. (ii)

Subtracting equation (i) from equation (ii),

5d = 10

d = 2

From equation (i), we can write it as;

a+3(2) = 7

a+ 6 = 7

a = 1

Hence, Sn = n/2[2a+(n-1)d] = n/2[2(1)+(n – 1)×2] = n/2(2+2n-2) = n/2(2n) = n2

## Solutions:

(i) an = 3+4n

a1 = 3+4(1) = 7

a2 = 3+4(2) = 3+8 = 11

a3 = 3+4(3) = 3+12 = 15

a4 = 3+4(4) = 3+16 = 19

We can see here, the common difference between the terms are; a2 − a1 = 11−7 = 4

a3 − a2 = 15−11 = 4

a4 − a3 = 19−15 = 4

Hence, ak + 1 − ak is the same value every time.

Therefore, this is an AP with common difference as 4 and first term as 7.

Now, we know, the sum of nth term is; Sn = n/2[2a+(n -1)d]

S15 = 15/2[2(7)+(15-1)×4] = 15/2[(14)+56] = 15/2(70) = 15×35 = 525

(ii) an = 9−5n

a1 = 9−5×1 = 9−5 = 4

a2 = 9−5×2 = 9−10 = −1

a3 = 9−5×3 = 9−15 = −6

a4 = 9−5×4 = 9−20 = −11

We can see here, the common difference between the terms are;

a2 − a1 = −1−4 = −5

a3 − a2 = −6−(−1) = −5

a4 − a3 = −11−(−6) = −5

Hence, ak + 1 − ak is same every time.

Therefore, this is an A.P. with common difference as −5 and first term as 4.

### YOU ARE READING: Arithmetic Progressions Chapter 5 NCERT Solutions For Class 10 CBSE Mathematics

Now, we know, the sum of nth term is; Sn = n/2 [2a +(n-1)d]

S15 = 15/2[2(4) +(15 -1)(-5)]

= 15/2[8 +14(-5)]

= 15/2(8-70)

= 15/2(-62)

= 15(-31) = -465

## Solution:

Given that, Sn = 4n− n2

First term, a = S1 = 4(1) − (1)2 = 4−1 = 3

Sum of first two terms = S2= 4(2)−(2)2 = 8−4 = 4

Second term, a2 = S2 − S1 = 4−3 = 1

Common difference, d = a2−a = 1−3 = −2

N th term, an = a+(n−1)d = 3+(n −1)(−2) = 3−2n +2 = 5−2n

Therefore, a3 = 5−2(3) = 5-6 = −1

a10 = 5−2(10) = 5−20 = −15

Hence, the sum of first two terms is 4. The second term is 1.

The 3 rd , the 10th, and the n th terms are −1, −15, and 5 − 2n respectively.

## Solution:

The positive integers that are divisible by 6 are 6, 12, 18, 24 ….

We can see here, that this series forms an A.P. whose first term is 6 and common difference is 6.

a = 6 d = 6 S40 = ?

By the formula of sum of n terms, we know, Sn = n/2 [2a +(n – 1)d]

Therefore, putting n = 40, we get, S40 = 40/2 [2(6)+(40-1)6] = 20[12+(39)(6)] = 20(12+234) = 20×246 = 4920

## Solution:

The multiples of 8 are 8, 16, 24, 32…

The series is in the form of AP, having first term as 8 and common difference as 8.

Therefore, a = 8 d = 8

S15 = ?

By the formula of sum of nth term, we know,

Sn = n/2 [2a+(n-1)d]

S15 = 15/2 [2(8) + (15-1)8] = 15/2[6 +(14)(8)] = 15/2[16 +112] = 15(128)/2 = 15 × 64 = 960

## Solution:

The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49.

Therefore, we can see that these odd numbers are in the form of A.P.

Hence, First term, a = 1 Common difference, d = 2 Last term, l = 49

By the formula of last term, we know,

l = a+(n−1) d

49 = 1+(n−1)2

48 = 2(n − 1)

n − 1 = 24

n = 25 = Number of terms

By the formula of sum of nth term, we know,

Sn = n/2(a +l) S25 = 25/2 (1+49) = 25(50)/2 =(25)(25) = 625

## Solution:

We can see, that the given penalties are in the form of A.P. having first term as 200 and common difference as 50. Therefore, a = 200 and d = 50

Penalty that has to be paid if contractor has delayed the work by 30 days = S30

By the formula of sum of nth term, we know,

Sn = n/2[2a+(n -1)d]

Therefore, S30= 30/2[2(200)+(30 – 1)50] = 15[400+1450] = 15(1850) = 27750

Therefore, the contractor has to pay Rs 27750 as penalty.

## Solution:

Let the cost of 1st prize be Rs. P. Cost of 2nd prize = Rs. P − 20 And cost of 3rd prize = Rs. P − 40

We can see that the cost of these prizes are in the form of A.P., having common difference as −20 and first term as P.

Thus, a = P and d = −20

Given that, S7 = 700

By the formula of sum of nth term, we know,

Sn = n/2 [2a + (n – 1)d]

7/2 [2a + (7 – 1)d] = 700

a + 3(−20) = 100

a −60 = 100

a = 160

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

## Solution:

It can be observed that the number of trees planted by the students is in an AP. 1, 2, 3, 4, 5………………..12 First term, a = 1 Common difference, d = 2−1 = 1

Sn = n/2 [2a +(n-1)d]

S12 = 12/2 [2(1)+(12-1)(1)] = 6(2+11) = 6(13) = 78

Therefore, number of trees planted by 1 section of the classes = 78

Number of trees planted by 3 sections of the classes = 3×78 = 234

Therefore, 234 trees will be planted by the students.

## Solution:

We know, Perimeter of a semi-circle = πr

Therefore, P1 = π(0.5) = π/2 cm

P2 = π(1) = π cm

P3 = π(1.5) = 3π/2 cm

Where, P1, P2, P3 are the lengths of the semi-circles.

Hence we got a series here, as, π/2, π, 3π/2, 2π, ….

P1 = π/2 cm

P2 = π cm Common difference, d = P2 – P1 = π – π/2 = π/2

First term = P1= a = π/2 cm

By the sum of n term formula, we know, Sn = n/2 [2a + (n – 1)d]

Therefore, Sum of the length of 13 consecutive circles is;

S13 = 13/2 [2(π/2) + (13 – 1)π/2] = 13/2 [π + 6π] =13/2 (7π) = 13/2 × 7 × 22/7 = 143 cm

## Solution:

We can see that the numbers of logs in rows are in the form of an A.P.20, 19, 18…

For the given A.P., First term, a = 20 and common difference, d = a2−a1 = 19−20 = −1

Let a total of 200 logs be placed in n rows.

Thus, Sn = 200 By the sum of nth term formula,

Sn = n/2 [2a +(n -1)d]

S12 = 12/2 [2(20)+(n -1)(-1)]

400 = n (40−n+1)

400 = n (41-n)

400 = 41n−n2

n2 −41n + 400 = 0

n2 −16n−25n+400 = 0

n(n −16)−25(n −16) = 0

(n −16)(n −25) = 0

Either (n −16) = 0 or n−25 = 0 n = 16 or n = 25 By the nth term formula,

an = a+(n−1)d

a16 = 20+(16−1)(−1)

a16 = 20−15

a16 = 5

### YOU ARE READING: Arithmetic Progressions Chapter 5 NCERT Solutions For Class 10 CBSE Mathematics

Similarly, the 25th term could be written as;

a25 = 20+(25−1)(−1)

a25 = 20−24 = −4

It can be seen, the number of logs in 16th row is 5 as the numbers cannot be negative.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.

## Solution:

The distances of potatoes from the bucket are 5, 8, 11, 14…,

which is in the form of AP. Given, the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.

Therefore, distances to be run w.r.t distances of potatoes, could be written as; 10, 16, 22, 28, 34,……….

Hence, the first term, a = 10 and d = 16−10 = 6 S10 =?

By the formula of sum of n terms, we know,

S10 = 12/2 [2(20)+(n -1)(-1)] = 5[20+54] = 5(74) = 370

Therefore, the competitor will run a total distance of 370 m.

## Solution:

Given the AP series is 121, 117, 113, . . .,

Thus, first term, a = 121

Common difference, d = 117-121= -4

By the nth term formula,

an = a+(n −1)d

Therefore, an = 121+(n−1)(-4) = 121-4n+4 =125-4n

To find the first negative term of the series, an < 0

Therefore, 125-4n < 0 125 < 4n

n>125/4 n>31.25

Therefore, the first negative term of the series is 32nd term.

Arithmetic Progressions

## Solution:

From the given statements, we can write,

a3 + a7 = 6 …………………………….(i)

And a3 ×a7 = 8 ……………………………..(ii)

By the nth term formula, an = a+(n−1)d

Third term, a3 = a+(3 -1)d

a3 = a + 2d………………………………(iii)

And Seventh term, a7= a+(7-1)d

a7 = a + 6d ………………………………..(iv)

From equation (iii) and (iv), putting in equation(i), we get,

a+2d +a+6d = 6

2a+8d = 6

a+4d=3 or a = 3–4d …………………………………(v)

Again putting the eq.(iii) and (iv), in eq. (ii), we get,

(a+2d)×(a+6d) = 8

Putting the value of a from equation (v), we get,

(3–4d +2d)×(3–4d+6d) = 8 (3 –2d)×(3+2d) = 8

32– 2d2= 8

9 – 4d2 = 8

4d2 = 1

d = 1/2 or -½

Now, by putting both the values of d, we get,

a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1,

when d = ½

a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5,

when d = -1/2 We know, the sum of nth term of AP is;

### YOU ARE READING: Arithmetic Progressions Chapter 5 NCERT Solutions For Class 10 CBSE Mathematics

Sn = n/2 [2a +(n – 1)d] So, when a = 1 and d=1/2 Then, the sum of first 16 terms are;

S16 = 16/2 [2 +(16-1)1/2] = 8(2+15/2) = 76

And when a = 5 and d= -1/2 Then, the sum of first 16 terms are;

S16 = 16/2 [2.5+(16-1)(-1/2)] = 8(5/2)=20

Arithmetic Progressions

## Solution:

Given, Distance between the rungs of the ladder is 25cm.

Distance between the top rung and bottom rung of the ladder is =  5/2 ×100cm = 250cm

Therefore, total number of rungs = 250/25 + 1 = 11

As we can see from the figure, the ladder has rungs in decreasing order from top to bottom.

Thus, we can conclude now, that the rungs are decreasing in an order of AP.

And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.

So, First term, a = 45 Last term, l = 25

Number of terms, n = 11

Now, as we know, sum of nth terms is equal to,

Sn= n/2(a+ l)

Sn= 11/2(45+25) = 11/2(70) = 385 cm

Hence, the length of the wood required for the rungs is 385cm.

Arithmetic Progressions

## Solution:

Given, Row houses are numbers from 1,2,3,4,5…….49.

Thus we can see the houses numbered in a row are in the form of AP.

So, First term, a = 1 Common difference, d=1

Let us say the number of xth houses can be represented as;

Sum of nth term of AP = n/2[2a+(n-1)d]

Sum of number of houses beyond x house = Sx – 1 = (x-1)/2[2.1+(x-1-1)1]

= (x-1)/2 [2+x-2] = x(x-1)/2 ………………………………………(i)

By the given condition, we can write,

S49 – Sx = {49/2[2.1+(49-1)1]}–{x/2[2.1+(x-1)1]} = 25(49) – x(x + 1)/2 ………………………………….(ii)

As per the given condition, eq.(i) and eq(ii) are equal to each other;

Therefore, x(x-1)/2 = 25(49) – x(x-1)/2 x = ±35

As we know, the number of houses cannot be a negative number.

Hence, the value of x is 35.

## Solution:

the first step is ½ m wide, 2nd step is 1m wide and 3rd step is 3/2m wide.

Thus we can understand that the width of step by ½ m each time when height is ¼ m.

And also, given length of the steps is 50m all the time.

So, the width of steps forms a series AP in such a way that;

½ , 1, 3/2, 2, ……..

Arithmetic Progressions

### YOU ARE READING: Arithmetic Progressions Chapter 5 NCERT Solutions For Class 10 CBSE Mathematics

Volume of steps = Volume of Cuboid = Length × Breadth × Height

Now, Volume of concrete required to build the first step = ¼ ×1/2 ×50 = 25/4

Volume of concrete required to build the second step =¼ ×1/×50 = 25/2

Volume of concrete required to build the second step = ¼ ×3/2 ×50 = 75/2

Now, we can see the volumes of concrete required to build the steps, are in AP series; 25/4 , 25/2 , 75/2 …..

Thus, applying the AP series concept,

First term, a = 25/4

Common difference, d = 25/2 – 25/4 = 25/4

As we know, the sum of n terms is;

Sn = n/2[2a+(n-1)d] = 15/2(2×(25/4 )+(15/2 -1)25/4)

Upon solving, we get,

Sn = 15/2 (100)

Sn= 750

Hence, the total volume of concrete required to build the terrace is 750 m3.

Arithmetic Progressions