# RS Aggarwal Class 10 Solutions Chapter 11 - Arithmetic Progressions Ex 11C (11.3)

## RS Aggarwal Class 10 Chapter 11 - Arithmetic Progressions Ex 11C (11.3) Solutions Free PDF

Q.1: The sum of three consecutive terms of an AP is 21 and the sum of the squares of those terms is 165. Find the terms.

Sol:

Let the required numbers be (a – d) , a , Â (a + d)

Sum of these numbers = (a – d) + a + (a + d) = 3a

sum of these squares = $(a-d)^{2}$ + $a^{2})$ Â $(a+d)^{2}$ = $3a^{2})$ Â Â + $2d^{2})$

Sum of three numbers = 21 , sum of squares of these numbers = 165

:. 3a = 21

a = 7

and Â $3a^{2})$ Â Â Â + Â $2d^{2})$ Â Â Â = 165

$\Rightarrow$ $3(7)^{2})$ Â Â Â Â + Â $2d^{2})$ Â Â Â = 165

$\Rightarrow$ Â Â $2d^{2})$ Â Â Â = 18

$\Rightarrow$ Â Â $d^{2})$ Â Â Â = 9

$\Rightarrow$ Â d = $\pm$3

Thus , a = 7 and d = $\pm$3

Hence , the required numbers are (4 , 7 , 10) or (10 , 7 , 4).

Q.2: The angles of a quadrilateral are in AP whose common difference is $10^{\circ}$. Find the angles.

Sol:

Let the required angles be Â $(a-3d)^{\circ}$ , $(a-d)^{\circ}$ , $(a+d)^{\circ}$ and $(a+3d)^{\circ}$

Common difference = (a – d) – (a – 3d) = a – d – a + 3d = 2d

Common difference = $(10)^{\circ}$

Therefore, Â 2d = $(10)^{\circ}$ Â Â = d = $(5)^{\circ}$

Sum of four angles of quadrilateral = $(360)^{\circ}$

$(a-3d)^{\circ}$ , $(a-d)^{\circ}$ , $(a+d)^{\circ}$ and $(a+3d)^{\circ}$ Â = $(360)^{\circ}$

4a = $(360)^{\circ}$

a = $(90)^{\circ}$

Therefore, Â Â a = $(90)^{\circ}$ Â d = $(5)^{\circ}$

First angle = (a – 3d)Â° = (90 – 3 x 5) Â° = 75Â°

Second angle = (a – d)Â° = (90 – 5) Â° = 85Â°

Third angle = (a + d)Â° = (90 + 5Â°) = 95Â°

Fourth angle = (a + 3d)Â° = (90 + 3 x 5)Â° = 105Â°

Q.3: Find four numbers is AP whose sum is 28 and the sum of whose square is 216.

Sol:

Let the required number be (a – 3d), (a – d), (a + d) and (a + 3d)

Sum of these numbers = (a – 3d) + (a – d) + (a + d) + (a + 3d)

Therefore, Â Â 4a = 28

$\Rightarrow$ Â a = 7

Sum of the squares of these numbers = Â $(a-3d)^{2})$ Â Â + $(a-d)^{2})$ Â +$(a+d)^{2})$ Â + $(a+3d)^{2})$ Â = 4($(a)^{2})$ + $5(d)^{2})$ Â )

Therefore, 4($(a)^{2})$ + $5(d)^{2})$ ) = 216

$\Rightarrow$ Â $(a)^{2})$ Â + ($5(d)^{2})$ Â = 54 Â [Therefore, Â Â a = 5]

$\Rightarrow$ Â ($5(d)^{2})$ Â = 54 – 49

$\Rightarrow$ ($5(d)^{2})$ Â = 5

$\Rightarrow$ Â ($(d)^{2})$ Â = 1

$\Rightarrow$ Â d = Â±1

Hence, the required numbers (4, 6, 8, 10) or (10, 8, 6, 4).

Q.4: Divide 32 into four parts which are the four terms of an AP such that the first and the fourth terms is the product of the second and the third terms as 7: 15.

Sol:

Let the four parts in AP be (a – 3d), (a – d), (a + d) and (a + 3d)

Then,

$(a – 3d) + (a – d) + (a + d) + (a + 3d) = 32$

$\Rightarrow 4a = 32$

$\Rightarrow a = 8$ â€¦â€¦â€¦(i)

Also,

$(a – 3d) (a + 3d) : (a – d) (a + d) = 7 : 15$

$\Rightarrow \frac{(a – 3d) (a + 3d)}{ (a – d) (a + d)} = \frac{7}{15}$

$\Rightarrow \frac{(8 – 3d) (8 + 3d)}{ (8 – d) (8 + d)} = \frac{7}{15}$ Â Â Â Â Â Â Â Â Â Â [From (i) ]

$\Rightarrow \frac{64 – 9d^{2}}{ 64 – d^{2}} = \frac{7}{15}$

$\Rightarrow 15 (64 – 9d^{2}) = 7(64 – d^{2})$

$\Rightarrow 960 – 135d^{2}= 448 – 7d^{2}$

$\Rightarrow 128d^{2} = 512$

$d^{2} = 4$

$\Rightarrow d = \pm 2$

When a = 8 and d = 2,

$a – 3d = 8 – 3 \times 2 = 8 – 6 = 2$

$a – d = 8 – 2 = 6$

$a + d = 8 + 2 = 10$

$a + 3d = 8 + 3 \times 2 = 8 + 6 = 14$

When a = 8 and d = -2

$a – 3d = 8 – 3 \times (-2) = 8 + 6 = 14$

$a – d = 8 – (-2) = 8 + 2 = 10$

$a + d = 8 – 2 = 6$

$a + 3d = 8 + 3 \times (-2) = 8 – 6 = 2$

Hence the four parts are 2, 6, 10 and 14.

Q.5: The sum of first terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12. Find the AP.

Sol:

Let the required number be (a – d) , a and (a + d)

Sum of these numbers = (a – d) + a + (a + d) = 3a

But sum = 48

Therefore, 3a = 48

$\Rightarrow$ a = 16

Now,

$\left ( a – d \right ) \times a = 4 \left ( a + d \right ) = 48$ Â (given)

$\Rightarrow (16 – d) \times 16 = 4 (16 + d ) + 12$

$\Rightarrow$ 256 – 16d = 64 + 4d + 12

$\Rightarrow 16d + 4d = 256 – 76$

$\Rightarrow 20d = 180$

$\Rightarrow d = 9$

When a = 16 and d = 9

$a – d = 16 – 9 = 7$

$a + d = 16 + 9 = 25$

Hence the first three terms of the AP are 7, 16 and 25.

## EXERCISE – 11C

Question-1:Â The first three terms of an AP are respectively (3y-1), (3y+5) and (5y+1), find the value of y.

Solution:

The terms (3y-1), (3y+5) and (5y+1) are in AP.

Therefore, (3y+5) – (3y-1) = (5y+1) – (3y+5)

=> 3y+5-3y+1 = 5y+1-3y-5

=> 6 = 2y-4

=> 2y = 10

=> y = 5

Hence, the value of y is 5.

Question-2:Â If k, (2k-1) and (2k+1) are the three successive terms of an AP, find the value of k.

Solution:

It is given that k, (2k-1) and (2k+1) are the three successive terms of an AP.

Therefore, (2k-1) -1 = (2k+1) – (2k-1)

=> k-1 = 2

=> k = 3

Hence, the value of k is 3.

Question-3:Â If 18, a, (b-3) are in AP, then find the value of (2a-b).

Solution:

It is given that 18, a, (b-3) are in AP.

Therefore, a -18 = (b-3) â€“ a

=> a+a+-b = 18-3

=> 2a-b = 15

Hence, the required value is 15.

Question-4:Â If the numbers a, 9, b, 25 form an AP, find a and b.

Solution:

It is given that the numbers a, 9, b, 25 form an AP.

Therefore, 9 – a = b – 9 = 25 – b

So,

b-9 = 25 – b

=> 2b = 34

=> b = 17

Also,

9-a = b-9

=> a = 18-b

=> a = 18-17 Â Â Â (b=17)

=> a = 1

Hence, the required values of a and b are 1 and 17, respectively.

Question-5:Â If the numbers (2n-1), (3n+2) and (6n-1) are in AP, find the value of n and the numbers.

Solution:

It is given that the numbers (2n-1), (3n+3) and (6n-1) are in AP.

Therefore, (3n+2) â€“ (2n-1) = (6n-1)-(3n+2)

=> 3n+2-2n+1 = 6n -1 -3n-2

=> n+3 = 3n-3

=> 2n =6

=> n = 3

When n = 3,

2n-1 = 2×3-1 = 6-1 = 5

3n+2 = 3×3+2 = 9+2 = 11

6n-1 = 6×3-1 = 18-1 = 17

Hence, the required value of n is 3 and the numbers are 5, 11 and 17.

Question-6:Â How many three-digit natural numbers are divisible by 7?

Solution:

The three-digit natural numbers divisible by 7 are 105, 112, 119,â€¦.., 994.

Clearly, these numbers are in AP.

Here, a = 105 and d = 112-105 = 7

Let this AP contains n terms. Then,

$a_{n}=994$

=> 105 + (n-1) x 7 = 994 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [$a_{n}=a+\left ( n-1 \right )d$]

=> 7n +98 = 994

=> 7n = 994-98 = 896

=> n = 128

Hence, there are 128 three-digit numbers divisible by 7.

Question-7:Â How many three-digit natural numbers are divisible by 9?

Solution:

The three-digit natural numbers divisible by 9 are 108, 117, 126,â€¦.., 999.

Clearly, these numbers are in AP.

Here, a = 108 and d = 117-108 = 7

Let this AP contains n terms. Then,

$a_{n}=999$

=> 108 + (n-1) x 9 = 999 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [$a {n}=a+\left ( n-1 \right )d$]

=> 9n +99 = 999

=> 9n = 999-99 = 900

=> n = 100

Hence, there are 100 three-digit numbers divisible by 9.

Question-8:Â If the sum of m terms of an AP is $\left ( 2m^{2} +3m\right )$ then what is its second term?

Solution:

Let $S_{m}$ denotes the sum of first m terms of the AP.

$Therefore, S_{m}=2m^{2}+3m$

$\Rightarrow S_{m-1}=2\left ( m-1 \right )^{2}+3\left ( m-1 \right )=2\left ( m^{2}-2m+1 \right )+3\left ( m-1 \right )=2m^{2}-m-1$

Now,

$m^{th}\:term\:of\:the\:AP,a_{m}=S_{m}-S_{m-1}$

$Therefore, a_{m}=\left ( 2m^{2}+3m \right )-\left ( 2m^{2}-m-1 \right )=4m+1$

Putting m = 2, we get

$a_{2}$ = 4×2+1 = 9

Hence, the second term of the AP is 9.

Question-9:Â What is the sum of first n terms of the AP a, 3a, 5a,â€¦

Solution:

The given AP is a, 3a, 5a, â€¦.

Here,

First term, A = a

Common difference, D = 3a-a = 2a

$Therefore, S_{n}$ i.e sum of first n terms

$=\frac{n}{2}\left [ 2\times a+\left ( n-1 \right )\times 2a \right ]\:\:\:\:\left \{ S_{n}=\frac{n}{2}\left [ 2A+\left ( n-1 \right )D \right ] \right \}$

$=\frac{n}{2}\left ( 2a+2an-2a \right )$

$=\frac{n}{2}\times 2an$

$=an^{2}$

Hence, the required sum is $an^{2}$

Question-10:Â What is the 5th term from the end of the AP 2, 7, 12, â€¦,47?

Solution:

The given AP is 2, 7, 12, â€¦.,47.

Let us re-write the given AP in reverse order i.e 47, 42, .., 12, 7, 2.

Now, the 5th term from the end of the given AP is equal to the 5th term from beginning of the AP Â 47, 42, .., 12, 7, 2.

Consider the AP 47, 42, .., 12, 7, 2.

Here, a = 47 and d = 42 â€“ 47 = -5

5th term of this AP

= 47 + (5-1) x (-5)

= 47-20

= 27

Hence, the 5th term from the end of the given AP is 27.

Question-11:Â The nth term of an AP is (3n+5). Find its common difference.

Solution:

nth term of AP = $T_{n}$ Â = 3n + 5

Put n = 1 , Â T1 = Â 3 + Â 5= Â 8

Put n = 2 , Â T2 = 3 x 2 + 5 â€“ 11

Common difference =$T_{2}$ Â – $T_{1}$ Â = 11 – 8 = 3

Thus, common difference = 3

Question-12:Â The nth term of an AP is (7-4n). Find its common difference.

Solution:

nth term of AP = $T_{n}$ Â = 7 – 4n

put n =1 , Â $T_{n}$ â€“ 7 â€“ 4 – 3

put n = Â 2 , Â T2 = 7 – Â 4 x Â 2 – Â 7 â€“ 8 = -1

Common difference =$T_{2}$ Â – $T_{1}$ = Â – 1 â€“ 3 = Â -4.

Question-13:Â Write the next term of the AP $\sqrt{8},\sqrt{18},\sqrt{32},…$

Solution:

The term AP is Â $\sqrt{8}$ , $\sqrt{18}$ , $\sqrt{32}$ ,â€¦â€¦.

Or Â Â $2\sqrt{2}$ , $3\sqrt{2}$ , $4\sqrt{2}$ â€¦â€¦..

common difference = $3\sqrt{2}$$2\sqrt{2}$ Â = $\sqrt{2}$

Next term Â = $4\sqrt{2}$ + $\sqrt{2}$ = $5\sqrt{2}$ = $\sqrt{25 X 2}$ = $\sqrt{50}$

Question-14:Â Write the next term of the AP $\sqrt{2},\sqrt{8},\sqrt{18},…$

Solution:

The given AP is$\sqrt{2}$ , $\sqrt{8}$ , $\sqrt{18}$ , â€¦.â€¦

Or Â $\sqrt{2}$ , $2\sqrt{2}$ , $3\sqrt{2}$ Â â€¦â€¦

Common difference d = Â $2\sqrt{2}$$\sqrt{2}$ = $\sqrt{2}$

Term next to $3\sqrt{2}$$3\sqrt{2}$ +d = Â $3\sqrt{2}$ + $\sqrt{2}$ = Â $4\sqrt{2}$$\sqrt{16 X 2}$ = $\sqrt{32}$.

#### Practise This Question

If energy is always conserved in a chemical or physical process then why is there an energy crisis?