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Class 10 RD Sharma Solutions- Chapter 9 Arithmetic Progressions – Exercise 9.5
• Last Updated : 03 Jan, 2021

### Question 1: Find the value of x for which (8x + 4), (6x – 2) and (2x + 7) are in A.P.

Solution:

Since (8x + 4), (6x – 2) and (2x + 7) are in A.P.

Now we know that condition for three number being in A.P.-

⇒2*(Middle Term)=(First Term)+(Last Term)

⇒2(6x-2)=(8x+4)+(2x+7)

⇒12x-4=10x+11

⇒(12x-10x)=11+4

⇒2x=15

⇒x=15/2

Mean value of x=15/2.

### Question 2: If x + 1, 3x and 4x + 2 are in A.P., find the value of x.

Solution:

Since x + 1, 3x and 4x + 2 are in A.P.

Now we know that condition for three number being in A.P.-

⇒2*(Middle Term)=(First Term)+(Last Term)

⇒2(3x)=(x+1)+(4x+2)

⇒6x=5x+3

⇒x=3

Mean value of x=3.

### Question 3: Show that (a – b)², (a² + b²), and (a + b)² are in A.P.

Solution:

We know that condition for three number being in A.P.-

⇒2*(Middle Term)=(First Term)+(Last Term)   ———-(1)

First term=(a-b)2

=a2-2ab+b2

Middle term=a2+b2

Last term=(a+b)2

=a2+2ab+b2

Now putting these values in equation(1)-

⇒2(a2+b2)=(a2-2ab+b2)+(a2+2ab+b2)

⇒2(a2+b2)=2a2+2b2

⇒2(a2+b2)=2(a2+b2)

Since L.H.S=R.H.S.

Hence proved.

### Question 4: The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceed the second term by 6, find three terms.

Solution:

Let the first term of the A.P. is=a

and common difference is=d

let three terms are=(a-d),a,(a+d)

Now according to first condition-

⇒(a-d)+a+(a+d)=21

⇒3a=21

⇒a=7

Now according to second condition-

⇒(a-d)(a+d)=a+6

⇒a2-d2=a+6

Putting a=7-

⇒49-d2=7+6

⇒d2=49-13

⇒d2=36

⇒d=6    or   d=-6

When a=7 and d=6

First number=a-d=1

Second number=a=7

Third number=a+d=13

When a=7 and d=-6

First number=a-d=13

Second number=a=7

Third number=a+d=1

So required numbers are 1,7 and 13 or 13,7 and 1.

### Question 5: Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.

Solution:

Let the first term of the A.P. is=a

and common difference is=d

let three terms are=(a-d),a,(a+d)

Now according to first condition-

⇒(a-d)+a+(a+d)=27

⇒3a=27

⇒a=9

Now according to second condition-

⇒(a-d)*a*(a+d)=648

⇒(a2-d2)*a=648

putting a=9-

⇒(81-d2)*9=648

⇒81-d2=648/9

⇒81-d2=72

⇒d2=81-72

⇒d2=9

⇒d=-3   or     d=3

When a=9 and d=-3

First term=a-d=12

Second term=a=9

Third term=a+d=6

When a=9 and d=3

First term=a-d=6

Second term=a=9

Third term=a+d=12

means required numbers are 6,9 and 12 or 12,9 and 6.

### Question 6: Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.

Solution:

Let the first term of the A.P. is=a

and common difference is=d

let three terms are=(a-3d), (a-d), (a+d), (a+3d)

Now according to first condition-

⇒(a-3d)+(a-d)+(a+d)+(a+3d)=50

⇒4a=50

⇒a=50/4

⇒a=25/2

Now according to second condition-

⇒(greatest number)=4*(least number)

⇒a+3d=4(a-3d)

⇒a+3d=4a-12d

⇒3d+12d=4a-a

⇒15d=3a

Dividing by 3-

⇒5d=a

⇒5d=25/2

⇒d=5/2

with the help of a=25/2 and d=5/2-

⇒a-3d=(25/2)-(15/2)=10/2=5

⇒a-d=(25/2)-(5/2)=20/2=10

⇒a+d=(25/2)+(5/2)=30/2=15

⇒a+3d=(25/2)+(15/2)=40/2=20

So required numbers are 5,10,15 and 20.

### Question 7: The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.

Solution:

Let the first term of the A.P. is=a

and common difference is=d

let three terms are=(a-d),a,(a+d)

Now according to first condition-

⇒(a-d)+a+(a+d)=12

⇒3a=12

⇒a=4

And according to second condition-

⇒(a-d)3+a3+(a+d)3=288

by using (A+B)3=A3+3AB(A+B)+B3

and        (A-B)3=A3-3AB(A-B)-B3

⇒{a3-3ad(a-d)-d3}+a3+{a3+3ad(a+d)+d3}=288

⇒3a3-3a2d+3ad2+3a2d+3ad2=288

⇒3a3+6ad2=288

dividing by 3-

⇒a3+2ad2=96

putting a=4-

⇒64+8d2=96

dividing by 8-

⇒8+d2=12

⇒d2=4

⇒d=-2  or   d=2

when a=4 and d=-2

a-d=6

a=4

a+d=2

when a=4 and d=2

a-d=2

a=4

a+d=6

so required numbers are 6,4 and 2  or    2,4 and 6.

### Question 8: Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.

Solution:

Let the first term of the A.P. is=a

and common difference is=d

let required four parts are=(a-3d),(a-d),(a+d),(a+3d)

now since (a-3d),(a-d),(a+d),(a+3d) are parts of 56.

so-

⇒(a-3d)+(a-d)+(a+d)+(a+3d)=56

⇒4a=56

⇒a=14

Now extreme parts are=(a-3d) and (a+3d)

and mean parts are=(a-d) and (a+d)

According to given condition-

⇒{(a-3d)(a+3d)}/{(a-d)(a+d)}=5/6

⇒6{(a-3d)(a+3d)}=5{(a-d)(a+d)}

by using (A-B)(A+B)=A2-B2

⇒6{a2-9d2}=5{a2-d2}

putting a=14

⇒6{196-9d2}=5{196-d2}

⇒6*196-54d2=5*196-5d2

⇒6*196-5*196=54d2-5d2

⇒196=49d2

⇒d2=4

⇒d=2    or    d=-2

When a=14  and d=2  –

a-3d=8

a-d=12

a+d=16

a+3d=20

When a=14  and  d=-2

a-3d=20

a-d=16

a+d=12

a+3d=8

So required parts are 8,12,16 and 20   or     20,16,12 and 8.

### Question 9: The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.

Solution:

Let the angles of quadrilateral are-

(a-3d), (a-d), (a+d), (a+3d)

Now we know that sum of angles in a quadrilateral is=360°

⇒(a-3d)+(a-d)+(a+d)+(a+3d)=360°

⇒4a=360°

⇒a=90°

Now common difference=10°

⇒(Second angle)-(First angle)=10°

⇒(a-d)-(a-3d)=10°

⇒a-d-a+3d=10°

⇒2d=10°

⇒d=5°

So when a=90° and d=5° –

a-3d=75°

a-d=85°

a+d=95°

a+3d=105°

So required angles are 75°,85°,95° and 105°.

### Question 10: Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.

Solution:

Let the first term of the A.P. is=a

and common difference is=d

let three parts are=(a-d),a,(a+d)

Since (a-d),a and (a+d) are parts of 207—

⇒(a-d)+a+(a+d)=207

⇒3a=207

⇒a=69

Now according to given condition-

⇒(a-d)*a=4623

⇒(69-d)*69=4623

dividing by 69-

⇒69-d=67

⇒d=2

So when a=69 and d=2 –

a-d=67

a=69

a+d=71

means required three parts of 207 are 67,69 and 71.

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