# Class 10 RD Sharma Solutions- Chapter 9 Arithmetic Progressions – Exercise 9.5

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

Solution:

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

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

â‡’2*(Middle Term)=(First Term)+(Last Term)

â‡’2(6x-2)=(8x+4)+(2x+7)

â‡’12x-4=10x+11

â‡’(12x-10x)=11+4

â‡’2x=15

â‡’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.-

â‡’2*(Middle Term)=(First Term)+(Last Term)

â‡’2(3x)=(x+1)+(4x+2)

â‡’6x=5x+3

â‡’x=3

Mean value of x=3.

### Question 3: Show that (a â€“ b)Â², (aÂ² + bÂ²), and (a + b)Â² are in A.P.

Solution:

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

â‡’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)-

â‡’2(a2+b2)=(a2-2ab+b2)+(a2+2ab+b2)

â‡’2(a2+b2)=2a2+2b2

â‡’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-

â‡’(a-d)+a+(a+d)=21

â‡’3a=21

â‡’a=7

Now according to second condition-

â‡’(a-d)(a+d)=a+6

â‡’a2-d2=a+6

Putting a=7-

â‡’49-d2=7+6

â‡’d2=49-13

â‡’d2=36

â‡’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-

â‡’(a-d)+a+(a+d)=27

â‡’3a=27

â‡’a=9

Now according to second condition-

â‡’(a-d)*a*(a+d)=648

â‡’(a2-d2)*a=648

putting a=9-

â‡’(81-d2)*9=648

â‡’81-d2=648/9

â‡’81-d2=72

â‡’d2=81-72

â‡’d2=9

â‡’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-

â‡’(a-3d)+(a-d)+(a+d)+(a+3d)=50

â‡’4a=50

â‡’a=50/4

â‡’a=25/2

Now according to second condition-

â‡’(greatest number)=4*(least number)

â‡’a+3d=4(a-3d)

â‡’a+3d=4a-12d

â‡’3d+12d=4a-a

â‡’15d=3a

Dividing by 3-

â‡’5d=a

â‡’5d=25/2

â‡’d=5/2

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

â‡’a-3d=(25/2)-(15/2)=10/2=5

â‡’a-d=(25/2)-(5/2)=20/2=10

â‡’a+d=(25/2)+(5/2)=30/2=15

â‡’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-

â‡’(a-d)+a+(a+d)=12

â‡’3a=12

â‡’a=4

And according to second condition-

â‡’(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

dividing by 3-

putting a=4-

â‡’64+8d2=96

dividing by 8-

â‡’8+d2=12

â‡’d2=4

â‡’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-

â‡’(a-3d)+(a-d)+(a+d)+(a+3d)=56

â‡’4a=56

â‡’a=14

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

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

According to given condition-

â‡’{(a-3d)(a+3d)}/{(a-d)(a+d)}=5/6

â‡’6{(a-3d)(a+3d)}=5{(a-d)(a+d)}

by using (A-B)(A+B)=A2-B2  â€“

â‡’6{a2-9d2}=5{a2-d2}

putting a=14

â‡’6{196-9d2}=5{196-d2}

â‡’6*196-54d2=5*196-5d2

â‡’6*196-5*196=54d2-5d2

â‡’196=49d2

â‡’d2=4

â‡’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 10Â°. 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=360Â°

â‡’(a-3d)+(a-d)+(a+d)+(a+3d)=360Â°

â‡’4a=360Â°

â‡’a=90Â°

Now common difference=10Â°

â‡’(Second angle)-(First angle)=10Â°

â‡’(a-d)-(a-3d)=10Â°

â‡’a-d-a+3d=10Â°

â‡’2d=10Â°

â‡’d=5Â°

So when a=90Â° and d=5Â° –

a-3d=75Â°

a-d=85Â°

a+d=95Â°

a+3d=105Â°

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

### 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 207â€”

â‡’(a-d)+a+(a+d)=207

â‡’3a=207

â‡’a=69

Now according to given condition-

â‡’(a-d)*a=4623

â‡’(69-d)*69=4623

dividing by 69-

â‡’69-d=67

â‡’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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