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

### 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}=a

^{2}-2ab+b^{2}Middle term=a

^{2}+b^{2}Last term=(a+b)

^{2}=a

^{2}+2ab+b^{2}Now putting these values in equation(1)-

⇒2(a

^{2}+b^{2})=(a^{2}-2ab+b^{2})+(a^{2}+2ab+b^{2})⇒2(a

^{2}+b^{2})=2a^{2}+2b^{2}⇒2(a

^{2}+b^{2})=2(a^{2}+b^{2})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

⇒a

^{2}-d^{2}=a+6Putting a=7-

⇒49-d

^{2}=7+6⇒d

^{2}=49-13⇒d

^{2}=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

⇒(a

^{2}-d^{2})*a=648putting a=9-

⇒(81-d

^{2})*9=648⇒81-d

^{2}=648/9⇒81-d

^{2}=72⇒d

^{2}=81-72⇒d

^{2}=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}+a^{3}+(a+d)^{3}=288by using (A+B)

^{3}=A^{3}+3AB(A+B)+B^{3}and (A-B)

^{3}=A^{3}-3AB(A-B)-B^{3}⇒{a

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

^{3}-3a^{2}d+3ad^{2}+3a^{2}d+3ad^{2}=288⇒3a

^{3}+6ad^{2}=288dividing by 3-

⇒a

^{3}+2ad^{2}=96putting a=4-

⇒64+8d

^{2}=96dividing by 8-

⇒8+d

^{2}=12⇒d

^{2}=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)=A

^{2}-B^{2 }–⇒6{a

^{2}-9d^{2}}=5{a^{2}-d^{2}}putting a=14

⇒6{196-9d

^{2}}=5{196-d^{2}}⇒6*196-54d

^{2}=5*196-5d^{2}⇒6*196-5*196=54d

^{2}-5d^{2}⇒196=49d

^{2}⇒d

^{2}=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.