Arithmetic progression

EXERCISE 19.3

QUESTION 1

The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the
second term by 6, find three terms.

Sol :

Let the three terms of the A.P. be  (a-d),  a , (a+d)

Then , we have

\Rightarrow (a-d)+a+(a+d)=21

\Rightarrow 3 a=21

\Rightarrow a=7 \quad \ldots .(i)

 

Also , (a-d)(a+d)=6

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

7^{2}-d^{2}=7+6  [\because a=7 \text{ from (i) }]

d^{2}=36

d=\pm 6

 

When d = 6 , a = 7 ,putting in (a – d) , a , (a + d) we get :

1 , 7 , 13

When d = -6 , a = 7 ,putting in (a – d) , a , (a + d) we get :

13 , 7 ,1

 

QUESTION 2

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

Sol :

Let the three terms of the A.P. be  (a-d),  a , (a+d)

Their sum = 27

\Rightarrow (a-d)+a+(a+d)=27

\Rightarrow 3 a=27

\Rightarrow a=9 \ldots(i)

 

Product =(a-d) a(a+d)=648

\Rightarrow a\left(a^{2}-d^{2}\right)=648

\Rightarrow 9\left(81-d^{2}\right)=648  [\because a=9 \text{ from (i) }]

\Rightarrow\left(81-d^{2}\right)=72

\Rightarrow d^{2}=9

\Rightarrow d=\pm 3

When d = 3 , a = 9 ,putting in (a – d) , a , (a + d) we get :

6 , 9 , 12

When d = -3 , a = 9 ,putting in (a – d) , a , (a + d) we get :

12 , 9 , 6

 

QUESTION 3

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

Sol :

Let the four numbers be (a-3 d), (a-d), (a+d), (a+3 d)

Their sum = 50

\Rightarrow (a-3 d)+(a-d)+(a+d)+(a+3 d)=50

\Rightarrow 4 a=50

\Rightarrow a=\frac{25}{2} \ldots(i)

 

Also, (a+3 d)=4(a-3 d)

\Rightarrow a+3 d=4 a-12 d

\Rightarrow 3 a=15 d

\Rightarrow a=5 d

\Rightarrow d=\dfrac{a}{5} 

\Rightarrow d=\dfrac{25}{2\times 5}   [\because~a=\dfrac{25}{2} \text{ from (i) }]

\Rightarrow d=\frac{5}{2}

 

When d = \dfrac{5}{2}, a = \dfrac{25}{2} ,putting in (a-3 d), (a-d), (a+d), (a+3 d)

So , the terms are as follows :  \left(\frac{25}{2}-3 \times \frac{5}{2}\right) , \left(\frac{25}{2}-\frac{5}{2}\right) , \left(\frac{25}{2}+\frac{5}{2}\right) , \left(\frac{25}{2}+3 \times \frac{5}{2}\right)

= 5 , 10 , 15 ,20

 

QUESTION 4

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

Sol :

Let the three terms of the A.P. be (a-d),  a , (a+d)

Their sum = 12

\Rightarrow (a-d)+a+(a+d)=12

\Rightarrow 3 a=12

\Rightarrow a=4

 

Also, sum of their cubes are

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

\Rightarrow a^{3}-d^{3}-3 a^{2} d+3 a d^{2}+a^{3}+a^{3}+d^{3}+3 a^{2} d+3 a d^{2}=288

\Rightarrow 3 a^{3}+6 a d^{2}=288

\Rightarrow 3(4)^{3}+6 \times 4 \times d^{2}=288

\Rightarrow 192+24 d^{2}=288

\Rightarrow 24 d^{2}=96

\Rightarrow d^{2}=4

\Rightarrow d=\pm 2

 

When d = 2 , a = 4 ,putting in (a – d) , a , (a + d) we get :

2 , 4 , 6

When d = -2 , a = 4 ,putting in (a – d) , a , (a + d) we get :

6 , 4 , 2

 

QUESTION 5

If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.

Sol :

Let the three numbers be (a-d), a,(a+d)

Their sum = 24

\Rightarrow(a-d)+a+(a+d)=24

\Rightarrow 3 a=24

\Rightarrow a=8 \ldots(i)

 

Also , their product is

a(a-d)(a+d)=440

\Rightarrow a\left(a^{2}-d^{2}\right)=440

\Rightarrow 8\left(64-d^{2}\right)=440  [\because~a=8 \text{ from (i) }]

\Rightarrow\left(64-d^{2}\right)=55

\Rightarrow d^{2}=9

\Rightarrow d=\pm 3

 

When d = 3 , a = 8 ,putting in (a – d) , a , (a + d) we get :

5 , 8 , 11

When d = -3 , a = 8 ,putting in (a – d) , a , (a + d) we get :

11 , 8 , 5

 

QUESTION 6

The angles of a quadrilateral are in A.P. whose common difference is 10^{\circ} . Find the angles.

Sol :

Let the four angles be (A)^{\circ},(A+d)^{\circ},(A+2 d)^{\circ},(A+3 d)^{\circ}

Given : d = 10

Sum of all the angles is {360}^{\circ}

(A)^{\circ}+(A+10)^{\circ}+(A+20)^{\circ}+(A+30)^{\circ}=360^{\circ}

\Rightarrow 4 A=360-60

\Rightarrow A=\frac{300}{4}=75^{\circ}

75^{\circ},(75+10)^{\circ},(75+20)^{\circ},(75+30)^{\circ} i.e. 75^{\circ}, 85^{\circ}, 95^{\circ}, 105^{\circ}

 

ALTERNATE METHOD

Let the four angle be (a-3d) , (a-d) , (a+d) , (a+3d)

Then , 

Sum of all the angle = 360^{\circ}

(a-3 d)+(a-d)+(a+d)+(a+3 d)=360^{\circ}

4a=360^{\circ}

a=90^{\circ}\dots(i)

AND

(a-d)-(a-3 d)=10

2 d=10

d=5

\therefore The angle of the given quadrilateral are 75^{\circ}, 85^{\circ}, 95^{\circ} and 105^{\circ}

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