Arithmetic progression

EXERCISE 19.6

QUESTION 1

Find the A.M. between:

(i) 7 and 13

Sol :

Let A_{1} be the A.M. between 7 and 13

A_{1}=\frac{a+b}{2}

=\frac{7+13}{2}

= 10

 

(ii) 12 and -8

Sol :

Let A_{1} be the A.M. between 12 and -8

A_{1}=\frac{a+b}{2}

=\frac{12+(-8)}{2}

= 2

 

(iii) (x – y) and (x + y)

Sol :

Let A_{1} be the A.M. between (x-y) and (x+y)

A_{1}=\frac{a+b}{2}

=\frac{(x-y)+(x+y)}{2}

= x

 

QUESTION 2

Insert 4 A.M.s between 4 and 19

Sol :

Let A_{1},A_{2},A_{3},A_{3} be the four A.M.s between 4 and 19 Then, 4, A_{1},A_{2},A_{3},A_{4} and 19 are in A.P. whose common difference is as follows:

a = 4 , a_n = 19 

n = 6

 

a_n=a+(n-1)d

19=4+(6-1)d

19-4=5d

d=\dfrac{19-4}{5}

d=\dfrac{15}{5}

d = 3

 

A_{1}=4+d =4+3=7

A_{2}=4+2 d =4+6=10

A_{3}=4+3 d =4+9=13

A_{4}=4+4 d =4+12=16

Hence, the required A.M.s are , 7 , 10 , 13 , 16

 

QUESTION 3

Insert 7 A.M.s between 2 and 17

Sol :
Let A_{1},A_{2},A_{3},A_{4},A_{5},A_{5}, A_{7} be the seven A.M.s between 2 and 17

Then, 2,A_{1},A_{2},A_{3},A_{4}, A_{5},A_{5}, A_{7} and 17 are in A.P. whose common difference is as
follows:

Here , we have 

a = 2 , a_n=17

n = 9

 

a_n=a+(n-1)d

17=2+(9-1)d

17-2=8d

d=\dfrac{15-2}{8}

d=\dfrac{15}{8}

 

A_{1}=2+d =2+\frac{15}{8}=\frac{31}{8}

A_{2}=2+2 d =2+\frac{15}{4}=\frac{23}{4}

A_{3}=2+3 d =2+\frac{45}{8}=\frac{61}{8}

A_{4}=2+4 d =2+\frac{15}{2}=\frac{19}{2}

A_{5}=2+5 d =2+\frac{75}{8}=\frac{91}{8}

A_{6}=2+6 d =2+\frac{45}{4}=\frac{53}{4}

A_{7}=2+7 d =2+\frac{105}{8}=\frac{121}{8}

Hence, the required A.M.s are \frac{31}{8},\frac{23}{4}, \frac{61}{8}, \frac{19}{2},\frac{91}{8},\frac{51}{4},\frac{121}{8},\frac{121}{8}

 

QUESTION 4

Insert six A.M.s between 15 and -13.

Sol :

Let A_{1}, A_{2},A_{3},A_{4},A_{5},A_{6} be the 6 A.M.s between 15 and -13 

Then, 15,A_{1},A_{2},A_{3}, A_{4}, A_{5}, A_{5} and -13 are in A.P. whose common difference is as follows:

We have :

a = 15 , a_n=-13

n = 8

 

a_n=a+(n-1)d

-13=15+(8-1)d

-13=15+7 d

-15-13=7d

-28=7d

d=\dfrac{-28}{7}

d = -4

 

A_{1}=15+d =15+(-4)=11

A_{2}=15+2 d =15+(-8)=7

A_{3}=15+3 d =15+(-12)=3

A_{4}=15+4 d =15+(-16)=-1

A_{5}=15+5 d =15+(-20)=-5

A_{6}=15+6 d =15+(-24)=-9

Hence, the required A.M.s are 11 , 7 , 3 , -1 , -5 , -9

 

QUESTION 5

There are n A.M.s between 3 and 17 . The ratio of the last mean to the first mean is 3 : 1 Find the value of n

Sol :

Let A_{1}, A_{2},A_{3},A_{4},\dots A_{n} be the n A.M.s between 3 and 17

Let be the common difference of the A.P 3 , A_{1}, A_{2},A_{3},A_{4},\dots A_{n} and 17 . Then we have

a = 3 , a_n

total terms = n

d=\dfrac{a_n-a}{n+1}

d=\dfrac{17-3}{n+1}

d=\dfrac{14}{n+1}

Now , A_{1}=3+d =3+\frac{14}{n+1}=\frac{3 n+17}{n+1}

A_{n}=3+n d =3+n\left(\frac{14}{n+1}\right)=\frac{17 n+3}{n+1}

\therefore ~\frac{A_{n}}{A_{1}}=\frac{3}{1}

\Rightarrow \frac{\left(\frac{17 n+3}{n+1}\right)}{\left(\frac{3 n+17}{n+1}\right)}=\frac{3}{1}

\Rightarrow \frac{17 n+3}{3 n+17}=\frac{3}{1}

\Rightarrow 17 n+3=9 n+51

\Rightarrow 8 n=48

\Rightarrow n=6

 

QUESTION 6

Insert A.M.s between 7 and 71 in such a way that the 5^{\text { th }} A.M. is 27 Find the number of A.M.s.

Sol :

Let A_{1}, A_{2},A_{3},A_{4}, 27 ,A_{6},\dots A_{n} be the n A.M.s between 7 and 71

 

Leave a Reply

Your email address will not be published. Required fields are marked *