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 $\frac{31}{8}$ , $\frac{23}{4}$ , $\frac{61}{8}$ , $\frac{19}{2}$ , $\frac{91}{8}$ , $\frac{51}{4}$ , $\frac{121}{8}$ , $\frac{121}{8}$