Question

insert the five arithmetic mean between -3 and 25​

Answer 1

Step-by-step explanation:

$\huge\bf\underline\blue{answer}$

given,

the series are in A.P

.: common difference is same

to find:

insert the five arithmetic mean between -3 and 25

$\huge\fcolorbox{black}{pink}{solution}$

the first term of an A.P (a) = -3

last term of an A.P (b) = 25

n = 5

.: we know that common difference (d) = b-a/n+1

d = 25-(-3)/5+1

= 25+3/6

= 28/6

= 14/3

let the series of the A.P be:

-3,A1,A2,A3,A4,A5,25

.: nth term of A.M = a+(n)d

.: A1 = -3+(1)14/3

= -3+14/3

= -9+14/3

= 5/3

.: A1 = 5/3

.: A2 = -3+(2)14/3

= -3+28/3

= -9+28/3

= 19/3

.: A2 = 19/3

A3 = -3+(3)14/3

= -3+42/3

= -9+42/3

= 33/3

= 11

.: A3 = 11

.: A4 = -3+(4)14/3

= -3+56/3

= -9+56/3

= 47/3

.: A4 = 47/3

.: A5 = -3+(5)14/3

= -3+70/3

= -9+70/3

= 61/3

.: A5 = 61/3

Answer 2

$\sf{Answer}$

Step-by-step-explanation:-

Question :-

insert the five arithmetic mean between -3 and 25

Solution:-

Hence we have to insert 5 A.M's B/W

-3 , 25

So, It is In

Let the 5 arthemetic means be

• a1
• a2
• a3
• a4
• a5

So,

-3 , a1 , a2 , a3 , a4 , a5 , 25

First term = a = -3

Last term = b = 25

n = 5 (No.of terms)

Common difference (d) = $\sf\dfrac{b- a}{n+1}$

So, common difference = $\sf\dfrac{25-(-3)}{5+1}$

Common difference = $\sf\dfrac{28}{6}$ = $\sf\dfrac{14}{3}$

Common difference = 14/3

Finding numbers :-

In an A.P

a1 = a + d

a2 = a + 2d

a3 = a + 3d

a4 = a + 4d

a5 = a + 4d

Plugging values of a, d

a1 = -3 + $\sf\dfrac{14}{3}$

a1 = $\sf\dfrac{-9+14}{3}$

a1 = $\sf\dfrac{5}{3}$

________________________

a2 = -3 + 2 × $\sf\dfrac{14}{3}$

a2 = -3 + $\sf\dfrac{28}{3}$

a2 = $\sf\dfrac{-9+28}{3}$

a2 = $\sf\dfrac{19}{3}$

__________________________

a3 = -3 + 3 × $\sf\dfrac{14}{3}$

a3 = -3 + $\sf\dfrac{42}{3}$

a3 = -3 + 14

a3 = 11

_________________________

a4 = -3 + 4 × $\sf\dfrac{14}{3}$

a4 = -3 + $\sf\dfrac{56}{3}$

a4 = $\sf\dfrac{56-9}{3}$

a4 = $\sf\dfrac{47}{3}$

_________________________

a5 = -3 + 5 × $\sf\dfrac{14}{3}$

a5 = -3 + $\sf\dfrac{70}{3}$

a5 = $\sf\dfrac{-9+70}{3}$

a5 = $\sf\dfrac{61}{3}$

_______________________________

So, Required 5 AM's are

• 5/3
• 19/3
• 11
• 47/3
• 61/3