Question

if A1,A2,A3,a4,a5 and a6 are six arithmetic mean between 3 and 31, then a6-a5 and A1+a6 are respectively equal to​

Answer 1

Step-by-step explanation:

Given :-

A1,A2,A3,a4,a5 and a6 are six arithmetic mean between 3 and 31.

To find :-

Find the values of a6-a5 and A1+a6 ?

Solution :-

Given that :

Given two numbers = 3 and 31

Let a = 3 and b = 31

Number of Arithmetic means = 6

They are A1,A2,A3,a4,a5 and a6

3, A1, A2, A3, a4, a5 ,a6 ,31

We know that

If 'n' arithmetic means are between a and b in an AP then the common difference d = (b-a)/(n+1)

The common difference of the given AP

=>d = (31-3)/(6+1)

=>d = 28/7

=> d = 4

So, Common difference = 4

We have First term (a)=3

Second term = A1 = a+d

=> A1 = 3+4

=> A1 = 7

and Third term = A2 = a+2d

=> A2 = 3+2(4)

=>A2 = 3+8

=> A2 = 11

Fourth term = A3 = a+3d

=> A3 = 3+3(4)

=> A3 = 3+12

=> A3 = 15

Fifth term = a4 = a+4d

=> a4 = 3+4(4)

=> a4 = 3+16

=> a4 = 19

Sixth term = a5 = a+5d

=> a5 = 3+5(4)

=> a5 = 3+20

=> a5 = 23

Seventh term = a6 = a+6d

=> a6 = 3+6(4)

=> a6 = 3+24

=> a6 = 27

Now the AP : 3,7,11,15,19,23,27,31

The arithmetic means are 7,11,15,19,23,27

Now

a6 - a5

=> 27-23

=> 4

a6-a5 = 4

and

A1+a6

=> 7+27

=> 34

A1+a6 = 34

Answer:-

The value of a6-a5 for the given problem is 4

The Value of A1+a6 for the given problem is 34

Used formulae:-

• If 'n' arithmetic means are between a and b in an AP then the common difference d = (b-a)/(n+1)

Points to know :-

• The succeeding number is obtained by adding a constant term to the preceding number in a series except first number is called An Arithmetic Progression. Shortly AP

• The common difference is same throughout the series in the AP.

• The common difference is denoted by d.

• The general form of an AP is a , a+d,a+2d ,...

• a is the first term.

• The arithmetic Mean of two terms a and b in the AP is (a+b)/2.