Question

If n arithmetic mean are inserted between 20 and 80 such that the ratio of firstmean to the last mean is 1:3 then find the value of n.​

Answer 1

Answer:

Let A

1

,A

2

,A

3

,...A

n

be n arithmetic means between 20 and 80

And let d be the common difference between the terms of the A.P.

So,d=

n+1

b−a

=

n+1

80−20

=

n+1

60

Now, A

1

=20+d

=20+

n+1

60

=20(

n+1

n+4

)

And A

n

=20+nd=20+

n+1

60n

=20(

n+1

4n+1

)

Thus,

A

n

A

1

=

3

1

(given)

20+nd

20+d

=

3

1

20(

n+1

4n+1

)

20(

n+1

n+4

)

=

3

1

3

1

⇒4n+1=3n+12

⇒n=11

∴ The value of n is 11

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