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.
Answers
Answered by
0
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
verified_toppr
Similar questions