Question

"n" number of arithmetic means are between 20 and 80 .if first mean:3rd mean=1:3,then find n.

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Answer 1

Given:-

• 1,A2 ,A3 ,...A n be n arithmetic means between 20 and 80

To Find:-

• $n$

Solution:-

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

So,d= n+1b−a

= n+180−20 = n+160

Now, A1 =20+d

=20+n+160

=20( n+1n+4 )

And

An =20+nd=20+n+160n =20( n+14n+1 )

Thus,An

A 1 = 31(given)

20+nd20+d

= 3120( n+14n+1 )20( n+1n+4 )

= 31

⇒4n+1n+4 = 31

⇒4n+1=3n+12

⇒n=11

Required answer:-

$n = 11$

Answer 2

$\huge\textit{Question :-}$

n" number of arithmetic means are between 20 and 80 .if first mean:3rd mean=1:3,then find n.

$\huge\textit{Answer:-}$

$\bold{\boxed{Given:}}$

arithmetic means are inserted between 20 and 780.ratio of the first and last arithmetic mean is 1:33

$\bold{\boxed{To\: find: }}$

Value of n

$\bold{\boxed{Solution:}}$

if n arithmetic means are inserted between 20 and 80 Then Series becomes

20 ,20+ d, 20+2d., 20 + nd , 80

total Terms = n +2

970 20+ (n +2-1) d = 20 + nd +d 20+ nd = 80 - d

ratio of first and last arithmetic mean is 1:3

(20+ d/(20 + nd) = 1/3

=> (20+ dy(80 - d) = 1/3

=>60+3d -80 - d

=>4d = 120 d=5

20+ nd = 80 d

>20+ n(5) = 7 60 -5

>n{5) = 55=>n = 11

11 arithmetic means are inserted between 20 and 70 Series becomes

20,25 ,30,.. 75,80

$\bold{\boxed{Series\: becomes}}$

20,25,30, .., 75, 80

We are all done :)

Thankyou :)