If `n` arithmetic means are inserted between `20 and 80` such that the ratio of first mean to the last mean is `1:3,` then find the value of `ndot`

Explanation

Let ` A_(1) , A_(2),….., A_(n) ` be n Ams between 20 and 80.
Then , ` 20, A_(1) , A_(2),…..,A_(n), 80` are in AP.
Let their common differnece be d. Then ,
` T_(n +2) = 80 Rightarrow 20 + ( n +2-1) d = 80 Rightarrow d= 60/((n +1))`
First mean , ` A_(1)=T_(2)= ( 20+d) = ( 20 60/( n+1))`
` ((20n +80))/ ((n+1))`
Last mean, ` A_(n)= T_(n+1)= [ 20 + ( n + 1 -1)d]`
` { 20+ ( nxx60)/((n +1)) }`
` = ((80n+ 20))/((n+1))`
` A_(1)/A_(n) = 1/3 Rightarrow = ((20n +80))/ ((n +1))xx ((n+1))/((80n +20)) = 1/3`
Hence, `3( 20n+80) = ( 80n +20) Rightarrow 20n =220 Rightarrow n=11`
Hence, n=11