Question

if an arithmetic means are inserted between 1 and 31 such that the ratio of 1st main and the nth mean is 3:29 then the value of n is

Answer 1

Hello,
the exercise in the figure:

bye :-)

Answer 2

Let the new AP after inserting n arithmetic means between 1 and 31 be as:

$AP: 1,a_1,a_2,a_3,......,a_{n-1},a_n,31\\common \: difference=d=\frac{31-1}{n+1}\\ \implies d=30/(n+1)\\\\a_1=1+1*d\\a_n=1+(n+1-1)d=1+n d\\\\Given \: a_1 : a_n=3 : 29\\ \implies 29*a_1=3*a_n \\\implies 29(1+d)=3(1+n d) \\ \implies (3n-29)d =26\\ \implies (3n-29)*\frac{30}{n+1}=26 \\\\ \implies 32 n = 448 \\ \implies n = 14$

We get answer =   n = 14 .

14 arithmetic means were inserted between 1 and 31.
So the common difference becomes 2.