If A1 , A2 , A3 ,............, An are nth arithmetic means inserted between a and b.
Then a , A1 , A2 , A3 ,............, An , b are in A P
here first term = a
Common difference is given by \[d=\frac{(b-a)}{n+1} \]
then b = a + (n + 2 - 1) d
=a+(n+1)*d
\[A_{1}=a+\frac{b-a}{n+1} \]
\[A_{2}=a+\frac{2(b-a)}{n+1} \]
\[A_{3}=a+\frac{3(b-a)}{n+1} \]
......
.....
......
\[A_{n}=a+\frac{n(b-a)}{n+1} \]
For example
Insert 5 Arithmetic mean between 1 and 11
Then AP what we get can be written as 1 , A1 , A2 , A3 , A 4, A5, 11
here total no of terms = 5+2 =7
\[d=\frac{(b-a)}{n+1} \] where n is no of arithmetic mean inserted
\[d=\frac{11-1}{5+1}=\frac{10}{6} \]
\[A_{1}=1+\frac{10}{6}=\frac{16}{6} \]
\[A_{2}=1+\frac{20}{6}=\frac{26}{6} \]
\[A_{3}=1+\frac{30}{6}=\frac{36}{6}=6 \]
\[A_{4}=1+\frac{40}{6}=\frac{46}{6} \]
\[A_{5}=1+\frac{50}{6}=\frac{56}{6} \]
