Question

a1, a2, a3,.........an are in arithmetic progression where a1 = 0, Find the value of :
[(a3/a2) + (a4/a3)......(an/an-1)] - a2[(1/a2) + (1/a3)....(1/an-2)

Answer 1

Do you think
\[\frac{a_3}{a_2}+\frac{a_4}{a_3}+\cdots+\frac{a_n}{a_{n-1}}-a_2\left(\frac{1}{a_2}+\frac{1}{a_3}+\cdots+\frac{1}{a_{n-2}}\right)\]

Answer 2

\[d=a_n-a_{n-1}=a_2-a_1=a_2\,\,,\,\,a_n=(n-1)d=(n-1)a_2\]
\[\frac{a_3}{a_2}+\frac{a_4}{a_3}+\cdots+\frac{a_n}{a_{n-1}}-a_2\left(\frac{1}{a_2}+\frac{1}{a_3}+\cdots+\frac{1}{a_{n-2}}\right)=\]
\[=1+\frac{a_2}{a_2}+1+\frac{a_2}{a_3}+\cdots+1+\frac{a_2}{a_{n-1}}-a_2\left(\frac{1}{a_2}+\frac{1}{a_3}+\cdots+\frac{1}{a_{n-2}}\right)=\]
\[=n-2+a_2\left(\frac{1}{a_2}+\frac{1}{a_3}+\cdots+\frac{1}{a_{n-1}}\right)-a_2\left(\frac{1}{a_2}+\frac{1}{a_3}+\cdots+\frac{1}{a_{n-2}}\right)=\]
\[=n-2+\frac{a_2}{a_{n-1}}=n-2+\frac{a_2}{(n-2)a_2}=n-2+\frac{1}{n-2}\,\,,\,\,n>2\]