Question

If \[{{a}_{1}},{{a}_{2}},{{a}_{3}}\,........\,,{{a}_{n}}\] are in arithmetic progression. Prove that$\dfrac{1}{{{a}_{1}}{{a}_{2}}}+\dfrac{1}{{{a}_{2}}{{a}_{3}}}+\,.........\,+\dfrac{1}{{{a}_{n-1}}{{a}_{n}}}=\dfrac{n-1}{{{a}_{n-1}}{{a}_{n}}}$.

Answer 1

Hint: If \[a,b,c\] are in arithmetic progression,
$b=\dfrac{a+c}{2}$
$b=a+d=c-d$
$c=b+d=a+2d$
Here, ‘d’ is called the common difference of the arithmetic progression.
$d=c-b=b-a$
The general term in an arithmetic progression can be written as ${{a}_{n}}=a+(n-1)d$.

Complete step by step answer:
We are given, \[{{a}_{1}},{{a}_{2}},{{a}_{3}}\,........\,,{{a}_{n}}\] are in arithmetic progression.
We have to prove that $\dfrac{1}{{{a}_{1}}{{a}_{2}}}+\dfrac{1}{{{a}_{2}}{{a}_{3}}}+\,.........\,+\dfrac{1}{{{a}_{n-1}}{{a}_{n}}}=\dfrac{n-1}{{{a}_{n-1}}{{a}_{n}}}$.
LHS$=\dfrac{1}{{{a}_{1}}{{a}_{2}}}+\dfrac{1}{{{a}_{2}}{{a}_{3}}}+\,.........\,+\dfrac{1}{{{a}_{n-1}}{{a}_{n}}}$
We know that, ${{a}_{2}}={{a}_{1}}+d,\,{{a}_{3}}={{a}_{2}}+d,\,{{a}_{4}}={{a}_{3}}+d$ and so on.
Hence,
LHS$=\dfrac{1}{{{a}_{1}}({{a}_{1}}+d)}+\dfrac{1}{{{a}_{2}}({{a}_{2}}+d)}+\,.........\,+\dfrac{1}{{{a}_{n-1}}({{a}_{n-1}}+d)}$
Each term in the LHS is of the form $\dfrac{1}{a(a+d)}$ . Multiplying numerator and denominator by $d$ we get each terms in the form $\dfrac{1}{d}\left( \dfrac{d}{a(a+d)} \right)$ . Now we can split each terms of the LHS as $\dfrac{1}{d}\left( \dfrac{d}{a(a+d)} \right)=\dfrac{1}{d}\left( \dfrac{1}{a}-\dfrac{1}{a+d} \right)$ .
Hence, the LHS becomes,
LHS$=\dfrac{1}{d}\left( \dfrac{1}{{{a}_{1}}}-\dfrac{1}{{{a}_{1}}+d} \right)+\dfrac{1}{d}\left( \dfrac{1}{{{a}_{2}}}-\dfrac{1}{{{a}_{2}}+d} \right)+.......+\dfrac{1}{d}\left( \dfrac{1}{{{a}_{n-1}}}-\dfrac{1}{{{a}_{n-1}}+d} \right)$
$\dfrac{1}{d}$is common for all the terms in the LHS. Hence, on taking the common term outside we get,
LHS$=\dfrac{1}{d}\left( \dfrac{1}{{{a}_{1}}}-\dfrac{1}{{{a}_{1}}+d}+\dfrac{1}{{{a}_{2}}}-\dfrac{1}{{{a}_{2}}+d}+.......+\dfrac{1}{{{a}_{n-1}}}-\dfrac{1}{{{a}_{n-1}}+d} \right)$
We know that, ${{a}_{2}}={{a}_{1}}+d,\,{{a}_{3}}={{a}_{2}}+d,\,{{a}_{4}}={{a}_{3}}+d$ and so on.
On substituting the above in the LHS, we get,
LHS\[=\dfrac{1}{d}\left( \dfrac{1}{{{a}_{1}}}-\dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{2}}}-\dfrac{1}{{{a}_{3}}}+.......+\dfrac{1}{{{a}_{n-2}}}-\dfrac{1}{{{a}_{n-1}}}+\dfrac{1}{{{a}_{n-1}}}-\dfrac{1}{{{a}_{n}}} \right)\]
\[=\dfrac{1}{d}\left( \dfrac{1}{{{a}_{1}}}-\dfrac{1}{{{a}_{n}}} \right)\]
On taking LCM, we get,
LHS\[=\dfrac{1}{d}\left( \dfrac{{{a}_{n}}-{{a}_{1}}}{{{a}_{1}}{{a}_{n}}} \right)\]
We know that, ${{a}_{n}}={{a}_{1}}+(n-1)d$
Substituting ${{a}_{n}}={{a}_{1}}+(n-1)d$ in LHS, we obtain,
LHS\[=\dfrac{1}{d}\left( \dfrac{{{a}_{1}}+(n-1)d-{{a}_{1}}}{{{a}_{1}}{{a}_{n}}} \right)\]
\[=\dfrac{1}{d}\left( \dfrac{(n-1)d}{{{a}_{1}}{{a}_{n}}} \right)\]
\[=\dfrac{n-1}{{{a}_{1}}{{a}_{n}}}\]
$\therefore $ LHS\[=\dfrac{n-1}{{{a}_{1}}{{a}_{n}}}\] = RHS
That is, LHS=RHS.
Hence, we have proved that $\dfrac{1}{{{a}_{1}}{{a}_{2}}}+\dfrac{1}{{{a}_{2}}{{a}_{3}}}+\,.........\,+\dfrac{1}{{{a}_{n-1}}{{a}_{n}}}=\dfrac{n-1}{{{a}_{n-1}}{{a}_{n}}}$.

Note: Similar questions can be asked in terms of the progression being even geometric and harmonic. In other words, the question can be, “If \[{{a}_{1}},{{a}_{2}},{{a}_{3}}\,........\,,{{a}_{n}}\] are in geometric progression” or “If \[{{a}_{1}},{{a}_{2}},{{a}_{3}}\,........\,,{{a}_{n}}\] are in harmonic progression”, solve the equation. In both cases, we can solve the question in a way similar to what we have already discussed.
We need only keep in mind some basic properties of geometric and harmonic progressions.
Some properties of geometric progression are:
Suppose $a,b,c$ are in G.P. Then,
${{b}^{2}}=ac$
$b=ar=\dfrac{c}{r}$
$c=br=a{{r}^{2}}$
Here ‘$r$’ is called the common ratio of the geometric.
Some properties of harmonic progression are:
Suppose $a,b,c$ are in H.P. Then,
$\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b}$
The reciprocals of the terms in H.P are in arithmetic progression.
$b=\dfrac{2ac}{a+c}$