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Question Number 4704 by lakshaysethi039 last updated on 25/Feb/16

If a_1 ,a_2 ,.........a_n be an arithmetic progression, then show that (1/(a_1 a_n )) + (1/(a_2 a_(n−1) )) + (1/(a_3 a_(n−2) )) +.............+(1/(a_n a_1 )) = (2/(a_1 +a_n ))((1/a_1 ) + (1/a_2 ) + ......+(1/a_n ))

$${If}\:{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,.........{a}_{{n}} {be}\:{an}\:{arithmetic}\:{progression}, \\ $$$${then}\:{show}\:{that} \\ $$$$\frac{\mathrm{1}}{{a}_{\mathrm{1}} {a}_{{n}} }\:+\:\frac{\mathrm{1}}{{a}_{\mathrm{2}} {a}_{{n}−\mathrm{1}} }\:+\:\frac{\mathrm{1}}{{a}_{\mathrm{3}} {a}_{{n}−\mathrm{2}} }\:+.............+\frac{\mathrm{1}}{{a}_{{n}} {a}_{\mathrm{1}} }\: \\ $$$$=\:\frac{\mathrm{2}}{{a}_{\mathrm{1}} +{a}_{{n}} }\left(\frac{\mathrm{1}}{{a}_{\mathrm{1}} }\:+\:\frac{\mathrm{1}}{{a}_{\mathrm{2}} }\:+\:......+\frac{\mathrm{1}}{{a}_{{n}} }\right) \\ $$$$ \\ $$$$ \\ $$

Commented by Yozzii last updated on 25/Feb/16

Interesting property of A.Ps of non−zero terms. I′ve never seen this before.

$${Interesting}\:{property}\:{of}\:{A}.{Ps}\:{of}\: \\ $$$${non}−{zero}\:{terms}.\:{I}'{ve}\:{never}\:{seen}\:{this} \\ $$$${before}. \\ $$

Answered by Yozzii last updated on 25/Feb/16

Let S_n =Σ_(i=1) ^n a_i ^(−1) and J=(1/(a_1 a_n ))+(1/(a_2 a_(n−1) ))+(1/(a_3 a_(n−2) ))...+(1/(a_n a_1 )). First calculate sums that yield the denominators of terms of J in terms of a_i . (1) a_1 ^(−1) +a_n ^(−1) =((a_1 +a_n )/(a_1 a_n )) (2) a_2 ^(−1) +a_(n−1) ^(−1) =((a_2 +a_(n−1) )/(a_2 a_(n−1) )) (3) a_3 ^(−1) +a_(n−2) ^(−1) =((a_3 +a_(n−2) )/(a_3 a_(n−2) )) (4) a_4 ^(−1) +a_(n−3) ^(−1) =((a_4 +a_(n−3) )/(a_4 a_(n−3) )) ⋮ (n) a_n ^(−1) +a_1 ^(−1) =((a_n +a_1 )/(a_1 a_n )) Adding lines (1)→(n) we see that all a_i ^(−1) are added twice. Hence, we may write 2(Σ_(i=1) ^n a_i ^(−1) )=((a_1 +a_n )/(a_1 a_n ))+((a_2 +a_(n−1) )/(a_2 a_(n−1) ))+((a_3 +a_(n−2) )/(a_3 a_(n−2) ))+((a_4 +a_(n−3) )/(a_4 a_(n−3) ))+...+((a_n +a_1 )/(a_1 a_n )). (∗) We are given that {a_i }_(i=1) ^n is an arithmetic progression. Therefore, a_2 −a_1 =a_3 −a_2 =a_4 −a_3 =...=a_(n−2) −a_(n−3) =a_(n−1) −a_(n−2) =a_n −a_(n−1) . Using these equations we obtain a_2 −a_1 =a_n −a_(n−1 ) , or a_1 +a_n =a_2 +a_(n−1) . Similarly we have a_(n−1) −a_(n−2) =a_3 −a_2 so that a_(n−2) +a_3 =a_(n−1) +a_2 =a_1 +a_n . Also, a_(n−2) −a_(n−3) =a_4 −a_3 ⇒a_4 +a_(n−3) =a_3 +a_(n−2) =a_1 +a_n . Upon repeated use of this process we obtain the numerators of the terms on the r.h.s of (∗) all being equal to a_1 +a_n . ∴(a_1 +a_n ){(1/(a_1 a_n ))+(1/(a_2 a_(n−1) ))+(1/(a_3 a_(n−2) ))+(1/(a_4 a_(n−3) ))+...+(1/(a_1 a_n ))}=2S_n or (a_1 +a_n )J=2S_n ⇒J=(2/(a_1 +a_n ))S_n This is valid iff a_i ≠0 for 1≤i≤n. □

$${Let}\:{S}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} ^{−\mathrm{1}} \:{and}\:{J}=\frac{\mathrm{1}}{{a}_{\mathrm{1}} {a}_{{n}} }+\frac{\mathrm{1}}{{a}_{\mathrm{2}} {a}_{{n}−\mathrm{1}} }+\frac{\mathrm{1}}{{a}_{\mathrm{3}} {a}_{{n}−\mathrm{2}} }...+\frac{\mathrm{1}}{{a}_{{n}} {a}_{\mathrm{1}} }. \\ $$$${First}\:{calculate}\:{sums}\:{that}\:{yield}\:{the}\: \\ $$$${denominators}\:{of}\:{terms}\:{of}\:{J}\:{in}\:{terms}\:{of}\:{a}_{{i}} . \\ $$$$\left(\mathrm{1}\right)\:{a}_{\mathrm{1}} ^{−\mathrm{1}} +{a}_{{n}} ^{−\mathrm{1}} =\frac{{a}_{\mathrm{1}} +{a}_{{n}} }{{a}_{\mathrm{1}} {a}_{{n}} } \\ $$$$\left(\mathrm{2}\right)\:{a}_{\mathrm{2}} ^{−\mathrm{1}} +{a}_{{n}−\mathrm{1}} ^{−\mathrm{1}} =\frac{{a}_{\mathrm{2}} +{a}_{{n}−\mathrm{1}} }{{a}_{\mathrm{2}} {a}_{{n}−\mathrm{1}} } \\ $$$$\left(\mathrm{3}\right)\:{a}_{\mathrm{3}} ^{−\mathrm{1}} +{a}_{{n}−\mathrm{2}} ^{−\mathrm{1}} =\frac{{a}_{\mathrm{3}} +{a}_{{n}−\mathrm{2}} }{{a}_{\mathrm{3}} {a}_{{n}−\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:{a}_{\mathrm{4}} ^{−\mathrm{1}} +{a}_{{n}−\mathrm{3}} ^{−\mathrm{1}} =\frac{{a}_{\mathrm{4}} +{a}_{{n}−\mathrm{3}} }{{a}_{\mathrm{4}} {a}_{{n}−\mathrm{3}} } \\ $$$$\vdots \\ $$$$\left({n}\right)\:{a}_{{n}} ^{−\mathrm{1}} +{a}_{\mathrm{1}} ^{−\mathrm{1}} =\frac{{a}_{{n}} +{a}_{\mathrm{1}} }{{a}_{\mathrm{1}} {a}_{{n}} } \\ $$$${Adding}\:{lines}\:\left(\mathrm{1}\right)\rightarrow\left({n}\right)\:{we}\:{see}\:{that}\: \\ $$$${all}\:{a}_{{i}} ^{−\mathrm{1}} \:{are}\:{added}\:{twice}.\:{Hence},\:{we}\:{may} \\ $$$${write}\: \\ $$$$\mathrm{2}\left(\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} ^{−\mathrm{1}} \right)=\frac{{a}_{\mathrm{1}} +{a}_{{n}} }{{a}_{\mathrm{1}} {a}_{{n}} }+\frac{{a}_{\mathrm{2}} +{a}_{{n}−\mathrm{1}} }{{a}_{\mathrm{2}} {a}_{{n}−\mathrm{1}} }+\frac{{a}_{\mathrm{3}} +{a}_{{n}−\mathrm{2}} }{{a}_{\mathrm{3}} {a}_{{n}−\mathrm{2}} }+\frac{{a}_{\mathrm{4}} +{a}_{{n}−\mathrm{3}} }{{a}_{\mathrm{4}} {a}_{{n}−\mathrm{3}} }+...+\frac{{a}_{{n}} +{a}_{\mathrm{1}} }{{a}_{\mathrm{1}} {a}_{{n}} }.\:\:\left(\ast\right) \\ $$$${We}\:{are}\:{given}\:{that}\:\left\{{a}_{{i}} \right\}_{{i}=\mathrm{1}} ^{{n}} \:{is}\:{an}\: \\ $$$${arithmetic}\:{progression}.\:{Therefore}, \\ $$$${a}_{\mathrm{2}} −{a}_{\mathrm{1}} ={a}_{\mathrm{3}} −{a}_{\mathrm{2}} ={a}_{\mathrm{4}} −{a}_{\mathrm{3}} =...={a}_{{n}−\mathrm{2}} −{a}_{{n}−\mathrm{3}} ={a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} ={a}_{{n}} −{a}_{{n}−\mathrm{1}} . \\ $$$${Using}\:{these}\:{equations}\:{we}\:{obtain} \\ $$$${a}_{\mathrm{2}} −{a}_{\mathrm{1}} ={a}_{{n}} −{a}_{{n}−\mathrm{1}\:} ,\:{or}\:{a}_{\mathrm{1}} +{a}_{{n}} ={a}_{\mathrm{2}} +{a}_{{n}−\mathrm{1}} . \\ $$$${Similarly}\:{we}\:{have}\:{a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} ={a}_{\mathrm{3}} −{a}_{\mathrm{2}} \\ $$$${so}\:{that}\:{a}_{{n}−\mathrm{2}} +{a}_{\mathrm{3}} ={a}_{{n}−\mathrm{1}} +{a}_{\mathrm{2}} ={a}_{\mathrm{1}} +{a}_{{n}} . \\ $$$${Also},\:{a}_{{n}−\mathrm{2}} −{a}_{{n}−\mathrm{3}} ={a}_{\mathrm{4}} −{a}_{\mathrm{3}} \\ $$$$\Rightarrow{a}_{\mathrm{4}} +{a}_{{n}−\mathrm{3}} ={a}_{\mathrm{3}} +{a}_{{n}−\mathrm{2}} ={a}_{\mathrm{1}} +{a}_{{n}} . \\ $$$${Upon}\:{repeated}\:{use}\:{of}\:{this}\:{process}\:{we} \\ $$$${obtain}\:{the}\:{numerators}\:{of}\:{the}\:{terms}\:{on} \\ $$$${the}\:{r}.{h}.{s}\:{of}\:\left(\ast\right)\:{all}\:{being}\:{equal}\:{to}\:{a}_{\mathrm{1}} +{a}_{{n}} . \\ $$$$\therefore\left({a}_{\mathrm{1}} +{a}_{{n}} \right)\left\{\frac{\mathrm{1}}{{a}_{\mathrm{1}} {a}_{{n}} }+\frac{\mathrm{1}}{{a}_{\mathrm{2}} {a}_{{n}−\mathrm{1}} }+\frac{\mathrm{1}}{{a}_{\mathrm{3}} {a}_{{n}−\mathrm{2}} }+\frac{\mathrm{1}}{{a}_{\mathrm{4}} {a}_{{n}−\mathrm{3}} }+...+\frac{\mathrm{1}}{{a}_{\mathrm{1}} {a}_{{n}} }\right\}=\mathrm{2}{S}_{{n}} \\ $$$${or}\:\left({a}_{\mathrm{1}} +{a}_{{n}} \right){J}=\mathrm{2}{S}_{{n}} \\ $$$$\Rightarrow{J}=\frac{\mathrm{2}}{{a}_{\mathrm{1}} +{a}_{{n}} }{S}_{{n}} \:\:\:\: \\ $$$${This}\:{is}\:{valid}\:{iff}\:{a}_{{i}} \neq\mathrm{0}\:{for}\:\mathrm{1}\leqslant{i}\leqslant{n}.\:\:\:\:\:\Box \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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