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Question Number 213797 by efronzo1 last updated on 17/Nov/24

Answered by Frix last updated on 17/Nov/24

Σ_(n=1) ^(20) (1/(a_n a_(n+1) )) =((20)/(a_1 (a_1 +20d)))=(4/9) Σ_(n=1) ^(21) a_n =21a_1 +210d=189 ⇒ { ((a_1 =3∧d=(3/5) ⇒ a_6 =6∧a_(16) =12)),((a_1 =15∧d=−(3/5) ⇒ a_6 =12∧a_(16) =6)) :} ⇒ a_6 a_(16) =72

$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{20}} {\sum}}\:\frac{\mathrm{1}}{{a}_{{n}} {a}_{{n}+\mathrm{1}} }\:=\frac{\mathrm{20}}{{a}_{\mathrm{1}} \left({a}_{\mathrm{1}} +\mathrm{20}{d}\right)}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{21}} {\sum}}\:{a}_{{n}} \:=\mathrm{21}{a}_{\mathrm{1}} +\mathrm{210}{d}=\mathrm{189} \\ $$$$\Rightarrow \\ $$$$\begin{cases}{{a}_{\mathrm{1}} =\mathrm{3}\wedge{d}=\frac{\mathrm{3}}{\mathrm{5}}\:\Rightarrow\:{a}_{\mathrm{6}} =\mathrm{6}\wedge{a}_{\mathrm{16}} =\mathrm{12}}\\{{a}_{\mathrm{1}} =\mathrm{15}\wedge{d}=−\frac{\mathrm{3}}{\mathrm{5}}\:\Rightarrow\:{a}_{\mathrm{6}} =\mathrm{12}\wedge{a}_{\mathrm{16}} =\mathrm{6}}\end{cases} \\ $$$$\Rightarrow \\ $$$${a}_{\mathrm{6}} {a}_{\mathrm{16}} =\mathrm{72} \\ $$

Commented by efronzo1 last updated on 17/Nov/24

Why Σ_(n=1) ^(20) (1/(a_n .a_(n+1) )) = ((20)/(a_1 .a_(21) )) =?

$$\:\mathrm{Why}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{20}} {\sum}}\:\frac{\mathrm{1}}{{a}_{{n}} .{a}_{{n}+\mathrm{1}} }\:=\:\frac{\mathrm{20}}{{a}_{\mathrm{1}} .{a}_{\mathrm{21}} }\:=? \\ $$

Commented by mehdee7396 last updated on 17/Nov/24

(1/(a_1 a_2 ))=(1/d)((1/a_1 )−(1/a_2 )) (1/(a_2 a_3 ))=(1/d)((1/a_2 )−(1/a_3 )) , ....,(1/(a_(20) a_(21) ))=(1/d)((1/a_(20) )−(1/a_(21) )) ⇒Σ_1 ^(20) (1/(a_n a_(n+1) ))=(1/d)((1/a_1 )−(1/a_(21) ))=((20d)/(da_1 a_(21) ))=((20)/(a_1 a_(21) ))

$$\frac{\mathrm{1}}{{a}_{\mathrm{1}} {a}_{\mathrm{2}} }=\frac{\mathrm{1}}{{d}}\left(\frac{\mathrm{1}}{{a}_{\mathrm{1}} }−\frac{\mathrm{1}}{{a}_{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{1}}{{a}_{\mathrm{2}} {a}_{\mathrm{3}} }=\frac{\mathrm{1}}{{d}}\left(\frac{\mathrm{1}}{{a}_{\mathrm{2}} }−\frac{\mathrm{1}}{{a}_{\mathrm{3}} }\right)\:,\:....,\frac{\mathrm{1}}{{a}_{\mathrm{20}} {a}_{\mathrm{21}} }=\frac{\mathrm{1}}{{d}}\left(\frac{\mathrm{1}}{{a}_{\mathrm{20}} }−\frac{\mathrm{1}}{{a}_{\mathrm{21}} }\right) \\ $$$$\Rightarrow\underset{\mathrm{1}} {\overset{\mathrm{20}} {\sum}}\frac{\mathrm{1}}{{a}_{{n}} {a}_{{n}+\mathrm{1}} }=\frac{\mathrm{1}}{{d}}\left(\frac{\mathrm{1}}{{a}_{\mathrm{1}} }−\frac{\mathrm{1}}{{a}_{\mathrm{21}} }\right)=\frac{\mathrm{20}{d}}{{da}_{\mathrm{1}} {a}_{\mathrm{21}} }=\frac{\mathrm{20}}{{a}_{\mathrm{1}} {a}_{\mathrm{21}} } \\ $$$$ \\ $$

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