Question Number 61117 by Tawa1 last updated on 29/May/19 $$\mathrm{The}\:\mathrm{2nd},\:\mathrm{4th}\:\mathrm{and}\:\mathrm{8th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{are}\:\mathrm{the}\:\mathrm{consecutive}\:\mathrm{term}\:\mathrm{of}\:\mathrm{a}\:\mathrm{GP}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{3rd}\:\mathrm{and}\:\mathrm{4th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{AP}\:\mathrm{is}\:\mathrm{20}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{first}\:\mathrm{four}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{AP}. \\ $$ Answered by Kunal12588 last updated on 29/May/19 $${given}\:\:{a}_{\mathrm{2}} ,{a}_{\mathrm{4}}…
Question Number 192186 by Tawa11 last updated on 11/May/23 $$\mathrm{If}\:\:\alpha,\:\beta\:\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\:\:\mathrm{x}^{\mathrm{3}} \:\:+\:\:\mathrm{px}\:\:+\:\:\mathrm{q}\:\:=\:\:\mathrm{0},\:\:\:\:\mathrm{find}\:\:\:\Sigma\alpha^{\mathrm{4}} . \\ $$ Answered by BaliramKumar last updated on 10/May/23 $$\mathrm{2p}^{\mathrm{2}} \\ $$ Commented…
Question Number 61111 by Tawa1 last updated on 29/May/19 $$\mathrm{Please}\:\mathrm{what}\:\mathrm{does}\:\mathrm{the}\:\mathrm{2}\:\mathrm{on}\:\mathrm{the}\:\mathrm{C}\:\mathrm{mean}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{C}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{2}\:\mathrm{C}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{3}\:\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} \:+\:…\:+\:\mathrm{n}\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2}} \:\:\:\:=\:\:\:\frac{\left(\mathrm{2n}\:−\:\mathrm{1}\right)!}{\left[\left(\mathrm{n}\:−\:\mathrm{1}\right)!\right]^{\mathrm{2}} } \\ $$$$\mathrm{Does}\:\mathrm{the}\:\mathrm{2}\:\mathrm{on}\:\mathrm{C}\:\mathrm{mean}\:\mathrm{square}\:??…
Question Number 192173 by universe last updated on 10/May/23 $${prove}\:{that} \\ $$$$\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{4}} \:=\:\left({x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{2}} {a}^{\mathrm{2}} +{a}^{\mathrm{4}} \right)^{\mathrm{2}} +\left(\mathrm{4}{x}^{\mathrm{3}} {a}−\mathrm{4}{xa}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$ Answered…
Question Number 192172 by mehdee42 last updated on 10/May/23 $${Q}\mathrm{1}\:\therefore\:\:{x}=<\mathrm{1}{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} >\in\mathbb{N}\:\:\&\:\:{y}=<{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} \mathrm{1}>\in\mathbb{N} \\ $$$${if}\:\:{y}=\mathrm{3}{x}\:\:{then}\:\:,\:{find}\:{the}\:{smallest}\: \\ $$$${value}\:{of}\:\:{x} \\ $$$${Q}\mathrm{2}\:\therefore\:{with}\:{the}\:{above}\:{conditions}\:,{what}\:{other}\:{values}\: \\ $$$${can}\:{be}\:{placed}\:\:{besides}\:{the}\:{number}\:“\:\mathrm{1}\:''\: \\…
Question Number 192171 by mathlove last updated on 10/May/23 $$\mathrm{2}^{{x}^{{x}^{{x}} } } =\mathrm{2}^{\sqrt{\mathrm{2}}} \\ $$$${x}=? \\ $$ Commented by Frix last updated on 11/May/23 $$\mathrm{2}^{{x}^{{x}^{{x}}…
Question Number 126619 by O Predador last updated on 22/Dec/20 $$\: \\ $$$$\:\:\mathrm{5}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{7}^{\boldsymbol{\mathrm{x}}} \:=\:\left(\mathrm{9}.\mathrm{8}\right)^{\boldsymbol{\mathrm{x}}} \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$$$\: \\ $$ Answered by…
Question Number 192149 by Shrinava last updated on 09/May/23 $$\mathrm{Find}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }\:+\:\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }\:+\:\frac{\mathrm{7}}{\mathrm{2}^{\mathrm{7}} }\:+\:… \\ $$ Answered by aleks041103 last updated on 09/May/23 $$\underset{{k}=\mathrm{1}}…
Question Number 192143 by Shrinava last updated on 09/May/23 $$\mathrm{Find}: \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}\:+\:\frac{\mathrm{77}}{\mathrm{22}}\:+\:\frac{\mathrm{777}}{\mathrm{222}}\:+\:\frac{\mathrm{7777}}{\mathrm{2222}}\:+…+\:\frac{\mathrm{77777777}}{\mathrm{22222222}} \\ $$ Commented by AST last updated on 09/May/23 $$=\frac{\mathrm{7}}{\mathrm{2}}×\mathrm{8}=\mathrm{28} \\ $$ Commented…
Question Number 192142 by mehdee42 last updated on 09/May/23 $${Question} \\ $$$${let}\:\:\:{x}=<{a}_{{n}} {a}_{{n}−\mathrm{1}} …{a}_{\mathrm{1}} {a}_{\mathrm{0}} >\:\in\mathbb{N}\:;\:{a}_{\mathrm{0}} \neq\mathrm{0}\:\:\&\: \\ $$$$\:{y}=<{a}_{{n}} {a}_{{n}−\mathrm{1}} …{a}_{\mathrm{1}} >\:\in\mathbb{N}\:\:{be}\: \\ $$$${two}\:{natural}\:{numbers}\: \\…