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Question Number 94710 by abony1303 last updated on 20/May/20

Commented by PRITHWISH SEN 2 last updated on 20/May/20

2019^(th) term will be 89 then tbe remainder is 9 ans-(5)

$$\mathrm{2019}^{\mathrm{th}} \mathrm{term}\:\mathrm{will}\:\mathrm{be}\:\mathrm{89} \\ $$$$\mathrm{then}\:\mathrm{tbe}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{9} \\ $$$$\boldsymbol{\mathrm{ans}}-\left(\mathrm{5}\right) \\ $$

Commented by abony1303 last updated on 20/May/20

correct answer 2019^(th ) term is 89 but how?

$${correct}\:{answer}\:\mathrm{2019}^{{th}\:} {term}\:{is}\:\mathrm{89}\:{but}\:{how}? \\ $$

Answered by PRITHWISH SEN 2 last updated on 20/May/20

the sequence comprises of the sets of similar terms (1st set only 1 term),(2nd set 3 terms), and so on... the no. of terms in each set of the sequence can be written as 1,3,5,... and it is a new sequence representing the no. of terms of every set then 2019 th term will be the sum of the sequence upto n^(th) set or the n^(th) number of new sequence(say) ∴ sum upto n^(th) set = n^2 ∵ the sum of odd natural numbers upto n term be n^2 ∴ n^2 =2019 ⇒n∽44 that is upto 44^(th) set that is 44^2 =1936 terms ∴ for remaining (2019−1936)=83 terms the 2019^(th) number will be in 45th set ∴ in 45th set the number is (2×45−1)=89 thus 89 when divided by 10 gives 9 as a remainder.

$$\mathrm{the}\:\mathrm{sequence}\:\mathrm{comprises}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sets}\:\mathrm{of}\:\mathrm{similar}\:\mathrm{terms} \\ $$$$\left(\mathrm{1st}\:\mathrm{set}\:\mathrm{only}\:\mathrm{1}\:\mathrm{term}\right),\left(\mathrm{2nd}\:\mathrm{set}\:\mathrm{3}\:\mathrm{terms}\right),\:\mathrm{and}\:\mathrm{so}\:\mathrm{on}... \\ $$$$\mathrm{the}\:\mathrm{no}.\:\mathrm{of}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{each}\:\mathrm{set}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\: \\ $$$$\mathrm{1},\mathrm{3},\mathrm{5},...\:\mathrm{and}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{new}\:\mathrm{sequence}\:\mathrm{representing} \\ $$$$\mathrm{the}\:\mathrm{no}.\:\mathrm{of}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{every}\:\mathrm{set} \\ $$$$\mathrm{then}\:\mathrm{2019}\:\mathrm{th}\:\mathrm{term}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sequence} \\ $$$$\mathrm{upto}\:\mathrm{n}^{\mathrm{th}} \:\mathrm{set}\:\mathrm{or}\:\mathrm{the}\:\mathrm{n}^{\mathrm{th}} \:\mathrm{number}\:\mathrm{of}\:\mathrm{new}\:\mathrm{sequence}\left(\mathrm{say}\right) \\ $$$$\therefore\:\mathrm{sum}\:\mathrm{upto}\:\mathrm{n}^{\mathrm{th}} \mathrm{set}\:=\:\mathrm{n}^{\mathrm{2}} \:\because\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{odd}\:\mathrm{natural} \\ $$$$\mathrm{numbers}\:\mathrm{upto}\:\mathrm{n}\:\mathrm{term}\:\mathrm{be}\:\mathrm{n}^{\mathrm{2}} \\ $$$$\therefore\:\mathrm{n}^{\mathrm{2}} =\mathrm{2019}\:\Rightarrow\mathrm{n}\backsim\mathrm{44} \\ $$$$\mathrm{that}\:\mathrm{is}\:\mathrm{upto}\:\mathrm{44}^{\mathrm{th}} \mathrm{set}\:\mathrm{that}\:\mathrm{is}\:\mathrm{44}^{\mathrm{2}} =\mathrm{1936}\:\mathrm{terms} \\ $$$$\therefore\:\mathrm{for}\:\mathrm{remaining}\:\left(\mathrm{2019}−\mathrm{1936}\right)=\mathrm{83}\:\mathrm{terms}\:\mathrm{the}\:\mathrm{2019}^{\mathrm{th}} \\ $$$$\mathrm{number}\:\mathrm{will}\:\mathrm{be}\:\mathrm{in}\:\mathrm{45th}\:\mathrm{set} \\ $$$$\therefore\:\mathrm{in}\:\mathrm{45th}\:\mathrm{set}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\left(\mathrm{2}×\mathrm{45}−\mathrm{1}\right)=\mathrm{89} \\ $$$$\mathrm{thus}\:\mathrm{89}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{10}\:\mathrm{gives}\:\mathrm{9}\:\mathrm{as}\:\mathrm{a}\:\mathrm{remainder}. \\ $$$$ \\ $$

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