Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 4485 by Rasheed Soomro last updated on 31/Jan/16

By murging two sequences a_1 ,a_2 ,...a_n and b_1 ,b_2 ,...,b_n a new sequence a_1 ,b_1 ,a_2 ,b_2 ,...,a_n ,b_n has been produced. Determine a single-formula general term of this new sequence in terms of a_n and b_n .

$${By}\:\:{murging}\:\:{two}\:{sequences} \\ $$$$\:\:\:\:{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,...{a}_{{n}} \:\:\:{and}\:\:{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,...,{b}_{{n}} \\ $$$${a}\:{new}\:{sequence}\: \\ $$$$\:\:\:\:{a}_{\mathrm{1}} ,{b}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,{b}_{\mathrm{2}} ,...,{a}_{{n}} ,{b}_{{n}} \:\:{has}\:{been} \\ $$$${produced}. \\ $$$$\:\:\:\:\:\:{Determine}\:{a}\:\mathrm{single}-\mathrm{formula} \\ $$$$\mathrm{general}\:\mathrm{term}\:{of}\:{this}\:{new}\:{sequence} \\ $$$${in}\:{terms}\:{of}\:{a}_{{n}} \:{and}\:{b}_{{n}} \:. \\ $$

Commented by prakash jain last updated on 31/Jan/16

c_n =(1/2)[a_n +(−1)^(n+1) a_n +b_n +(−1)^n b_n ]

$${c}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left[{a}_{{n}} +\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {a}_{{n}} +{b}_{{n}} +\left(−\mathrm{1}\right)^{{n}} {b}_{{n}} \right] \\ $$

Commented by FilupSmith last updated on 01/Feb/16

How did you come to this solution?

$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{come}\:\mathrm{to}\:\mathrm{this}\:\mathrm{solution}? \\ $$

Commented by prakash jain last updated on 01/Feb/16

My solution is wrong. I gave solution for it was a_1 ,b_2 ,a_3 ,c_4

$$\mathrm{My}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{wrong}.\:\mathrm{I}\:\mathrm{gave}\:\mathrm{solution}\:\mathrm{for} \\ $$$$\mathrm{it}\:\mathrm{was}\:\mathrm{a}_{\mathrm{1}} ,\mathrm{b}_{\mathrm{2}} ,\mathrm{a}_{\mathrm{3}} ,\mathrm{c}_{\mathrm{4}} \\ $$

Commented by Yozzii last updated on 01/Feb/16

c_r =a_(⌊((r+1)/2)⌋) ∣sin(((rπ)/2))∣+b_(⌊((r+1)/2)⌋) ∣sin((((r+1)π)/2))∣ r=1,2,3,4,...,2n−3,2n−2,2n−1,2n. c_1 =a_1 ×1+b_1 ×0=a_1 c_2 =a_1 ×0+b_1 ×1=b_1 c_3 =a_2 ×1+b_2 ×0=a_2 c_4 =a_2 ×0+b_2 ×1=b_2 ⋮ c_(2n−2) =a_(n−1) ×0+b_(n−1) ×1=b_(n−1) c_(2n−1) =a_n ×1+b_n ×0=a_n c_(2n) =a_n ×0+b_n ×1=b_n

$${c}_{{r}} ={a}_{\lfloor\frac{{r}+\mathrm{1}}{\mathrm{2}}\rfloor} \mid{sin}\left(\frac{{r}\pi}{\mathrm{2}}\right)\mid+{b}_{\lfloor\frac{{r}+\mathrm{1}}{\mathrm{2}}\rfloor} \mid{sin}\left(\frac{\left({r}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)\mid \\ $$$${r}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},...,\mathrm{2}{n}−\mathrm{3},\mathrm{2}{n}−\mathrm{2},\mathrm{2}{n}−\mathrm{1},\mathrm{2}{n}. \\ $$$${c}_{\mathrm{1}} ={a}_{\mathrm{1}} ×\mathrm{1}+{b}_{\mathrm{1}} ×\mathrm{0}={a}_{\mathrm{1}} \\ $$$${c}_{\mathrm{2}} ={a}_{\mathrm{1}} ×\mathrm{0}+{b}_{\mathrm{1}} ×\mathrm{1}={b}_{\mathrm{1}} \\ $$$${c}_{\mathrm{3}} ={a}_{\mathrm{2}} ×\mathrm{1}+{b}_{\mathrm{2}} ×\mathrm{0}={a}_{\mathrm{2}} \\ $$$${c}_{\mathrm{4}} ={a}_{\mathrm{2}} ×\mathrm{0}+{b}_{\mathrm{2}} ×\mathrm{1}={b}_{\mathrm{2}} \\ $$$$\vdots \\ $$$${c}_{\mathrm{2}{n}−\mathrm{2}} ={a}_{{n}−\mathrm{1}} ×\mathrm{0}+{b}_{{n}−\mathrm{1}} ×\mathrm{1}={b}_{{n}−\mathrm{1}} \\ $$$${c}_{\mathrm{2}{n}−\mathrm{1}} ={a}_{{n}} ×\mathrm{1}+{b}_{{n}} ×\mathrm{0}={a}_{{n}} \\ $$$${c}_{\mathrm{2}{n}} ={a}_{{n}} ×\mathrm{0}+{b}_{{n}} ×\mathrm{1}={b}_{{n}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com