Could the double-slit experiment negate the no-communication theorem?
Answers
Could the double-slit experiment negate the no-communication theorem?
In quantum physics, the no-communication theorem states that it is not possible to transmit information from one observer to another observer, whether entangled or not, by making a measurement of a subsystem of a total state, common to both observers.
Most people think that the theorem is important because it limits quantum entanglement, that separated events cannot be correlated in any way to lead to the possibility of communication.
However, the double-slit experiment [1] says that what one observer does (e.g. turn on a detector) influences what is detected at the other observer (e.g. the electron did not pass here).
Explanation:
Given that , In an Airthmetic Progression [ A.P. ] sum of it's n th terms is 3n²+5n & it's k th terms is 164 .
Exigency To Find : The value of k ?
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\begin{gathered}\maltese\:\:\: \sf Let's \:say \:that \:, \:\pmb{\bf S_n } \: be \:the \:sum \:of \;n \:terms \:of \:an \:A.P\: & \:,\:\\ \sf \pmb{\sf a_n } \:be \: n^{th} \:term \:of \:an \:A.P \: \:[ \:Airthmetic \:Progression \:] \:\\\\\end{gathered}✠Let′ssaythat,SnSnbethesumofntermsofanA.PananbenthtermofanA.P[AirthmeticProgression],
Given that ,
Sum of n terms of an A.P is 3n²+5n .
\begin{gathered}\qquad \sf \therefore \;\: S_n \: = \: 3n^2 \: + \: 5n \: \\\\\qquad \dashrightarrow \:\sf \;\: S_n \: = \: 3n^2 \: + \: 5n \: \\\\\end{gathered}∴Sn=3n2+5n⇢Sn=3n2+5n
Therefore,
Sum of n - 1 terms of an Airthmetic Progression [ A.P. ] .
\begin{gathered}\qquad \sf \therefore \;\: S_{ n- 1 } \: = \: 3( n - 1 )^2 \: + \: 5( n - 1 ) \: \\\\\qquad \dashrightarrow \sf \;\: S_{ n- 1 } \: = \: 3( n - 1 )^2 \: + \: 5( n - 1 ) \: \\\\\qquad \dashrightarrow \sf \;\: S_{ n- 1 } \: = \: 3n^2 - \: n - 2 \: \\\\\end{gathered}∴Sn−1=3(n−1)2+5(n−1)⇢Sn−1=3(n−1)2+5(n−1)⇢Sn−1=3n2−n−2
Now ,
As , We know that ,
\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{ a_n \: =\: S_n - \: S_{n-1} }\bigg\rgroup \\\\\qquad \dashrightarrow \sf a_n \: =\: S_n - \: S_{n-1} \:\:\\\\\end{gathered}†⎩⎪⎪⎪⎧an=Sn−Sn−1⎭⎪⎪⎪⎫⇢an=Sn−Sn−1
⠀⠀⠀⠀⠀⠀\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\end{gathered}⋆NowBySubstitutingtheknownValues:
\begin{gathered}\qquad \dashrightarrow \sf a_n \: =\: S_n - \: S_{n-1} \:\:\\\\\qquad \dashrightarrow \sf a_n \: =\: \Big\{ 3n^2 + 5n \:\Big\} - \: \Big\{ 3n^2 - n - 2 \Big\} \:\:\\\\\qquad \dashrightarrow \sf a_n \: =\: 3n^2 + 5n \: - \: 3n^2 + n + 2 \:\:\\\\\qquad \dashrightarrow \sf a_n \: =\: 6n + 2 \:\:\\\\\dashrightarrow \underline {\boxed{\pmb{\frak{\purple { a_n \: =\: 6n + 2 }}}}}\:\\\\\end{gathered}⇢an=Sn−Sn−1⇢an={3n2+5n}−{3n2−n−2}⇢an=3n2+5n−3n2+n+2⇢an=6n+2⇢an=6n+2an=6n+2
Therefore,
\sf k^{th}kth term will be 6k + 2 .
AND ,
In an Airthmetic Progression [ A.P. ] \sf k^{th}kth term is 164 .
\begin{gathered}\qquad \qquad \sf \leadsto \;\: a_k \: = \: 6k +2 \: \& ,\:\\\\\qquad \sf \leadsto \;\: a_k \: = \: 164 \: \\\\\qquad \sf \therefore \;\: 164 \: = \: 6k +2 \: \\\\\qquad \sf \dashrightarrow \;\: 164 \: = \: 6k +2 \: \\\\\qquad \sf \dashrightarrow \;\: 164 - 2 \: = \: 6k \: \\\\\qquad \sf \dashrightarrow \;\: 162 \: = \: 6k \: \\\\\qquad \sf \dashrightarrow \;\: k \: = \: \dfrac{162 }{6} \: \\\\\qquad \sf \dashrightarrow \;\: k \: = \: 27 \: \\\\\dashrightarrow \underline {\boxed{\pmb{\frak{\purple { k \: =\: 27 }}}}}\:\\\\\end{gathered}⇝ak=6k+2&,⇝ak=164∴164=6k+2⇢164=6k+2⇢164−2=6k⇢162=6k⇢k=6162⇢k=27⇢
