8 Anu wants to fence the garden in front of her
house (Fig 11.5), on three sides with lengths
20 m, 12 m and 12 m. Find the cost of fencing
at the rate of 150 per metre.
The length of the fence required is the perimeter
of the garden (excluding one side) which is
equal to 20 m + 12 m+ 12 m, i.e., 44 m.
Cost of fencing = * 150 x 44 = 6,600.
Answers
Answer:
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⇢