Question

♠ QUESTION :

PROPER ANSWER NEEDED.

Answer 1

$\boxed{\mathsf{ Solution : }}$

$\mathsf{ Formula \: to \: be \: used \: for \: an \: A.P. \: : \: } \\ \\ \\ \mathsf{ \implies T_n \: = \: a \: + \: ( \: n \: - \: 1 \: )d \: } \\ \\ \\ \mathsf{ Here , } \\ \\ \\\mathsf{\implies T_n \: = \: Nth \: term \: of \: an \: A.P. }\\ \\ \\ \mathsf{ \implies a \: = \: First \: term \: of \: an \: A.P. } \\ \\ \\ \mathsf{ \implies n \: = \: Terms \: of \: an \: A.P. } \\ \\ \\ \mathsf{\implies d \: = \: Common \: difference } \: \:$

$\mathsf{ Now : } \\ \\ \\ \mathsf{\implies T_p \: = \: a \: + \: ( \: p \: - \:1 \: )d } \\ \\ \\ \mathsf{\implies T_q \: = \: a \: + \:( \: q \: - \: 1 \: )d } \\ \\ \\ \mathsf{\implies T_r \: = \: a \: + \:( \: r \: - \: 1 \: )d } \\ \\ \\ \mathsf{\implies T_s \: = \:a \:+\:(\:s\: - \:1\:)d }$

$\mathsf{ According \: \: to \: \: question , \: } \\ \\ \\ \mathsf{ \implies \dfrac{T_q}{T_p} \: = \: \dfrac{T_r}{T_q} \: = \: \dfrac{T_s}{T_r}} \\ \\ \\ \mathsf{\therefore {({T_q})}^{2} \: = \: {T_p}\: \times \:{T_r}} \\ \\ \\ \mathsf{ Now, } \\ \\ \\\mathsf{ \implies \dfrac{T_q}{T_p} \: = \: \dfrac{T_r}{T_q} } \: \\ \\ \\ \mathsf{ Subtracting \: 1 \: to \: both \: sides : } \\ \\ \\\mathsf{ \implies \dfrac{T_q}{T_p} \: - 1 \: = \: \dfrac{T_r}{T_q} \: - \: 1 }$

$\mathsf{\implies \dfrac{ T_q \: - \: T_p }{T_p} \: = \: \dfrac{T_r \: - \: T_q }{T_q } \qquad...(1) }$

$\mathsf{And, } \\ \\ \\\mathsf{\implies \dfrac{T_s}{T_r} \: = \: \dfrac{T_r}{T_q} } \\ \\ \\\mathsf{ Subtracting \: 1 \: to \: both \: sides , } \\ \\ \\ \\ \mathsf{ \implies \dfrac{T_s}{T_r} \: - \: 1 \: = \: \dfrac{T_r}{T_q} \: - \: 1 } \\ \\\\ \mathsf{ \implies \dfrac{T_s \: - \: T_r}{T_r} \: = \: \dfrac{T_r\: - \: T_q}{T_q} \qquad...(2)}$

$\mathsf{Multiply \:(1) \: to \:( 2 ), \: }$

$\mathsf{ \implies \dfrac{T_q \: - \:T_p }{T_p} \: \times \: \dfrac{T_s \: - \:T_r}{T_r} \: = \: \dfrac{ T_r \: - \:T_q }{T_q} \: \times \: \dfrac{ T_r \: - \:T_q}{T_q}}$

$\mathsf{\implies \dfrac{(\: T_q \: - T_p \: )( \: T_s \: - T_r \: )}{ \cancel{T_p \: \times \:T_r}} \: = \: \dfrac{ {{( \:T_r \: - \:T_q \: )} \: }^{2} }{ \cancel{{( \: T_q \: )}^{2} } }}$

$\mathsf{\implies (\: T_q \: - \:T_p \: )(\: T_s \: - \:T_r \: ) \: = \: { ( \:T_r \: - \: T_q \: )}^{2}}$

$\mathsf{ \implies [\: ( \: a \: + \:( \: q \: - \:1 \:)d \:) \: - \:( \: a \: + \:( \: p \: - \: 1 \: )d\:) \: ][( \: a \: +\:( \: s \: - \: 1 \:)d \: ) \: - \:( \: a \: + \:( \: r \: - \: 1 \: )d \: ) \: ] \: =} \\ \mathsf{ \:{[ \:( \: a \: + \:( \: r \: - \: 1 \: )d \: ) \: - \: ( \: a \: + \:( \: q \: - \: 1 \: )d \: ) \: ] }^{2}}$

$\mathsf{ \implies [ \:( \: a \: + \: qd \: - \:d \: ) \: - \: ( \: a \: + \:pd \: - \:d \: ) \: ][ \:( \: a \: + \: sd \: - \:d \: ) \: - \:( \: a \: + \: rd \: - \:d \: ] \: = }\\\mathsf{[ \: ( \: a \: + \: rd \: - \:d \: ) \: - \: ( \: a \: + \:qd \: - \:d \: )\:]}^{2} } \:$

$\mathsf{ \implies( \: \cancel{a} \: + \:qd \: - \: \cancel{d } \: - \: \cancel{ a} \: - \: pd \: + \: \cancel{d} \: )( \: \cancel{a} \: + \: sd \: - \: \cancel{d} \: - \: \cancel{a} \: - \: rd \: + \: \cancel{d} \: ) \: = }\\ \mathsf{\: {( \: \cancel{a} \: + \: rd \: - \: \cancel{d} \: - \: \cancel{a} \: - \: qd \: + \: \cancel{d} \: )}^{2} }$

$\mathsf{ \implies( \: qd \: - \: pd \: )( \: sd \: - \: rd \: ) \: = \: { ( \: rd \: - \: qd \: ) }^{2} }$

$\mathsf{ \implies \: (- d)( p\: \: - \: q \: )( \: r \: - \: s \: )( - d) \: = \: {[ \:( -d )( \: q \: - \:r \: ) \: ]}^{2} }$

$\mathsf{ \implies \: {d}^{2}( \: p \: - \: q \: )( \: r \: - \: s \: ) \: = \: { (- d)}^{2} {( \: q \: - \: r \: )}^{2}}$

$\mathsf{ \implies \cancel{ {d}^{2} }( \: p \: - \: q \: )( \: r \: - \: s \: ) \: = \: \cancel{ {d}^{2}} {( \: q \: - \: r \: )}^{2} }$

$\mathsf{ \implies \dfrac{( \: p \: - \: q \: )}{( \: q \: - \: r \: )} \: = \: \dfrac{( \: q \: - \: r \: )}{ ( \: r \: - \: s \: )}}$

$\mathsf{ \therefore ( \: p \: - \:q \: ) , ( \: q \: - \: r \: ) \: and \: ( \: r \: - \:s \: ) \: are \: in \: G.P. }$

$\boxed{\mathsf{ \underline {Proved \: !! }}}$

Answer 2

A bullet of mass 50 g is fired from a stationary rifle into a wooden block of mass 1.95 kg placed on a horizontal wooden surface. On striking, the bullet is embedded into the block and both together cover a distance of 200 m in a straight path before coming to rest. Given the coefficient of friction between the block and the surface is 0.2 and the mass of the rifle from which the bullet is triggered is 14 kg, find the magnitude of the velocity of the recoil of the rifle. ( Neglect the resistance of air on the bullet ).