Question

$\huge \color{red}Question -$
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

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Answer 1

$\bf{Given \::}\begin {cases} & \sf { a_{12} \:or \:Third \:term\:of\:A.P \: is\:12 }\\\\ & \sf{a_{50}\;or\:50th\:term\:or\:Last\:term \:of\:A.P\:is\:106\:}\end{cases}$

Exigency To Find : The 29 th term of A.P .

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$\dag\frak{\underline {We,\:know\:that\;;}}$

$\dag\:\:\boxed {\sf{ a_{n} = a + (n-1)d}}$

Where,

• a is the first term of A.P.
• n is the n th term of A.P.
• d is the Common difference

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Given that ,

• Third Term of A.P is 12 .

Or ,

• $:\implies \sf { a_3 \:or\:12\:= a + ( 3 -1 ) d}$

⠀⠀⠀⠀⠀$:\implies \sf {\:12\:= a + ( 3 -1 ) d}$

⠀⠀⠀⠀⠀$:\implies \sf { \:12\:= a + 2 d}$

⠀⠀⠀⠀⠀$:\implies \sf { \:12- 2d\:= a }$

⠀⠀⠀⠀⠀$:\implies \bf { \bigg(\:a= 12 - 2d \bigg) \qquad \longrightarrow Eq.1 }$

Given that ,

• Last Term or 50 th Term of A.P is 106 .

Or ,

• $:\implies \sf { a_{50} \:or\:106\:= a + ( 50 -1 ) d}$

⠀⠀⠀⠀⠀$:\implies \sf {\:106\:= a + ( 50 -1 ) d}$

⠀⠀⠀⠀⠀$:\implies \sf { \:106\:= a + 49 d}$

⠀⠀⠀⠀⠀$:\implies \sf { \:106- 49d\:= a }$

⠀⠀⠀⠀⠀$:\implies \bf { \bigg(\:a= 106 - 49d \bigg) \qquad \longrightarrow Eq.2 }$

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⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: From\: Eq.1 \: and \: Eq.2 \: \::}}$

• $:\implies \bf { \bigg(\:a= 12 - 2d \bigg) \qquad \longrightarrow Eq.1 }$

• $:\implies \bf { \bigg(\:a= 106 - 49d \bigg) \qquad \longrightarrow Eq.2 }$

Therefore,

⠀⠀⠀⠀⠀$:\implies \sf { \: 12 - 2d = 106- 49d }$

⠀⠀⠀⠀⠀$:\implies \sf { \: - 2d = 106-12- 49d }$

⠀⠀⠀⠀⠀$:\implies \sf { \: - 2d + 49 d = 106- 12 }$

⠀⠀⠀⠀⠀$:\implies \sf { \: 47d = 94 }$

⠀⠀⠀⠀⠀$:\implies \sf { \: d =\cancel {\dfrac{94}{47}} }$

⠀⠀⠀⠀⠀$:\implies \bf { \bigg(\:d = 2 \bigg) \qquad }$

⠀⠀⠀⠀⠀⠀$\underline {\mathrm{\star\:Now \: By \: Substituting \:d=2\:in\: the \: Equation \: 1 \::}}$

⠀⠀⠀⠀⠀$:\implies \bf { \bigg(\:a= 12 - 2d \bigg) \qquad \longrightarrow Eq.1 }$

⠀⠀⠀⠀⠀$:\implies \sf { \:a= 12 - 2(2) }$

⠀⠀⠀⠀⠀$:\implies \sf { \:a= 12 - 4 }$

⠀⠀⠀⠀⠀$:\implies \bf { \bigg(\:a = 8 \bigg) \qquad }$

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❍ Finding 29 th term of A.P.

• The 29 th Term of A.P is $a_{29}$

Or ,

• $:\implies \sf { a_{29} \:\:= a + ( 29 -1 ) d}$

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

⠀⠀⠀⠀⠀$:\implies \sf { a_{29} \:\:= 8 + ( 29 -1 ) 2}$

⠀⠀⠀⠀⠀$:\implies \sf { a_{29} \:\:= 8 + ( 28) 2}$

⠀⠀⠀⠀⠀$:\implies \sf { a_{29} \:\:= 8 + 56}$

⠀⠀⠀⠀⠀$:\implies \bf { \bigg(\:a_{29} = 64 \bigg) \qquad }$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm { Hence,\: The\:29^{th}\;Term \:of\:A.P\:is\:\bf{64\: }}}}$

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Answer 2

Answer:

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