Question

An arthmetic progression consists of 37 terms.The sum of first 3 terms of it is 12 and sum of its last 3 terms is 318 then find first & last term of progression​

Answer 1

Given: An AP (Arithmetic Progression) consists of 37 terms. The sum of first three terms is 12 and sum of its last three terms is 318.

Need to find: The first 'a' and last term 'l' of AP.

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❍ Let's say, that first three terms of the AP be a, a + d and a + 2d.

⠀

¤ For any Arithmetic Progression ( AP ), the formula of of nth terms is Given by :

⠀⠀⠀⠀⠀

$\bigstar\;\large{\underline{\boxed{\sf{a_n = a +\Big(n - 1\Big)d}}}}$

where:

• an = nth term
• a = first term
• d = common difference

⠀

$\underline{\bigstar\:\boldsymbol{According\:to\:the\:Question\::}}$

$:\implies\sf a + a + d + a + 2d = 12 \\\\\\:\implies\sf 3(a + d) = 12 \\\\\\:\implies\sf a + d = \cancel\dfrac{12}{3} \\\\\\:\implies\sf a + d = 4 \\\\\\:\implies\sf a = 4 - d\qquad\qquad\bigg\lgroup\pmb eq^{n}\;(1)\bigg\rgroup$

⠀

Also,

• Arithmetic progression consists total 37 terms. It means there are total 37 terms therefore the last three terms would be 37th, 36th and 35th. & sum of last three terms is 318.

⠀

$\dashrightarrow\sf a + \Big\{37 - 1\Big\}d + a + \Big\{ 36 - 1\Big\} + a + \Big\{35 - 1\Big\}d = 318 \\\\\\\dashrightarrow\sf 3a + 36d + 35d + 34d = 318\\\\\\\dashrightarrow\sf 3\Big\{4 - d\Big\} + 36d + 35d + 34d = 318\\\\\\\dashrightarrow\sf 12 - 3d + 36d + 35d + 34d = 318\\\\\\\dashrightarrow\sf 12 - 3d + 105d = 318\\\\\\\dashrightarrow\sf 102d = 318 - 12\\\\\\\dashrightarrow\sf 102d = 306\\\\\\\dashrightarrow\sf d = \cancel\dfrac{306}{102}\\\\\\\dashrightarrow\underline{\boxed{\pmb{\frak{d = 3}}}}\;\bigstar$

⠀⠀

⠀⠀⠀⠀$\bf{\dag}\;{\underline{\frak{Putting\;value\;of\;d\;in\;eq^{n}\;(1) :}}}$

⠀

⠀⠀$\longrightarrow\sf a = 4 - d \\\\\\\longrightarrow\sf a = 4 - 3 \\\\\\\longrightarrow\underline{\boxed{\pmb{\frak{a = 1}}}}\;\bigstar$

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⠀⠀

Now, finding last term of the AP, 'l' :

⠀⠀

$\frak{we \;have}\begin{cases} &\sf{a = \bf{1}} \\ & \sf{d = \bf{3}} \\ & \sf{l = \bf{37}}\end{cases}$

$\twoheadrightarrow\sf a_{37} = a + 36d \\\\\\\twoheadrightarrow\sf a_{37} = 1 + 36 \times 3 \\\\\\\twoheadrightarrow\sf a_{37} = 1 + 108\\\\\\\twoheadrightarrow\underline{\boxed{\pmb{\frak{\pink{a_{37} = 109}}}}}\;\bigstar$

⠀

$\therefore{\underline{\textsf{Hence, the first and last terms of the AP are \textbf{1} \sf{and} \textbf{109} respectively.}}}$

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$\qquad\quad\boxed{\underline{\underline{\bigstar{ \pmb{\mathbb { \: \: MORE \ FORMULAS\: \: \bigstar}}}}}}$

⠀

$\sf ( I )\;Sum\;of\;n\;term\;of\;an\;AP\; = \; {\pmb{\sf{S_n = \dfrac{n}{2} \bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup}}}}$⠀⠀⠀⠀⠀⠀⠀

$\sf ( II )\;Sum\;of\;all\;terms\;of\;AP\;having\;last\:term\;as\;'l'\; = \; {\pmb{\sf{{\dfrac{n}{2}(a + l)}}}}$⠀⠀⠀⠀

Answer 2

Given that , An arthmetic progression consists of 37 terms . The sum of first 3 terms of it is 12 and sum of its last 3 terms is 318 .

Exigency To Find : First & last term of progression ?

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⠀⠀⠀⠀⠀Given that ,

• The sum of first 3 terms of it is 12 ,
• The sum of Last three term is 318 &
• An arthmetic progression consists of 37 terms

❍ Let's Consider first three terms of an Airthmetic Progression is a , a + d & a + 2d .

• The sum of first 3 terms of it is 12 .

$\qquad \therefore \sf \:\bigg( \: a \bigg) \: + \:\bigg( \: a + d \bigg) \: + \bigg( \: a + 2d \bigg) \: = 12 \:$

$\qquad \dashrightarrow \sf \:\bigg( \: a \bigg) \: + \:\bigg( \: a + d \bigg) \: + \bigg( \: a + 2d \bigg) \: = 12 \:$

$\qquad \dashrightarrow \sf \:\:a + a + d + a + 2d \: = 12 \:$

$\qquad \dashrightarrow \sf \:\:a + a + a + d + 2d \: = 12 \:$

$\qquad \dashrightarrow \sf \: 3a + 3d \: = 12 \:$

⠀⠀⠀ [ Canceling out each term by 3 ]

$\qquad \dashrightarrow \sf \: 3a + 3d \: = 12 \:$

$\qquad \dashrightarrow \sf \: a + d \: = 4 \:$

$\qquad \dashrightarrow \bf \: a \: = 4 - d \:\qquad \:\bigg\lgroup \sf{ Equation \: 1 \:}\bigg\rgroup$

⠀⠀⠀⠀AND ,

• The sum of Last three term is 318 &
• An arthmetic progression consists of 37 terms

As , We know that ,

• Formula for nth term of an A.P :

$\qquad \dag\:\:\bigg\lgroup \pmb{\bf \: a_n \:=\: a + \:( n - 1 ) d }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀Here , n is the nth term of an A.P ( Airthmetic Progression ) , a is the first term d is the Common Difference.

⠀⠀Last three terms are : 37 th , 36 th & 35 th .

Therefore,

• Last 3 terms will be a + ( 37 - 1 ) d , a + ( 36 - 1 ) d & a + ( 35 - 1 ) d .

$\qquad \dashrightarrow \sf a + \bigg( 37 - 1 \bigg) \: d \:+ a + \bigg( 36 - 1 \bigg) \: d \: + a + \bigg( 35 - 1 \bigg) \: d \:=\:318 \:$

$\qquad \dashrightarrow \sf 3a + 36d + 35d + 34d \:=\:318 \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Equation \:1 \:as \: Value\:of\:a\::}}$

$\qquad \dashrightarrow \sf 3a + 36d + 35d + 34d \:=\:318 \:$

$\qquad \dashrightarrow \sf 3( 4 - d ) + 36d + 35d + 34d \:=\:318 \:$

$\qquad \dashrightarrow \sf 12 - 3d + 36d + 35d + 34d \:=\:318 \:$

$\qquad \dashrightarrow \sf 102d \:=\:318 - 12 \:$

$\qquad \dashrightarrow \sf 102d \:=\:306 \:$

$\qquad \dashrightarrow \sf d \:=\:\dfrac{306}{102} \:$

$\qquad \dashrightarrow \sf d \:=\:3 \:$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\pink{ d \:op(\:or \:Common \:Difference \:)\:=\:3 \:}}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value\:of\:d \::}}$

$\qquad \dashrightarrow \bf \: a \: = 4 - d \:\qquad \:\bigg\lgroup \sf{ Equation \: 1 \:}\bigg\rgroup$

$\qquad \dashrightarrow \sf \: a \: = 4 - d$

$\qquad \dashrightarrow \sf \: a \: = 4 - (3)$

$\qquad \dashrightarrow \sf \: a \: = 1$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\pink{ a \:(\:or \:first \:term \:) \:=\:3 \:}}}}}\:\:\bigstar$

$\qquad \underline {\purple {\bf \:\bigstar \: Finding \: Last \:Term\:(\:or \:a_{37}) \:of \:an \:A.P \:\::\:}}$

• The Last term of an A.P is 37 .

$\qquad \dashrightarrow \sf \: a_n \:=\: a + \:( n - 1 ) d \:$

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

$\qquad \dashrightarrow \sf \: a_n \:=\: a + \:( n - 1 ) d \:$

$\qquad \dashrightarrow \sf \: a_{37} \:=\: 1 + \:( 37 - 1 ) 3 \:$

$\qquad \dashrightarrow \sf \: a_{37} \:=\: 1 + \:( 36) 3 \:$

$\qquad \dashrightarrow \sf \: a_{37} \:=\: 1 + \:108 \:$

$\qquad \dashrightarrow \sf \: a_{3}7 \:=\: 1 09 \:$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\pink{ Last \:Term \: :(\:or \:a_{37})\: =\:109 \:}}}}}\:\:\bigstar$

$\qquad \therefore \:\underline {\sf Hence, \:first \:and \:last \:term \:of \:an \:A.P \:are \:\pmb{\bf 1 \:}\: \& \:\pmb{\bf 109 }\:,respectively \:}$