Question

A A.P Cansits of 37 terms the sun of first
3 terms is 12 and sum of last 3 term is 31
then find first and 10th term of A.P.​

Answer 1

Question:-

An AP consists of 37 terms . The sum of first three terms is 12 and sum of last three terms is 318. Find first and last terms.

Answer:-

Given:-

Number of terms in an AP (n) = 37

Sum of first three terms = 12.

First three terms of an AP are a , a + d , a + 2d where a is the first term and d is common difference.

So,

⟹ a + a + d + a + 2d = 12

⟹ 3a + 3d = 12

⟹ 3(a + d) = 12

⟹ a + d = 12/3

⟹ a + d = 4 -- equation (1).

Also given that,

Sum of last three terms = 318.

That is;

⟹ 35th term + 36th term + 37th term = 318.

We know that,

nth term of an AP (aₙ) = a + (n - 1)d

So,

⟹ a + (35 - 1)d + a + (36 - 1)d + a + (37 - 1)d = 318

⟹ 3a + 34d + 35d + 36d = 318

⟹ 3a + 105d = 318

⟹ 3(a + 35d) = 3 × 106

⟹ a + 35d = 106 -- equation (2)

Now,

Subtract equation (1) from (2).

⟹ a + 35d - (a + d) = 106 - 4

⟹ a + 35d - a - d = 102

⟹ 34d = 102

⟹ d = 102/34

⟹ d = 3

Substitute d = 3 in equation (1).

⟹ a + 3 = 4

⟹ a = 4 - 3

⟹ a = 1

Hence,

Last term = 37th term = a + 36d

⟹ a₃₇ = 1 + 36(3)

⟹ a₃₇ = 1 + 108

⟹ a₃₇ = 109

∴

• First term of the AP = a = 1

• Last term of the AP = a₃₇ = 109.

Answer 2

Question :

⠀⠀⠀▪︎ An A.P consists 37 terms the sum of first 3 terms is 12 and sum of last 3 term is ⠀⠀⠀⠀⠀31 then find first and last term of A.P .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Given : An A.P consists 37 terms the sum of first 3 terms is 12 and sum of last 3 term is 31 .

Exigency To Find : First and last term of an A.P .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀CASE I : The sum of first 3 terms is 12 .

$\qquad :\implies \sf 1^{st} \:Term \: + \:2^{nd} \: Term \:+ \: 3^{rd} \: Term \: =\: 12 \:$

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad \maltese \: \bf Formaula \: for \: n^{th} \:term \::$

$\qquad \dag\:\: \bigg\lgroup \sf{ a_{n} = a + ( n - 1 ) d }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , a first term of an A.P , d is the Common Difference & n is the $n^{th}$ term of an A.P .

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

$\qquad :\implies \sf 1^{st} \:Term \: + \:2^{nd} \: Term \:+ \: 3^{rd} \: Term \: =\: 12 \:$

$\qquad :\implies \sf a \: + \: a + (2-1)d \: \:+ \: a + ( 3- 1) d \: =\: 12 \:$

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

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

$\qquad :\implies \sf \: 3(a + d) \: =\: 12 \:$

$\qquad :\implies \sf \: a + d \: =\: \dfrac{12}{3} \:$

$\qquad :\implies \sf \: a + d \: =\:\cancel {\dfrac{12}{3} } \:$

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

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

$\qquad :\implies \bf \: a \: =\: 4 - d \qquad \:\bigg\lgroup \sf{ eq^n \: 1 }\bigg\rgroup$

⠀⠀⠀⠀⠀CASE II : The sum of last 3 term is 31 .

$\qquad :\implies \sf 35^{th} \:Term \: + \:36^{th} \: Term \:+ \: 37^{th} \: Term \: =\: 318 \:$

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad \maltese \: \bf Formaula \: for \: n^{th} \:term \::$

$\qquad \dag\:\: \bigg\lgroup \sf{ a_{n} = a + ( n - 1 ) d }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , a first term of an A.P , d is the Common Difference & n is the $n^{th}$ term of an A.P .

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

$\qquad :\implies \sf 35^{st} \:Term \: + \:36^{nd} \: Term \:+ \: 37^{rd} \: Term \: =\: 318 \:$

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

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

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

$\qquad :\implies \sf 3a + 105d \: =\: 318 \:$

$\qquad :\implies \sf 3(a + 35d) \: =\: 318 \:$

$\qquad :\implies \sf a + 35d \: =\: \dfrac{318}{3} \:$

$\qquad :\implies \sf a + 35d \: =\: 106 \:$

$\qquad :\implies \bf \: a + 35d \: =\: 106 \qquad \:\bigg\lgroup \sf{ eq^n \: 2 }\bigg\rgroup$

⠀⠀⠀⠀⠀ Finding value of First Term :

From Equation 2 :

$\qquad :\implies \bf \: a + 35d \: =\: 106 \qquad \:\bigg\lgroup \sf{ eq^n \: 2 }\bigg\rgroup$

$\qquad :\implies \sf a + 35d \: =\: 106 \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: eq^n\; 1 \: \::}}$

$\qquad :\implies \bf \: a \: =\: 4 - d \qquad \:\bigg\lgroup \sf{ eq^n \: 1 }\bigg\rgroup$

$\qquad :\implies \sf a + 35d \: =\: 106 \:$

$\qquad :\implies \sf 4 - d + 35d \: =\: 106 \:$

$\qquad :\implies \sf - d + 35d \: =\: 106 - 4 \:$

$\qquad :\implies \sf 34d \: =\: 104 \:$

$\qquad :\implies \sf d \: =\: \dfrac{104}{34} \:$

$\qquad :\implies \bf d \: =\: 3 \:$

$\qquad :\implies \frak{\underline{\purple{\:d = 3 }} }\:\:\bigstar$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: value \:of \: d \:in \: eq^n \: 1 \::}}$

$\qquad :\implies \bf \: a \: =\: 4 - d \qquad \:\bigg\lgroup \sf{ eq^n \: 1 }\bigg\rgroup$

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

$\qquad :\implies \sf \: a \: =\:4 - 3 \:$

$\qquad :\implies \bf \: a \: =\:1 \:$

$\qquad:\implies \frak{\underline{\purple{\:a = 1 }} }\:\:\bigstar$

⠀⠀⠀⠀⠀ Finding value of Last Term [ Last term is 37 ] :

$\qquad \dag\:\: \bigg\lgroup \sf{ a_{n} = a + ( n - 1 ) d }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , a first term of an A.P , d is the Common Difference & n is the $n^{th}$ term of an A.P .

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

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

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

$\qquad:\implies \sf a_{37} = 1 + 36 \times 3$

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

$\qquad:\implies \bf a_{37} = 109$

$\qquad:\implies \frak{\underline{\purple{\:a_{37} = 109 }} }\:\:\bigstar$