Question

i) Four numbers are in A.P. Their sum is 16 and the sum of their squares is 84. Find the
numbers.​

Answer 1

❍ Let's say that the four terms be a - 3d , a - d , a + d & a + 3d .

Given that ,

》 The sum of four terms of an A.P is 16 .

$\qquad \therefore \sf \bigg( \:a - 3d \:\bigg) + \bigg( \:a - d\:\bigg) + \bigg( \:a + d \:\bigg) + \bigg( \: a + 3d \:\bigg)\:\:=\:\:16\:\:\\\\\dashrightarrow \sf \bigg( \:a - 3d \:\bigg) + \bigg( \:a - d\:\bigg) + \bigg( \:a + d \:\bigg) + \bigg( \: a + 3d \:\bigg)\:\:=\:\:16\:\:\\\\\dashrightarrow \sf \:a - 3d \: + \:a - d\: + \:a + d \: + \: a + 3d \:\:\:=\:\:16\:\:\\\\\dashrightarrow \sf \:a + a + a + a - 3d + 3d \: - d\: + d \:\:\:=\:\:16\:\:\\\\\dashrightarrow \sf \:4a - 3d + 3d \: - d\: + d \:\:\:=\:\:16\:\:\\\\\dashrightarrow \sf \:4a \:\:\:=\:\:16\:\:\\\\\dashrightarrow \sf a \:=\:\dfrac{16}{4} \:\\\\ a \:=\: 4\:\: \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:\:\:\:a\:=\:\:\:4\:\:}}}}}$

AND ,

》The sum of the square of the four terms of an A.P is 84 .

$\qquad \therefore \sf \bigg( \:a - 3d \:\bigg)^2 + \bigg( \:a - d\:\bigg)^2 + \bigg( \:a + d \:\bigg)^2 + \bigg( \: a + 3d \:\bigg)^2\:\:=\:\:84\:\:\\\\ \dashrightarrow \sf \bigg( \:a - 3d \:\bigg)^2 + \bigg( \:a - d\:\bigg)^2 + \bigg( \:a + d \:\bigg)^2 + \bigg( \: a + 3d \:\bigg)^2\:\:=\:\:84\:\:$

As , we know that ,

• Algebraic Indentity :

$\qquad \dag\:\:\bigg\lgroup \sf{ ( a + b )^2 \:=\: a^2 + b^2 + 2ab }\bigg\rgroup \\\\\qquad \dag\:\:\bigg\lgroup \sf{ ( a - b )^2 \:=\: a^2 + b^2 - 2ab }\bigg\rgroup$

$\dashrightarrow \sf \bigg( \:a - 3d \:\bigg)^2 + \bigg( \:a - d\:\bigg)^2 + \bigg( \:a + d \:\bigg)^2 + \bigg( \: a + 3d \:\bigg)^2\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \bigg(\:a^2 - 6ad \:+ 9d^2 \:\bigg) \: + \bigg( \:a^2 -2 ad +d^2\:\bigg)+ \bigg( \:a^2 + 2a d + d^2 \:\bigg) + \bigg( \: a^2 + 6ad + 9d^2 \:\bigg)\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \:a^2 - 6ad \:+ 9d^2 \: \: + \:a^2 -2 ad +d^2\:+ \:a^2 + 2a d + d^2 \: + \: a^2 + 6ad + 9d^2 \:\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \:4a^2 + 20 d \:\:\:=\:\:84\:\:\\\\ \: \underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value\:of \: a \::}}\\\\\dashrightarrow \sf \:4a^2 + 20 d ^2 \:\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \:4(4)^2 + 20 d^2 \:\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \:4(16) + 20 d^2 \:\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \:64 + 20 d^2 \:\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \: 20 d^2 \:\:\:=\:\:84\:- 64\:\\\\ \: \dashrightarrow \sf \: 20 d^2 \:\:\:=\:\:20\:\:\\\\ \: \dashrightarrow \sf \: d^2 \:\:\:=\:\:1\:\:\\\\ \: \dashrightarrow \sf \: d \:\:\:=\:\:\pm \:1\:\:\\\\ \: \dashrightarrow \sf \: d \:\:\:=\:\:1 \:or \: -1\:\:\\\\ \: \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:\:\:\:a\:=\:\:\:1 \:\:or \:-1\:\:}}}}}$

⠀⠀⠀⠀⠀¤ Finding four terms of an A.P ( Airthmetic Progression) :

As , We know that ,

• Four Terms of an A.P are :

$\qquad \dag\:\:\bigg\lgroup \sf{ \bigg( \:a - 3d \:\bigg) , \bigg( \:a - d\:\bigg) , \bigg( \:a + d \:\bigg) , \bigg( \: a + 3d \:\bigg)\:\: }\bigg\rgroup \\\\\dashrightarrow \sf \bigg( \:a - 3d \:\bigg) , \bigg( \:a - d\:\bigg) , \bigg( \:a + d \:\bigg) , \bigg( \: a + 3d \:\bigg)\:\: \\\\\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value \:of \: a \: and \: d \: \::}}$

• When d is equal to 1 .

$\dashrightarrow \sf \bigg( \:a - 3d \:\bigg) , \bigg( \:a - d\:\bigg) , \bigg( \:a + d \:\bigg) , \bigg( \: a + 3d \:\bigg)\:\: \\\\\dashrightarrow \sf \bigg( \:4 - 3(1) \:\bigg) , \bigg( \:4 - 1\:\bigg) , \bigg( \:4 + 1 \:\bigg) , \bigg( \: 4 + 3(1) \:\bigg)\:\: \\\\\dashrightarrow \sf \bigg( \:4 - 3 \:\bigg) , \bigg( \:4 - 1\:\bigg) , \bigg( \:4 + 1 \:\bigg) , \bigg( \: 4 + 3 \:\bigg)\:\: \\\\\dashrightarrow \sf \bigg( \:1 \:\bigg) , \bigg( \:3\:\bigg) , \bigg( \:5 \:\bigg) , \bigg( \: 7 \:\bigg)\:\:$

$\dashrightarrow \sf \:1 \:\ , \:3\: , \:5 \: , \: 7 \:\:\: \\\\\dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:Four \:Terms \: \:_{\:(\:d\:=\:1\:)}\:= \: 1, \:3\:,\:5\:,\:7\:\:}}}}}$

⠀⠀⠀⠀⠀AND ,

• When d is equal to -1 .

$\dashrightarrow \sf \bigg( \:a - 3d \:\bigg) , \bigg( \:a - d\:\bigg) , \bigg( \:a + d \:\bigg) , \bigg( \: a + 3d \:\bigg)\:\: \\\\\dashrightarrow \sf \bigg( \:4 - 3(-1) \:\bigg) , \bigg( \:4 - (-1)\:\bigg) , \bigg( \:4 + (-1) \:\bigg) , \bigg( \: 4 + 3(-1) \:\bigg)\:\: \\\\\dashrightarrow \sf \bigg( \:4 + 3 \:\bigg) , \bigg( \:4 + 1\:\bigg) , \bigg( \:4 - 1 \:\bigg) , \bigg( \: 4 - 3 \:\bigg)\:\: \\\\\dashrightarrow \sf \bigg( \:7\:\bigg) , \bigg( \:5\:\bigg) , \bigg( \:4 \:\bigg) , \bigg( \: 1 \:\bigg)\:\: \\\\\dashrightarrow \sf \:7 \:\ , \:5\: , \:3 \: , \: 1 \:\:\: \\\\ \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:Four \:Terms \: \:_{\:(\:d\:=\:-1\:)}\:= \: 7, \:5\:,\:3\:,\:1\:\:}}}}}\\\\\qquad \therefore \:\underline {\sf \:Hence \:The \:four \:terms \:can \; be \;\bf 1, 3 , 5 , 7 \:\:\sf or \:\bf 7, 5 ,3 , 1 \:}$

Answer 2

❍ Let's say that the four terms be a - 3d , a - d , a + d & a + 3d .

$\sf \pmb{Answer :}$

Given that ,

》 The sum of four terms of an A.P is 16 .

$\qquad \therefore \sf \bigg( \:a - 3d \:\bigg) + \bigg( \:a - d\:\bigg) + \bigg( \:a + d \:\bigg) + \bigg( \: a + 3d \:\bigg)\:\:=\:\:16\:\:\\\\\dashrightarrow \sf \bigg( \:a - 3d \:\bigg) + \bigg( \:a - d\:\bigg) + \bigg( \:a + d \:\bigg) + \bigg( \: a + 3d \:\bigg)\:\:=\:\:16\:\:\\\\\dashrightarrow \sf \:a - 3d \: + \:a - d\: + \:a + d \: + \: a + 3d \:\:\:=\:\:16\:\:\\\\\dashrightarrow \sf \:a + a + a + a - 3d + 3d \: - d\: + d \:\:\:=\:\:16\:\:\\\\\dashrightarrow \sf \:4a - 3d + 3d \: - d\: + d \:\:\:=\:\:16\:\:\\\\\dashrightarrow \sf \:4a \:\:\:=\:\:16\:\:\\\\\dashrightarrow \sf a \:=\:\dfrac{16}{4} \:\\\\ a \:=\: 4\:\: \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:\:\:\:a\:=\:\:\:4\:\:}}}}}$

AND ,

》The sum of the square of the four terms of an A.P is 84 .

$\qquad \therefore \sf \bigg( \:a - 3d \:\bigg)^2 + \bigg( \:a - d\:\bigg)^2 + \bigg( \:a + d \:\bigg)^2 + \bigg( \: a + 3d \:\bigg)^2\:\:=\:\:84\:\:\\\\ \dashrightarrow \sf \bigg( \:a - 3d \:\bigg)^2 + \bigg( \:a - d\:\bigg)^2 + \bigg( \:a + d \:\bigg)^2 + \bigg( \: a + 3d \:\bigg)^2\:\:=\:\:84\:\:$

As , we know that ,

Algebraic Indentity :

$\qquad \dag\:\:\bigg\lgroup \sf{ ( a + b )^2 \:=\: a^2 + b^2 + 2ab }\bigg\rgroup \\\\\qquad \dag\:\:\bigg\lgroup \sf{ ( a - b )^2 \:=\: a^2 + b^2 - 2ab }\bigg\rgroup$

$\dashrightarrow \sf \bigg( \:a - 3d \:\bigg)^2 + \bigg( \:a - d\:\bigg)^2 + \bigg( \:a + d \:\bigg)^2 + \bigg( \: a + 3d \:\bigg)^2\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \bigg(\:a^2 - 6ad \:+ 9d^2 \:\bigg) \: + \bigg( \:a^2 -2 ad +d^2\:\bigg)+ \bigg( \:a^2 + 2a d + d^2 \:\bigg) + \bigg( \: a^2 + 6ad + 9d^2 \:\bigg)\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \:a^2 - 6ad \:+ 9d^2 \: \: + \:a^2 -2 ad +d^2\:+ \:a^2 + 2a d + d^2 \: + \: a^2 + 6ad + 9d^2 \:\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \:4a^2 + 20 d \:\:\:=\:\:84\:\:\\\\ \: \underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value\:of \: a \::}}\\\\\dashrightarrow \sf \:4a^2 + 20 d ^2 \:\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \:4(4)^2 + 20 d^2 \:\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \:4(16) + 20 d^2 \:\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \:64 + 20 d^2 \:\:\:=\:\:84\:\:\\\\ \: \dashrightarrow \sf \: 20 d^2 \:\:\:=\:\:84\:- 64\:\\\\ \: \dashrightarrow \sf \: 20 d^2 \:\:\:=\:\:20\:\:\\\\ \: \dashrightarrow \sf \: d^2 \:\:\:=\:\:1\:\:\\\\ \: \dashrightarrow \sf \: d \:\:\:=\:\:\pm \:1\:\:\\\\ \: \dashrightarrow \sf \: d \:\:\:=\:\:1 \:or \: -1\:\:\\\\ \: \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:\:\:\:a\:=\:\:\:1 \:\:or \:-1\:\:}}}}}$

⠀⠀⠀⠀⠀¤ Finding four terms of an A.P ( Airthmetic Progression) :

As , We know that ,

Four Terms of an A.P are :

$\qquad \dag\:\:\bigg\lgroup \sf{ \bigg( \:a - 3d \:\bigg) , \bigg( \:a - d\:\bigg) , \bigg( \:a + d \:\bigg) , \bigg( \: a + 3d \:\bigg)\:\: }\bigg\rgroup \\\\\dashrightarrow \sf \bigg( \:a - 3d \:\bigg) , \bigg( \:a - d\:\bigg) , \bigg( \:a + d \:\bigg) , \bigg( \: a + 3d \:\bigg)\:\: \\\\\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value \:of \: a \: and \: d \: \::}}$

When d is equal to 1 .

$\dashrightarrow \sf \bigg( \:a - 3d \:\bigg) , \bigg( \:a - d\:\bigg) , \bigg( \:a + d \:\bigg) , \bigg( \: a + 3d \:\bigg)\:\: \\\\\dashrightarrow \sf \bigg( \:4 - 3(1) \:\bigg) , \bigg( \:4 - 1\:\bigg) , \bigg( \:4 + 1 \:\bigg) , \bigg( \: 4 + 3(1) \:\bigg)\:\: \\\\\dashrightarrow \sf \bigg( \:4 - 3 \:\bigg) , \bigg( \:4 - 1\:\bigg) , \bigg( \:4 + 1 \:\bigg) , \bigg( \: 4 + 3 \:\bigg)\:\: \\\\\dashrightarrow \sf \bigg( \:1 \:\bigg) , \bigg( \:3\:\bigg) , \bigg( \:5 \:\bigg) , \bigg( \: 7 \:\bigg)\:\:$

$\dashrightarrow \sf \:1 \:\ , \:3\: , \:5 \: , \: 7 \:\:\: \\\\\dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:Four \:Terms \: \:_{\:(\:d\:=\:1\:)}\:= \: 1, \:3\:,\:5\:,\:7\:\:}}}}}$

⠀⠀⠀⠀⠀AND ,

When d is equal to -1 .

$\dashrightarrow \sf \bigg( \:a - 3d \:\bigg) , \bigg( \:a - d\:\bigg) , \bigg( \:a + d \:\bigg) , \bigg( \: a + 3d \:\bigg)\:\: \\\\\dashrightarrow \sf \bigg( \:4 - 3(-1) \:\bigg) , \bigg( \:4 - (-1)\:\bigg) , \bigg( \:4 + (-1) \:\bigg) , \bigg( \: 4 + 3(-1) \:\bigg)\:\: \\\\\dashrightarrow \sf \bigg( \:4 + 3 \:\bigg) , \bigg( \:4 + 1\:\bigg) , \bigg( \:4 - 1 \:\bigg) , \bigg( \: 4 - 3 \:\bigg)\:\: \\\\\dashrightarrow \sf \bigg( \:7\:\bigg) , \bigg( \:5\:\bigg) , \bigg( \:4 \:\bigg) , \bigg( \: 1 \:\bigg)\:\: \\\\\dashrightarrow \sf \:7 \:\ , \:5\: , \:3 \: , \: 1 \:\:\: \\\\ \dashrightarrow \:\:\underline {\boxed{\purple {\pmb{\frak{\:Four \:Terms \: \:_{\:(\:d\:=\:-1\:)}\:= \: 7, \:5\:,\:3\:,\:1\:\:}}}}}\\\\\qquad \therefore \:\underline {\sf \:Hence \:The \:four \:terms \:can \; be \;\bf 1, 3 , 5 , 7 \:\:\sf or \:\bf 7, 5 ,3 , 1 \:}$