Question

Q6. Prove that nth term of an A.P. can't be n² + n + 1.​​

Answer 1

Answer:

Answer

Step-by-step explanation:

1+1=2

and

2/2=1

Hence proved .

Answer 2

Given that , n² + n + 1 as nth term of an A.P ( Airthmetic Progression ) .

Exigency To Prove : n² + n + 1 cannot be nth term of an A.P ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's say that the nth term of an A.P , $\sf a_n \:= \: n^2 + n + 1 \:$

Now ,

$\qquad \bigstar \:\sf \underline { By , \:Substituting \:The \:value \:of \: \bf{n} \: as \:\bf{ 1 \: , \: , 2 \:, 3 \:, 4 \:...}}$

⠀⠀⠀⠀⠀We get ,

$\qquad \qquad \leadsto \sf \: First \:Term \: = \: a_1 \:=\: 1^2 + 1 + 1 \: = 1 + 1 + 1 = \pmb{\bf 3 } \:,$

$\qquad \qquad \leadsto \sf \: Second \:Term \: = \: a_2 \:=\: 2^2 + 2 + 1 \: = 4 + 3 = \pmb{\bf 7 } \:,$

$\qquad \qquad \leadsto \sf \: Third \:Term \: = \: a_3 \:=\: 3^2 + 3 + 1 \: = 9 + 4 = \pmb{\bf 13 } \:\&$

$\qquad \qquad \leadsto \sf \: Fourth \:Term \: = \: a_4 \:=\: 4^2 + 4 + 1 \: = 16 + 5 = \pmb{\bf 21 } \:$

Therefore,

$\qquad \therefore \underline {\sf The \:Obtained \:Airthmetic \:Progression \:is \: \pmb{\bf 3 \:,7\:,\:13\:\& \:21 \:}\:.}$

Now ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ It's Common Difference ( d ) :

As , We know that ,

• The Common Difference of an , A.P are always equal :

$\qquad \dag\:\:\bigg\lgroup \bf{ \:\: Common \:Difference \:(d)\:=\:\Big\{ a_2 \: - \:a_1\Big\}\:=\Big\{\: a_3 \: - \:a_2 \Big\}\: = \Big\{\: a_4 \:- \:a_3 \: \Big\}\:}\bigg\rgroup$

$\qquad \dashrightarrow \sf \:\: Common \:Difference \:(d)\:=\:\Big\{ a_2 \: - \:a_1\Big\}\:=\Big\{\: a_3 \: - \:a_2 \Big\}\: = \Big\{\: a_4 \:- \:a_3 \: \Big\} \:$

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

$\qquad \dashrightarrow \sf \:\: Common \:Difference \:(d)\:=\:\Big\{ a_2 \: - \:a_1\Big\}\:=\Big\{\: a_3 \: - \:a_2 \Big\}\: = \Big\{\: a_4 \:- \:a_3 \: \Big\} \:$

$\qquad \dashrightarrow \sf \:\: Common \:Difference \:(d)\:=\:\Big\{ 7 \: - \:3\Big\}\:=\Big\{\: 13 \: - \:7 \Big\}\: = \Big\{\: 21 \:- \:13 \: \Big\} \:$

$\qquad \dashrightarrow \sf \:\: Common \:Difference \:(d)\:=\:\Big\{ 4\Big\}\:=\Big\{\: 13 - 7 \Big\}\: = \Big\{\: 21 \:- \:13 \: \Big\} \:$

$\qquad \dashrightarrow \sf \:\: Common \:Difference \:(d)\:=\:\Big\{ 4\Big\}\:\neq\Big\{\: 6 \Big\}\: = \Big\{\: 21 \:- \:13 \: \Big\} \:$

$\qquad \dashrightarrow \sf \:\: Common \:Difference \:(d)\:=\:\Big\{ 4\Big\}\:\neq\Big\{\: 6 \Big\}\: \neq \Big\{\: 8\: \Big\} \:$

$\qquad \dashrightarrow \sf \:\: Common \:Difference \:(d)\:=\:4\:\neq\: 6 \: \neq \: 8\: \:$

⠀⠀⠀⠀⠀Since , Here , The Common Difference (d) of an A.P ( Airthmetic Progression ) are not Equal.

$\qquad \therefore \: \underline {\sf Hence , \:\pmb{\bf n^2 \:+ \: n + \:1 }\: is \:not \:the \:n^{th} \:term\: of \:an \: A.P \:}$