Question

6. The value of p if x, 2x + p and 3x + 6 are
in A.P.

valu of p = ------------​

Answer 1

Required Answer:-

In an AP, the difference between any two consecutive terms is constant. Hence, it is known as constant difference (d).

• That means, the difference between the first and second term = difference between the second and third term$.$

• Or we can consider the opposite. The difference will remain constant throughout.

Hence:-

➛ 2nd - 1st = 3rd - 2nd

➛ 2x + p - x = 3x + 6 - (2x + p)

➛ x + p = 3x + 6 - 2x - p

➛ x + p = x - p + 6

➛ x + p - x + p - 6 = 0

➛ 2p - 6 = 0

➛ 2p = 6

➛ p = 3(Ans)

Answer 2

Given : x , 2x + p and 3x + 6 are in A.P [ Airthmetic Progression ] .

Exigency To Find : The value of p .

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⠀⠀⠀⠀⠀Given that ,

$\qquad \dag\:\:\bigg\lgroup \sf{ x \:, \:2x\: + \:p \:and \:3x\: + \:6\: are\: in \:A.P\: [\ Airthmetic\: Progression \ ]\: . }\bigg\rgroup$

⠀⠀⠀⠀Here ,

⠀⠀⠀▪︎⠀ $\sf 1^{st} \:Term\: =\: \bf \: x \:$

⠀⠀⠀▪︎⠀ $\sf 2^{nd} \:Term\: =\: \bf \: 2x + p \:$

⠀⠀⠀▪︎ ⠀$\sf 3^{rd} \:Term\: =\: \bf \: 3x + 6 \:$

⠀⠀⠀⠀⠀▪︎⠀It is an A.P ( Airthmetic Progression ) . The difference ( d ) between two consecutive term is constant .

Therefore,

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

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

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

$\qquad:\implies \sf 2x + p\:- \: x\: = 3x + 6\: - \:(2x + p) \:\:$

$\qquad:\implies \sf 2x + p\:- \: x\: = 3x + 6\: - \:2x - p \:\:$

$\qquad:\implies \sf 2x - x+ p\:\: = 3x- 2x + 6\: - p \:\:$

$\qquad:\implies \sf x+ p\:\: = x - p + 6 \:\:$

$\qquad:\implies \sf \cancel {x}\:+ p\:\: =\cancel {x }+ 6\: - p \:\:$

$\qquad:\implies \sf p\:\: = 6\: - p \:\:$

$\qquad:\implies \sf p\:\: = 6\: - p \:\:$

$\qquad:\implies \sf p\:+ p\: = 6\: \:\:$

$\qquad:\implies \sf 2p\: = 6\: \:\:$

$\qquad:\implies \sf p\: = \dfrac{ 6}{2}\: \:\:$

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

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

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The\:value \:of\:p \:is\:\bf{3\:}}}}$

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