find the value of P, if the number x,2x+p,3x+6 are the three conjuctive terms of an arithmetic progestion
Anonymous:
Answer: p=3
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Answered by
287
given x, 2x+p, 3x+6 is in arithmetic progression
a = x , d₁ = a₂-a₁ = 2x+p-x
d₁= x+p
d₂ = a₃-a₂ = 3x+6-2x-p
= x-p+6
in an AP, d₁ = d₂
x+p = x-p+6
x+p-x+p = 6
2p = 6
p = 3
a = x , d₁ = a₂-a₁ = 2x+p-x
d₁= x+p
d₂ = a₃-a₂ = 3x+6-2x-p
= x-p+6
in an AP, d₁ = d₂
x+p = x-p+6
x+p-x+p = 6
2p = 6
p = 3
pls mrk as best
Answered by
73
x , 2x+p , 3x+6 are in A.P
so we know that ,
t₂ = (t₁+t₃)/2
or, 2x+p = (x+3x+6)/2
or, 2x+p = 2(2x+3)/2
or, p = 2x+3-2x
therefore p = 3
so we know that ,
t₂ = (t₁+t₃)/2
or, 2x+p = (x+3x+6)/2
or, 2x+p = 2(2x+3)/2
or, p = 2x+3-2x
therefore p = 3
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