1/x+2, 1/x+3, 1/x+5 are in a.p. . find the value of x
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Answered by
5
Since we know that if a , b , c are in A.P. then
2b= a+c
ACCORDING TO QUESTION
2/(x+3)=1/(x+2)+1/(x+5)
FROM THIS WE GET x=1
2b= a+c
ACCORDING TO QUESTION
2/(x+3)=1/(x+2)+1/(x+5)
FROM THIS WE GET x=1
Answered by
4
a1=1/x+2
a2=1/x+3
a3=1/x+5
common difference=a2 - a1=a3 - a2
2(a2)=a3+a1
2(1/x+3)=1/x+2 + 1/x+5
2/x+3=(x+5+x+2)/(x+2)(x+5)
2/x+3=(2x+7)/x^2+7x+10
2x^2+14x+20=2x^2+13x+21
14x - 13x=20 - 21
x=1
a2=1/x+3
a3=1/x+5
common difference=a2 - a1=a3 - a2
2(a2)=a3+a1
2(1/x+3)=1/x+2 + 1/x+5
2/x+3=(x+5+x+2)/(x+2)(x+5)
2/x+3=(2x+7)/x^2+7x+10
2x^2+14x+20=2x^2+13x+21
14x - 13x=20 - 21
x=1
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