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प्रश्न
If \[\frac{1}{x + 2}, \frac{1}{x + 3}, \frac{1}{x + 5}\] are in A.P. Then, x =
पर्याय
5
3
1
2
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उत्तर
Here, we are given three terms,
First term (a1) = `1/(x + 2)`
Second term (a2) = `1/(x + 3)`
Third term (a3) = `1/(x + 5)`
We need to find the value of x for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,
d = a2 - a1
`d = (1/(x + 3)) - (1/(x + 2 ))`
`d = ((x + 2) - (x - 3))/((x + 2)(x + 3))`
`d =(x +2-x - 3)/((x + 2)(x + 3))`
`d =( -1) /((x + 2)(x +3 ))` ...............(1)
Also,
`d = a_3 - a_2`
`d = (1/(x +5 ) ) - (1/(x + 3))`
`d = (( x + 3) - ( x + 5) ) /((x + 5)(x +3 ))`
`d = (x +3 - x - 5)/((x + 5)(x + 3))`
`d = (-2)/((x + 5)(x + 3))` .............(2)
Now, on equating (1) and (2), we get,
`(-2)/((x +5)(x + 3)) = (-1)/((x + 3)(x +2 ))`
2(x +3 )( x + 2) = 1 (x +5 ) ( x +3 )
2x + 4 = x +5
2x - x = 5 - 4
x = 1
Therefore, for x = 1 , these three terms will form an A.P.
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