Advertisements
Advertisements
प्रश्न
If x, y, z are in continued proportion, prove that: `(x + y)^2/(y + z)^2 = x/z`.
Advertisements
उत्तर
x, y, z are in continued proportion
Let `x/y = y/z = k`
Then y = kz
x = yk
= kz × k
= k2z
Now L.H.S.
= `(x + y)^2/(y + z)^2`
= `(k^2 z + kz)^2/(kz + z)^2`
= `{kz(k + 1)}^2/{z(k + 1)}^2`
= `(k^2z^2(k + 1)^2)/(z^2(k + 1)^2)`
= k2
R.H.S. = `x/z`
= `(k^2z)/z`
= k2
∴ L.H.S. = R.H.S.
APPEARS IN
संबंधित प्रश्न
Check whether the following numbers are in continued proportion.
9, 12, 16
Find the value of the unknown in the following proportion :
3 : 4 : : p : 12
Find the fourth proportion to the following:
(p2q - qr2 ), (pqr - pr2 ) and (pq2 - pr2)
If y is the mean proportional between x and z, show that :
xyz (x+y+z)3 =(xy+yz+xz)3
Find the smallest number that must be subtracted from each of the numbers 20, 29, 84 and 129 so that they are in proportion.
If a : b : : c : d, then prove that
2a+7b : 2a-7b = 2c+7d : 2c-7d
Find the value of x in each of the following proportions:
55 : 11 : : x : 6
If a, b, c and d are in proportion, prove that: `(a^2 + ab)/(c^2 + cd) = (b^2 - 2ab)/(d^2 - 2cd)`
If a, b, c and d are in proportion, prove that: `(a^2 + b^2)/(c^2 + d^2) = "ab + ad - bc"/"bc + cd - ad"`
Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
2 kg : 80 kg and 25 g : 625 g
