Advertisements
Advertisements
प्रश्न
If a, b, c and d are in proportion, prove that: (5a + 7b) (2c – 3d) = (5c + 7d) (2a – 3b).
Advertisements
उत्तर
It is given that
a, b, c, d are in proportion
Consider `a/b = c/d = k`
a = bk, c = dk
LHS = (5a + 7b)(2c – 3d)
LHS = (5bk + 7b)(2dk – 3d) ...[Substituting the values]
LHS = b(5k + 7) d(2k – 3) ...[Taking out the common terms]
LHS = bd (5k + 7)(2k - 3)
LHS = bd [5k (2k - 3) + 7(2k - 3)]
LHS = bd (10k2 - 15k + 14k - 21)
LHS = bd (10k2 - k - 21) ... (I)
RHS = (5c + 7d)(2a – 3b)
RHS = (5dk + 7d)(2bk – 3b) ...[Substituting the values]
RHS = d(5k + 7) b(2k – 3) ...[Taking out the common terms]
RHS = bd (5k + 7)(2k – 3)
RHS = bd [5k(2k - 3) + 7(2k - 3)]
RHS = bd (10k2 - 15k + 14k - 21)
LHS = bd (10k2 - k - 21) ... (II)
From (I) and (II),
Therefore, LHS = RHS.
APPEARS IN
संबंधित प्रश्न
If x, y, z are in continued proportion, prove that `(x + y)^2/(y + z)^2 = x/z`
Find the mean proportional between `6 + 3sqrt(3)` and `8 - 4sqrt(3)`
if `a/b = c/d` prove that each of the given ratio is equal to: `((8a^3 + 15c^3)/(8b^3 + 15d^3))^(1/3)`
Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?
What least number must be added to each of the numbers 5, 11, 19 and 37, so that they are in proportion?
What number must be added to each of the numbers 16, 26 and 40 so that the resulting numbers may be in continued proportion?
The 1st, 3rd, and 4th terms of a proportion are 12, 8, and 14 respectively. Find the 2nd term.
Show that the following numbers are in continued proportion:
16, 84, 441
If 40 men can finish a piece of work in 26 days, how many men will be required to finish it in 20 days?
In the proportional statement p : q :: r : s, which pair of terms represents the means, and which pair represents the extremes?
