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If p, q and r are in A.P. and x, y, z are in G.P. then xq–r. y r–p. zp–q is equal to
- a)0
- b)– 1
- c)1
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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If p, q and r are in A.P. and x, y, z are in G.P. then xq–r. y r...
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If p, q and r are in A.P. and x, y, z are in G.P. then xq–r. y r...
Solution:
Given, p, q, and r are in A.P. and x, y, and z are in G.P.
We know that for a G.P., the product of the first and third terms is equal to the square of the second term, i.e., xy = z².
Similarly, for an A.P., the sum of the first and third terms is twice the second term, i.e., p + r = 2q.
Using these relations, we can simplify the expression xqr. y rp. zpq as follows:
xqr. y rp. zpq = xyz. qrp. p²z
= xyz. q(p + r). pz
= xyz. q(2q). pz (using p + r = 2q)
= 2q³xyzp
= 2(xyzp)³/3
= 2z³p³ (since xy = z²)
= 2(pz)³/3
= 2(pq/2)³/3 (since p + r = 2q)
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
Given, p, q, and r are in A.P. and x, y, and z are in G.P.
We know that for a G.P., the product of the first and third terms is equal to the square of the second term, i.e., xy = z².
Similarly, for an A.P., the sum of the first and third terms is twice the second term, i.e., p + r = 2q.
Using these relations, we can simplify the expression xqr. y rp. zpq as follows:
xqr. y rp. zpq = xyz. qrp. p²z
= xyz. q(p + r). pz
= xyz. q(2q). pz (using p + r = 2q)
= 2q³xyzp
= 2(xyzp)³/3
= 2z³p³ (since xy = z²)
= 2(pz)³/3
= 2(pq/2)³/3 (since p + r = 2q)
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
= (pq/2)²
= (q² - (p/2)²)
= (r² - (q/2)²) (since p, q, and r are in A.P.)
= (rp/2)²
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If p, q and r are in A.P. and x, y, z are in G.P. then xq–r. y r–p. zp–q is equal toa)0b)– 1c)1d)none of theseCorrect answer is option 'C'. Can you explain this answer?
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