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Example 24 - If p, q, r are in GP and equations px2 + 2qx + r = 0 - AP and GP mix questions

 

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Example 24 If p,q,r are in G.P. and the equations, px2 + 2qx + r = 0 and dx2 + 2ex + f = 0 have a common root, then show that (d )/p, (e )/q, (f )/r are in A.P It is given that p, q, r are in G.P So, their common ratio is same ๐‘ž/๐‘ = ๐‘Ÿ/๐‘ž q2 = pr Solving the equation px2 + 2qx + r = 0 For ax2 + bx + c roots are x = (โˆ’๐‘ ยฑโˆš(๐‘2 โˆ’ 4๐‘Ž๐‘))/2๐‘Ž Here a = p, b = 2q & c = r Hence the roots of equation px2 + 2qx + r = 0 are x = (โˆ’2q ยฑโˆš(4q2 โˆ’4rp))/2p Putting q2 = pr from (1) x = (โˆ’2q ยฑโˆš(4q2 โˆ’4rp))/2p = (โˆ’2q ยฑโˆš(4pr โˆ’4rp))/2p = (โˆ’2๐‘žยฑ0)/2๐‘ = (โˆ’2๐‘ž)/2๐‘ = (โˆ’q )/p Thus x = (โˆ’q )/p is the root of the equation px2 + 2qx + r = 0 Also, it is given that equations px2 + 2qx + r = 0 & dx2 + 2ex + f = 0 have a common root So, (โˆ’q )/p is a root of dx2 + 2ex + f = 0 Putting x = (โˆ’q )/p in dx2 + 2ex + f = 0 d ((โˆ’q )/p)^2 + 2e ((โˆ’q )/p) + f = 0 d ((โˆ’q)^2 )/p2 โˆ’ ((2eq )/p) + f = 0 (๐‘‘๐‘ž^2 )/p2 โˆ’ ((2eq )/p) + f = 0 (๐‘‘๐‘ž2 โˆ’2๐‘’๐‘ž๐‘+๐‘“๐‘2)/๐‘2 = 0 dq2 โ€“ 2eap + fp2 = 0 We need to show (d )/p, (e )/q, (f )/r are in AP i.e. we need to show their common difference is some i.e. to show :- (e )/q โ€“ (d )/p = ๐‘“/๐‘Ÿ โ€“ ๐‘’/๐‘ž i.e. to show (e )/q + ๐‘’/๐‘ž = ๐‘“/๐‘Ÿ + (d )/p To show :- 2๐‘’/๐‘ = ๐‘“/๐‘Ÿ + ๐‘‘/๐‘ Now, from (2) dq2 โ€“ 2eap + fp2 = 0 Dividing this by pq2 dq2/pq2 โ€“ 2epq/pq2 + fp2/pq2 = 0/pq2 (d )/p โ€“ (2e )/q + (fp )/q2 = 0 (d )/p + (fp )/q2 = (2e )/q Putting q2 = pr from (2) d/p + fp/pr = 2e/q d/p + f/r = 2e/q which is what we have to prove โˆด (d )/p, (e )/q, (f )/r are in A.P Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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