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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 𝑞/𝑝 = 𝑟/𝑞 q2 = pr Solving the equation px2 + 2qx + r = 0 For ax2 + bx + c roots are x = (−𝑏 ±√(𝑏2 − 4𝑎𝑐))/2𝑎 Here a = p, b = 2q & c = r Hence the roots of equation px2 + 2qx + r = 0 are x = (−2q ±√(4q2 −4rp))/2p Putting q2 = pr from (1) x = (−2q ±√(4q2 −4rp))/2p = (−2q ±√(4pr −4rp))/2p = (−2𝑞±0)/2𝑝 = (−2𝑞)/2𝑝 = (−q )/p Thus x = (−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, (−q )/p is a root of dx2 + 2ex + f = 0 Putting x = (−q )/p in dx2 + 2ex + f = 0 d ((−q )/p)^2 + 2e ((−q )/p) + f = 0 d ((−q)^2 )/p2 − ((2eq )/p) + f = 0 (𝑑𝑞^2 )/p2 − ((2eq )/p) + f = 0 (𝑑𝑞2 −2𝑒𝑞𝑝+𝑓𝑝2)/𝑝2 = 0 dq2 – 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 – (d )/p = 𝑓/𝑟 – 𝑒/𝑞 i.e. to show (e )/q + 𝑒/𝑞 = 𝑓/𝑟 + (d )/p To show :- 2𝑒/𝑝 = 𝑓/𝑟 + 𝑑/𝑝 Now, from (2) dq2 – 2eap + fp2 = 0 Dividing this by pq2 dq2/pq2 – 2epq/pq2 + fp2/pq2 = 0/pq2 (d )/p – (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 ∴ (d )/p, (e )/q, (f )/r are in A.P Hence proved

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