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Worked Examples · Example 10

Q.The sum of three numbers in A.P. is 24 and their product is 440. Find the numbers.

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Writing the three A.P. numbers symmetrically as a−d,a,a+da-d,a,a+d turns both conditions into simple equations, giving the numbers 5, 8, 11.

Three numbers in A.P. can always be written symmetrically as:

a−d,a,a+da-d,\quad a,\quad a+d

where aa is the middle term and dd is the common difference.

  1. Let the three numbers in A.P. be a−da-d, aa, a+da+d.
  2. Sum condition: (a−d)+a+(a+d)=24⇒3a=24⇒a=8(a-d)+a+(a+d) = 24 \Rightarrow 3a=24 \Rightarrow a=8.
  3. Product condition: (a−d)(a)(a+d)=440⇒a(a2−d2)=440(a-d)(a)(a+d) = 440 \Rightarrow a(a^2-d^2) = 440.
  4. Substitute a=8a=8: 8(64−d2)=4408(64-d^2) = 440.
  5. Divide both sides by 8: 64−d2=5564-d^2 = 55. …

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