A.P

Algebra Level pending

The sum of three positive numbers in an arithmetic progression is 27, and the sum of their squares is 293. Let λ , μ , ξ \lambda, \mu, \xi denotes these three numbers in that order. Submit your answer as λ 2 − ξ + 2 μ \lambda^2 - \xi + 2 \mu .


The answer is 20.

Syed Baqir by Syed Baqir , 24, Cyprus

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1 solution

Syed Baqir
Aug 27, 2015

a + a-d + a+d = 27 => 3a = 27 => a=27/3 = 9

and (9-d)^2 + 81 + (9+d)^2 = 293 => d = +- 5

Hence , 9 - 5 = 4 ,

Similarly, Second term = 9 and third = a + d = 9 + 5 = 14

Hence: (\lambda )^{ 2 }-(\xi )+2 \mu = (4)^2 - 14 + 2 (9) = 20

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It is already given in the question that their sum is 27. Why bother working out the values?

Pi Han Goh - 5 years, 9 months ago

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The main emphasis was the first three terms , I will edit my question, thanks

Syed Baqir - 5 years, 9 months ago

Could the answer be both 20 and 210? It did not say anything about whether d = 5 or -5.

Saya Suka - 4 years, 7 months ago

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