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Question Number 134553 by liberty last updated on 05/Mar/21

$$$$ If three numbers are in Arithmetic progression prove that the square of middle number is equal to the sum of the product of the first and the last numbers and square of the common difference? \n

Answered by MJS_new last updated on 05/Mar/21

$$\mathrm{the}\mathrm{numbers}\mathrm{are}x-d;x;x+d$$ $$x^2=(x-d)(x+d)+d^2$$ $$\mathrm{true}$$

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