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In the following question . enter image description here

How do they get to know that the possible values of $a$ are only $3$ and $4$?

Robert Z
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Koolman
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  • Because, since the terms are positive interger, then common diference must be interger. So $5-a$ must divide $16$ – Arnaldo Dec 06 '16 at 18:07

2 Answers2

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$\lambda$ has to be an integer, so $a$ cannot be $2$. It has to be positive, so $a$ cannot be $1$ or greater than $5$, and it has to be finite, so $a$ cannot be $5$. Finally, $a$ itself must be positive. Thus $a$ can only be $3$ or $4$.

Brian Tung
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Well, we know that a, b, c, and d are positive integers. If we try a=4 or anything greater than 4, we get the common difference to be negative, which cannot be so since we are given the arithmetic progression a

MathGuy
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