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Let the terms of the AP be $$a-3d, a-d, a+d, a+3d$$

So their addition will give 50 $$4a=50$$ This is where the whole thing breaks down. I realize that the fourth term is 4 times that of first. So some may say that fourth term will be $4a-12d$. But I disagree. Since it’s an AP, the fact that it is four times should already be included in the series. Since they are variables, their values should be according to whatever the question is about. It’s a controversial topic, since according to many it is extremely easy, and I get that too, but I just need this one conceptual problem cleared.

P.S. This was an MCQ, so I got the right answer anyway (5,10,15,20) but I still want a proper method to solve it.

Thanks

Aditya
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  • Why do you assume the step size in your arithmetic sequence is even? (I.e., $2d$ is always even but not all arithmetic sequences have even step sizes.) – Eric Towers Sep 09 '19 at 14:36
  • +1 This question was asked in my test today –  Oct 27 '21 at 02:30

2 Answers2

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Here's an arithmetic progression with your weirdly assumed even step size that does not agree with "the fact that it is four times should already be included in the series": $$ -3, -1, 1, 3 $$ Here's another: $$ 4, 6, 8, 10 $$ Here's another: $$ 1000, 1100, 1200, 1300 $$

The claim "the fact that it is four times should already be included in the series" only holds if the first member of the progression is $1$ times the step size.

So, a proper method to solve it... First, let's not make the weird assumption that the step size is even and let's start at $a$ so our final answer is easy to write down. Our arithmetic progression is $$ a, a + d, a + 2d, a+3d \text{.} $$ Their sum, $4a+6d = 50$, and the last is four times the first, $a + 3d = 4a$. From the second of these, $a = d$. Substituting for $a$ in the first, $10d = 50$, so $d = 5$ and then $a = d = 5$. Our arithmetic progression is $$ a = 5, 10, 15, 20 = a + 3d \text{.} $$

Eric Towers
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$a$ and $d$ are not the values you need. $a$ itself is not part of the sequence. Call the first term $x$ and the progression is 2d. So we have x,x+2d,x+4d,x+6d. The sum equals 50 and x+6d = 4x. If we simultaneously solve these equations, we get x = 5 and d = $\frac{5}{2}$ . So the terms are 5,10,15,20.

Joel Pereira
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