Math: (Finite) Arithmetic Series Problem

I found this interesting problem, well someone asked me, and I answered that on a whiteboard. But because I ran out of marker, I type the steps here. So that it will last longer.
The problem goes like this (paraphrased):

The sum of first 10 members of an Arithmetic progression is 110.

The sum of 11^th and 13^th terms is -4.

What is the sum of the first and second members?

It looks easy, maybe it can be done under 100 seconds. BUT, if we choose the wrong path, we should be be super sweating.

Let's Translate That to Math Expressions

First, Denotation

• Sum of n-finite series is usually denoted by $ S_n $.
• The $ n^{th} $ member (or term) is denoted by $ a_n $.
• The 1^st term is denoted by $ a $ (or to be consistent, $ a_1 $).
  To make it simpler, let's use $ a $.
• The difference between consecutive terms is denoted by $ d $ (as in difference).

Second, the Finite Arithmetic Formulas

• Arithmetic series (sum):

  $ S_n = n/2(2a + (n-1)d) $

• The progression member/term:

  $ a_n = a + (n-1)d $

Third, the Concept

We need to expand the known variables using the formulas above, so then we'll generate a system of linear equations in two variables.

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The Problem Translated Into MATH EXPRESSION

• $ S_10 = 110 $
• $ a_11 + a_13 = -4 $
• The question: $ a_1 + a_2 = ? $

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➡️ We need to find the $ a $ and $ d $, so then:

Expand the $ S_10 = 110 $

$ S_10 = 110 $

$ 10/2(2a + (10 - 1)d) = 110 $

$ 5(2a + 9d) = 110 $

$ 2a + 9d = 110/5 $

$ 2a + 9d = 22 ... (1) $ Our first linear equation

Expand the $ a_11 + a_13 = -4 $

$ a_11 + a_13 = -4 $

$ (a + (11 - 1)d) + (a + (13 - 1)d) = -4 $

$ (a + 10d) + (a + 12d) = -4 $

$ 2a + 22d = -4 ...(2)$ Second linear equation

Now we have 2 equations in 2 variables; we can find the value of each thing using basic elimination

$ 2a + 9d = 22 ... (1) $

$ 2a + 22d = -4 ... (2)$

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$ -13d = 26 $

$ d = -2 $

And then the $ a $; using substitution; let's take the 1^st equation

$ 2a + 9d = 22...(1)$

$ 2a + 9(-2) = 22 $

$ 2a + -18 = 22 $

$ 2a = 22 + 18 $

$ a = 40/2 $

$ a = 20 $ COMPLETE!

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FINALLY

The question is $ a_1 + a_2 = ?$

Expand that into:

$ ⇔ (a + (1 - 1)d) + (a + (2 - 1)d)$

$ ⇔ (a) + (a + d)$

$ ⇔ 2a + d $

Substitute $ a = 20 $ and $ d = -2 $

$ ⇔ 2(20) + (-2) $

$ ⇔ 38 $

And the answer is 38.

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Additional

• The other series/progression is called Geometric. Instead of (static) difference, it has (static) ratio between consecutive members.
• There are also sequences with difference or ratio "stages", sort of mutating but with finite steps. Those cannot use the formulas above.