# Arithmetic and Geometric Progressions Problem Solving

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## Problem Solving - Basic

This section contains basic problems based on the notions of arithmetic and geometric progressions. Starting with an example, we will head into the problems to solve.

I have an arithmetic progression such that the initial term is 5 and the common difference is 10. What is minimum value of $n$ such that the $n^\text{th}$ term is larger than 100?

We can just start by listing out the numbers: $5,15,25,35,45,55,65,75,85, 95,105.$ We can clearly see that the $11^\text{th}$ number is larger than 100, and thus $n=11.$

However, note that this will become impractical if the common difference becomes smaller and/or the number we are looking for becomes larger.

A practical way to solve it is via applying the $n^\text{th}$ term formula. With $a = 5, d = 10$, we have $T_n = a + (n-1) d > 100$. Then $5 + (n-1) \cdot 10 > 100$. Solving for $n$ yields $n > 10.5$. So the $11^\text{th}$ term is the smallest term that satisfies the condition. $_\square$

Here comes the problems for you to solve.

One side of an equilateral triangle is 24 cm. The midpoints of its sides are joined to form another triangle whose midpoints, in turn, are joined to form still another triangle. This process continues indefinitely.

Find the sum of the perimeters of all these triangles that are defined above.

Once a man did a favor to a king that made the king very happy. Out of joy the king told the man to wish for anything and he would be granted. The man wanted to ask for the whole kingdom which was worth 1500 trillion dollars, but obviously that would make the king mad and he would never be granted that wish.

The man who happened to be a mathematician thought a little bit and said the following:

"Bring in a big piece of rug with an $8\times 8$ grid in it. Starting from the top left square, put one dollar in that square. Put two dollars in the square next to it and then double of that, four dollars, in the next square and so on. When you reach the end of the first row, continue on to the next row, doubling the amount every time as you move to the next square, all the way until the $64^\text{th}$ square at the bottom right."

The king thought for a second. The first square will take one dollar, the second two dollars, the third, four dollars, and next 8 dollars, and then 16 dollars, and then 32 dollars, 64 dollars, 128 dollars, 256 dollars, and so on. That's not too bad. I can do it.

The king agreed. What happened next?

## Problem Solving - Intermediate

This section contains a bit harder problems than the previous section. But all these can be solved using arithmetic and geometric problems. Here we go:

We have three numbers in an arithmetic progression, and another three numbers in a geometric progression. Adding the corresponding terms of the two series, we get $120 , 116 , 130$. If the sum of all the terms in the geometric progression is $342$, what is the largest term in the geometric progression?

$$

**Details and Assumptions:**

- If the terms of the AP are A, B, C, and the terms of the GP are X, Y, Z, then
**adding the corresponding terms**will give us A+X, B+Y, C+Z.

## Problem Solving - Advanced

This section has problems which need advanced understanding of the notions and generally get solved on using multiple notions at a time. Let's give these problems an attempt.

In JEE examination the paper consists of 90 questions. The marks are awarded in such a way that if a person gets a question correct, he gets $+4$ marks; if he does it wrong, he gets $-2$ marks; if he leaves the question unanswered, he gets $0$ marks (as per 2015). Find the sum of all possible marks that a student can get in JEE.

Let $A=\{a_1, a_2, \ldots, a_n\}$ be a set of the first $n$ terms of an arithmetic progression. Similarly, let $B=\{b_1, b_2, \ldots, b_n\}$ be a set of the first $n$ terms of a geometric progression.

If a new set $C=A+B=\{a_1+b_1, a_2+b_2, \ldots, a_n+b_n\}$ and the first four terms of $C$ are $\{0, 0, 1, 0\},$ what is the $11^\text{th}$ term of $C?$

$\begin{aligned} &\displaystyle\sum_{n=0}^{7}\log_{3}(x_{n}) &= 308 \\ 56 \leq & \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right )& \leq 57 \\ \end{aligned}$

The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of 3. If they satisfy the two conditions above, find $\log_{3}(x_{14}).$

Suppose $2015$ people of different heights are arranged in a straight line from shortest to tallest such that

(i) the tops of their heads are collinear, and

(ii) for any two successive people, the horizontal distance between them is equal to the height of the shorter of the two people.

If the shortest person is $49$ inches tall and the tallest is $81$ inches tall, then how tall is the person at the middle of the line, (in inches)?

Given that $a_1,a_2,a_3$ is an arithmetic progression in that order so that $a_1+a_2+a_3=15$ and $b_1,b_2,b_3$ is a geometric progression in that order so that $b_1b_2b_3=27$.

If $a_1+b_1, a_2+b_2, a_3+b_3$ are positive integers and form a geometric progression in that order, determine the maximum possible value of $a_3$.

The answer is of the form $\dfrac{a+b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers and the fraction is in its simplest form and $c$ is square free. Submit the value of $a + b + c + d$.

## See Also

**Cite as:**Arithmetic and Geometric Progressions Problem Solving.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/arithmetic-and-geometric-progressions-problem/