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AMC 10 Geometry Math Olympiad USA Math Olympiad

Arithmetic sequence | AMC 10A, 2015 | Problem 7

Try this beautiful problem from Algebra: Arithmetic sequence from AMC 10A, 2015, Problem.

Arithmetic sequence – AMC-10A, 2015- Problem 7


How many terms are in the arithmetic sequence $13$, $16$, $19$, $\dotsc$, $70$, $73$?

  • \(20\)
  • \(21\)
  • \(24\)
  • \(60\)
  • \(61\)

Key Concepts


Algebra

Arithmetic sequence

Check the Answer


Answer: \(21\)

AMC-10A (2015) Problem 7

Pre College Mathematics

Try with Hints


The given terms are $13$, $16$, $19$, $\dotsc$, $70$, $73$. We have to find out the numbers of terms…..

If you look very carefully then the distance between two digits is \(3\).Therefore this is an Arithmetic Progression where the first term is \(13\) and common difference is \(3\)

Can you now finish the problem ……….

$a+(n-1) d \Longrightarrow 13+(n-1) 3=73$

\(\Rightarrow n=21\)

The number of terms=\(21\)

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AMC-8 Math Olympiad USA Math Olympiad

AP GP Problem | AMC-10A, 2004 | Question 18

Try this beautiful problem from Algebra : AP & GP from AMC 10A, 2004.

AP GP – AMC-10A, 2004- Problem 18


A sequence of three real numbers forms an arithmetic progression with a first term of \(9\). If \(2\) is added to the second term and \(20\) is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression?

  • \(1\)
  • \(4\)
  • \(36\)
  • \(49\)
  • \(81\)

Key Concepts


Algebra

AP

GP

Check the Answer


Answer: \(1\)

AMC-10A (2003) Problem 18

Pre College Mathematics

Try with Hints


We assume the common difference in the AP series is \(d\) …..Therefore the numbers will be \( 9, (9+d),(9+2d)\) .Therefore according to the condition if we add \(2\) with \(2\)nd term and add \(20\) to the third term the numbers becomes in Geometric Progression……..\(9\) , \((9+d+2)=11+d\) , \((9+2d+20)=29+2d\)

can you finish the problem……..

Now according to the Geometric Progression , \(\frac{11+d}{9}=\frac{29+d}{11+d}\)

\(\Rightarrow (11+d)^2 =9(29+2d)\)

\(\Rightarrow d^2 +4d-140=0\)

\(\Rightarrow (d+14)(d-10)=0\)

\(\Rightarrow 10 ,-14\)

can you finish the problem……..

Therefore we choose the value of \(d=-14\) (as smallest possible value for the third term)

The third term will be \( 2(-14)+29=1\)

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AMC 10 Math Olympiad USA Math Olympiad

Sectors in Circle | AMC-10A, 2012 | Problem 10

Try this beautiful problem from Geometry based on Sectors in Circle.

Sectors in Circle – AMC-10A, 2012- Problem 10


Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?

  • \(6\)
  • \(12\)
  • \(14\)
  • \(8\)
  • \(16\)

Key Concepts


Geometry

Circle

AP

Check the Answer


Answer: \(8\)

AMC-10A (2012) Problem 10

Pre College Mathematics

Try with Hints


We have to find out  the degree measure of the smallest possible sector angle.Let $x$ be the smallest sector angle and $r$ be the difference between consecutive sector angles,

Therefore the angles are $x, x+r, a+2r, \cdots. x+11r$. Now we know that sum of the angles of all sectors of a circle is \(360^{\circ}\).Can you find out the values of \(x\) and \(r\)?

can you finish the problem……..

Therefore using the AP formula we will get ,

\(\frac{x+x+11r}{2} . 12=360\)

\(\Rightarrow x=\frac{60-11r}{2}\)

can you finish the problem……..

Since all sector angles are integers so $r$ must be a multiple of 2. Now an even integers for $r$ starting from 2 to minimize $x.$ We find this value to be 4 and the minimum value of $x$ to be \(\frac{60-11(4)}{2}=8\)

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AMC 10 Math Olympiad USA Math Olympiad

Sum of digits | AMC-10A, 2020 | Problem 8

Try this beautiful problem from Algebra based on sum of digits

Sum of digits – AMC-10A, 2020- Problem 8


What is the value nof

\(1+2+3-4+5+6+7-8+……+197+198+199-200\)?

  • \(9800\)
  • \(9900\)
  • \(10000\)
  • \(10100\)
  • \(10200\)

Key Concepts


Algebra

Arithmetic Progression

Series

Check the Answer


Answer: \(9900\)

AMC-10A (2020) Problem 8

Pre College Mathematics

Try with Hints


The given sequence is \(1+2+3-4+5+6+7-8+……+197+198+199-200\). if we look very carefully then notice that \(1+2+3-4\)=\(2\), \(5+6+7-8=10\),\(9+10+11-12=18\)…..so on.so \(2,10,18……\) which is in A.P with common difference \(8\). can you find out the total sum which is given….

can you finish the problem……..

we take four numbers in a group i.e \((1+2+3-4)\),\((5+6+7-8)\),\((9+10+11-12)\)……,\((197+198+199-200)\). so there are \(\frac{200}{4}=50\) groups. Therefore first term is\((a)\)= \(2\) ,common difference\((d)\)=\(8\) and numbers(n)=\(50\). the sum formula of AP is \(\frac{n}{2}\{2a+(n-1)d\}\)

can you finish the problem……..

\(\frac{n}{2}\{2a+(n-1)d\}\)=\(\frac{50}{2}\{2.8+(50-1)8\}\)=\(9900\)

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AMC 10 USA Math Olympiad

Co-ordinate Geometry – AMC 10B – 2019 – Problem No – 4

Co-ordinate Geometry – AMC 10B – 2019 – Problem No – 4


This problem on co-ordinate geometry is from AMC 10B, 2019. First try it yourself.

All lines with equation ax+by= c such that a,b,c form an arithmetic progression pass through a common point . What are the coordinates of that point?

  • (-1,2)
  • (0,1)
  • (1,-2)
  • (1,0)

Key Concepts


Arithmetic Progression

2D – Co- ordinate Geometry

Cartesian System of Points

Check the Answer


Answer: (-1,2)

AMC 10B – 2019 – Problem No – 4

Challenges and Thrills in Pre-College Mathematics

Try with Hints


If all lines satisfy the condition, then we can just plug in values for a,b,c that form an arithmetic progression. Let’s use a =1,b= 2,c=3 and a= 1,b=3,c =5. Then the two lines we get are:

x+2y =3

x+3y = 5 so

y= 2

Now plug the value of y into one of the previous equations. We get :

x+4=3

x = -1

Ans is (-1,2)

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