Question

I have two questions. Please try to answer both!

How many non-similar triangles have angles whose degree measures are distinct positive integers in an arithmetic progression?


In an arithmetic sequence a7 −2a4 = 1, a3 = 0. Find the common difference.


Thank you!

Answer 1

Answer:

(a)59

(b)$d=\dfrac{1}{2}$

Step-by-step explanation:

Part A

The nth term of an arithmetic sequence, $T_n=a+(n-1)d$

Let the first angle of the triangle =a

Let the common difference = d.

Since the measures of the angles are in an arithmetic progression

We have:

a + (a + d) + (a + 2d) = 180

3a+3d=180

a+d=60

Since a or d cannot be equal to zero, the minimum possible values of a and d is 1 and the maximum possible values of a and d is 59.

Therefore, there are 59 non-similar triangles.

Part B

In an arithmetic sequence

$a_7 -2a_4 = 1, a_3 = 0.$

$2a_4=a_7-1\\a_4=\dfrac{a_7-1}{2}$

Common difference,

$d=a_4-a_3\\=\dfrac{a_7-1}{2}-0\\d=\dfrac{a_7-1}{2}$

$a_7=a_4+3d\\=\dfrac{a_7-1}{2}+3(\dfrac{a_7-1}{2})\\=\dfrac{a_7-1}{2}+\dfrac{3a_7-3}{2}\\=\dfrac{a_7-1+3a_7-3}{2}\\a_7=\dfrac{4a_7-4}{2}\\ Cross multiply\\2a_7=4a_7-4\\2a_7-4a_7=-4\\-2a_7=-4\\a_7=2$

Therefore:

$d=\dfrac{2-1}{2}\\d=\dfrac{1}{2}$