# Arithmatic Progression

Arithmetic and Geometric progressions.

### Arithmatic Progression

find the four numbers in AP whose sum is 50 and which the greatest number is 4 times the least.

I request you to give me solution for the above problem.

Thanks & Regards,
Shrinidhi

shrinidhi

Posts: 2
Joined: Mon Aug 17, 2009 8:08 am

Hallo,

Ok, let's define the elements of the progression with a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, a<sub>4</sub>. We know that the sum of the first and last elements of the progression is equal to the sum of two equal distant elements =>
a1 + a4 = a2 + a3; a4 = a1 => a1 + a4 = 5a1;
5a1 = a2 + a3 => 10a1 is equal to the whole sum => 10a1 = 50; => a1 = 5

a4 = 4a1 = 4.5 =20; => a2 = 10; a3 = 15;

q.e.d.
masterfromkardjali

Posts: 2
Joined: Fri Aug 21, 2009 2:41 pm

### Another method

Lets suppose, 4 terms are, a, a+d, a+2d, a+3d.

Then as per given condition, 4a=a+3d, thereby we have 3a=3d, meaning a=d.

So if we consider the Sum of N terms, Sn= 1/2(a+a+3d)*4=50, then we have:

1/2(2a+3a)*4=50
2(5a)=50
a=5

a=5, a+d= 5+5= 10, a+2d=15, a+3d=20

5,10.15.20.

Regds,
Prasanna
Prasanna Chandran
prasanna_chandran

Posts: 3
Joined: Thu Nov 05, 2009 9:11 am
Location: India