Question 754592: In an arithmetic progression, the 3rd term is equal to three times the 1st term, and the sum of the first two terms is equal to 3.
find the nth term of this series.
Hence find the sum of the terms from the 1st to the nth term inclusive.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In an arithmetic progression,
the 3rd term is equal to three times the 1st term, and the sum of the first two terms is equal to 3.
find the nth term of this series.
------
General arithmetic series term:
a(n) = a(1) +(n-1)d
-------------------------
Therefore a(3) = a(1) + 2d
------
Equation:
From the statement of the problem,
a(1) + 2d = 3*a(1)
2a(1) = 2d
a(1) = d
---------------------
From the statement of the problem,
Equation:
a(1) + a(2) = 3
a(2) = 3-d
Hence find the sum of the terms from the 1st to the nth term inclusive.
1st term: d
2nd term: 3-d
3rd term: 3-d + (3-2d) = 6-3d = 3(2-d)
4th term: 6-3d + (3-2d) = 9-5d
etc.
----
S(n) = (n/2)(a(1)+a(n)) = (n/2)(d+(d+(n-1)d)) = (n/2)((n+1)d)
=================
Cheers,
Stan H.
=================
|
|
|
