The sum of 3 consecutive terms of an arithmetic sequence is 165. The second of these terms is 11 times the first. Find the 3 terms
From what I gathered:
S(3) = 165 t(2)= 11(t1)
The given formulas are:
Sn= n/2[2a+(n-1)d] and tn=a+(n-1)d
I have no idea how to continue.
All you need to find is a and d
he first thing to do is use the formulas correctly, and use all the data given:
S3 = 3/2 (2a + 2d) = 165
T2 = a+d = 11a
So, d = 10a
3/2 (2a+20a) = 165
22a = 110
a = 5
d = 50
The sequence starts 5 55 105
To solve the problem, we need to use the formulas you provided and set up a system of equations.
Let's break down the problem and use the following variables:
- t1: the first term of the arithmetic sequence
- d: the common difference between the terms
We're given two pieces of information:
1. The sum of three consecutive terms is 165, which can be written as:
t1 + (t1 + d) + (t1 + 2d) = 165
2. The second term is 11 times the first term, which can be written as:
t1 + d = 11t1
Now, we can solve the system of equations:
From equation 2, we can rewrite it as:
d = 10t1
Substituting this value in equation 1, we get:
t1 + (t1 + 10t1) + (t1 + 20t1) = 165
4t1 + 30t1 = 165
34t1 = 165
t1 = 165/34
Now that we have the value of t1, we can find the common difference (d):
d = 10t1
d = 10 * (165/34)
Finally, we can find the three terms by plugging in the values of t1 and d:
First term: t1
Second term: t1 + d
Third term: t1 + 2d
Substituting the values:
First term: 165/34
Second term: 165/34 + 10 * (165/34)
Third term: 165/34 + 2 * 10 * (165/34)
Simplify these expressions, and you will have the three terms of the arithmetic sequence.