In this guide, we will explore the method of finding the first term and the common difference in an arithmetic progression (AP) given specific conditions. Specifically, we will solve two problems step-by-step, demonstrating the detailed process of using summation formulas and relationships between terms in an AP.
Problem 1: Sum of the First Six Terms
Given that the sum of the first six terms of an arithmetic progression is 18 and the sixth term is twice the third term, we need to find the first term and the common difference.
Solution:
Let the first term be (a) and the common difference be (d).
The sum of the first 6 terms is given by:S_6 18
The sum of the first (n) terms of an AP is given by:
S_n frac{n}{2} (2a (n-1)d)
For (n 6):
(frac{6}{2} (2a 5d) 18)
(6a 15d 18 quad text{(Equation 1)})
The sixth term is given by:
(T_6 a 5d)
The third term is:
(T_3 a 2d)
We are given that the sixth term is twice the third term:
(a 5d 2(a 2d))
(a 5d 2a 4d)
(5d - 4d 2a - a)
(d a quad text{(Equation 2)})
Substituting Equation 2 into Equation 1:
(6a 15a 18)
(21a 18)
(a frac{18}{21} frac{6}{7})
This gives us:
(d a frac{6}{7})
Therefore, the first term (a frac{6}{7}) and the common difference (d frac{6}{7}).
(P.S.) Derivation of the General Term
The arithmetic sequence can be written as:
(frac{6}{7}, frac{12}{7}, frac{18}{7}, frac{24}{7}, frac{30}{7}, frac{36}{7}, ldots)
The general term of the AP can be expressed as:
(t_n a (n-1)d frac{6}{7} (n-1) frac{6}{7} frac{6n}{7}))
Problem 2: Sum of the First Six Terms with Another Condition
Given that the sum of the first six terms of an arithmetic progression is 18 and the sixth term is two times the third term, we need to find the first term and the common difference.
Solution:
Let the first term be (a) and the common difference be (d).
The sum of the first 6 terms is given by:S_6 18
The sum of the first (n) terms of an AP is given by:
S_n frac{n}{2} (2a (n-1)d)
For (n 6):
(frac{6}{2} (2a 5d) 18)
(6a 15d 18 quad text{(Equation 1)})
The sixth term is given by:
(T_6 a 5d)
The third term is:
(T_3 a 2d)
We are given that the sixth term is twice the third term:
(a 5d 2(a 2d))
(a 5d 2a 4d)
(5d - 4d 2a - a)
(d a quad text{(Equation 2)})
Substituting Equation 2 into Equation 1:
(6a 15a 18)
(21a 18)
(a frac{18}{21} frac{6}{7})
This gives us:
(d a frac{6}{7})
Therefore, the first term (a frac{6}{7}) and the common difference (d frac{6}{7}).
Summary
We have solved two problems involving arithmetic progressions. In both cases, we used the sum formula of an AP and the relationship between specific terms to find the first term and the common difference. The detailed steps demonstrated how to manipulate and solve the given equations to find the required values.