Problem: In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by 30 . Find the sum of the four terms.
Solution:
Here is a comprehensive lecture note on the given problem, complete with explanations of the required concepts and theorems. We'll break down each step for clarity and understanding.
We are given an increasing sequence of four positive integers where:
- The first three terms form an arithmetic progression (AP).
- The last three terms form a geometric progression (GP).
- The difference between the first and fourth terms is 30.
Our task is to find the sum of these four terms.
ΒΆ Concepts and Theorems Needed
- An arithmetic progression is a sequence of numbers where the difference between consecutive terms is constant.
- If we denote the first term by a and the common difference by d, then the terms are a,a+d,a+2d,β¦.
- A geometric progression is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
- If we denote the first term by b and the common ratio by r, then the terms are b,br,br2,β¦.
Letβs denote the four numbers as a,a+d,a+2d,x. According to the problem, the first three terms a,a+d,a+2d form an AP, and the last three terms a+d,a+2d,x form a GP.
Given:
Plugging in x=a+d(a+2d)2β, we have:
a+d(a+2d)2ββa=30.
Multiply through by a+d to eliminate the fraction:
(a+2d)2βa(a+d)=30(a+d).
Expanding both sides:
a2+4ad+4d2βa2βad=30a+30d.
Simplifying:
3ad+4d2=30a+30d.
Rearrange the equation:
2d(2dβ15)=3a(10βd).
Both a and d are positive integers, implying 2dβ15 and 10βd must have the same sign.
To satisfy both conditions 2dβ15>0 and 10βd>0, we analyze possible values:
-
For 10βd>0:
-
For 2dβ15>0:
- d>215β=7.5.
Therefore, valid integer values for d are 8 and 9.
-
d=8:
- Substitute d=8 into the equation:
2β
8(2β
8β15)=3a(10β8)
16β
1=6a
- a=616β, not an integer. So dξ =8.
-
d=9:
- Substitute d=9 into the equation:
2β
9(2β
9β15)=3a(10β9)
18β
3=3a
54=3a
- a=18, which is an integer.
Using a=18 and d=9, the sequence is 18,27,36,48.
- AP: 18,27,36 β Common difference =9.
- GP: 27,36,48 β Common ratio r=2736β=34β and consistency 3648β=34β.
Difference Between First and Fourth:
48β18=30.
The sequence satisfies all conditions. Therefore, the sum is:
18+27+36+48=129.
Hence, the sum of the four terms is 129β.