Analyzing the Setup
We are given a geometric progression (G.P.) with distinct positive terms: a1,a2,a3,a4. Let the first term be a and the common ratio be r.
Given the constraints that terms are distinct and positive, we have a>0, r>0, and $r
eq 1$. The terms are defined as:
a1=a, a2=ar, a3=ar2, and a4=ar3.
We construct a new sequence bn using the partial sums of this G.P.:
b1=a
b2=a(1+r)
b3=a(1+r+r2)
b4=a(1+r+r2+r3)
The Arithmetic Progression Test
For a sequence to be in Arithmetic Progression (A.P.), the difference between consecutive terms must be constant. We check if b2−b1=b3−b2.
Calculating the differences:
b2−b1=a(1+r)−a=ar
b3−b2=a(1+r+r2)−a(1+r)=ar2
For the sequence to be in A.P., we require ar=ar2. Since a>0 and $r
eq 1$, we divide by ar to obtain 1=r.
This contradicts our initial constraint that $r
eq 1$. Therefore, the sequence is definitively not in A.P.
The Geometric Progression Test
For a sequence to be in Geometric Progression (G.P.), the ratio of consecutive terms must be constant. We check if b1b2=b2b3.
Substituting our expressions:
b1b2=aa(1+r)=1+r
b2b3=a(1+r)a(1+r+r2)=1+r1+r+r2
Equating these ratios:
(1+r)=1+r1+r+r2
Cross-multiplying yields (1+r)2=1+r+r2. Expanding the left side gives 1+2r+r2=1+r+r2.
Subtracting 1+r+r2 from both sides results in r=0. Since our terms are positive, r must be greater than zero, meaning the sequence is not in G.P.
The Harmonic Progression Test
For
b1,b2,b3 to be in Harmonic Progression (H.P.), their reciprocals must be in A.P., satisfying the condition:
b22=b11+b31
Substituting our expressions and canceling
a:
1+r2=1+1+r+r21
Combining the right side:
1+r2=1+r+r21+r+r2+1=1+r+r22+r+r2
Cross-multiplying gives:
2(1+r+r2)=(2+r+r2)(1+r)
2+2r+2r2=2+3r+2r2+r3
Simplifying the equation leads to r+r3=0, or r(1+r2)=0. Since r>0, this equation has no real solution.
Thus, the sequence is not in H.P. We have systematically proven that the sequence bn does not follow the standard progression rules of A.P., G.P., or H.P.