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What is the proof for the pth, qth and rth terms of a GP identity?
In algebra, if the p-th, q-th and r-th terms of a geometric progression (GP) are given as a, b and c respectively, you are usually asked to prove that:
a^(q - r) × b^(r - p) × c^(p - q) = 1
The key to this proof is using distinct variables for the sequence properties so you do not confuse them with the given terms. Let the first term of the progression be A and the common ratio be R. The n-th term of a geometric sequence is given by a_n = a r^(n-1) 1.
Using this formula, write out the three given terms:
a = A × R^(p - 1)
b = A × R^(q - 1)
c = A × R^(r - 1)
Next, substitute these expressions into the left-hand side of the identity: (A × R^(p - 1))^(q - r) × (A × R^(q - 1))^(r - p) × (A × R^(r - 1))^(p - q)
By applying the power rule of exponents, expand the terms to separate the base A from the base R: A^(q - r) × R^((p - 1)(q - r)) × A^(r - p) × R^((q - 1)(r - p)) × A^(p - q) × R^((r - 1)(p - q))
Group the terms with base A together, and the terms with base R together. When you multiply terms with the same base, you add their exponents.
For the base A, the exponent sum is: (q - r) + (r - p) + (p - q) = 0 This simplifies A^0 to 1.
For the base R, the exponent sum requires expanding the brackets: (p - 1)(q - r) = pq - pr - q + r (q - 1)(r - p) = qr - qp - r + p (r - 1)(p - q) = rp - rq - p + q
Add these three lines together: pq - pr - q + r + qr - qp - r + p + rp - rq - p + q
Notice that every term has a matching opposite (for example, pq cancels with -qp, and -pr cancels with rp). The entire sum simplifies to 0. This makes the second part of our expression R^0, which is also 1.
Multiplying them together gives 1 × 1 = 1, which completes the proof.
How do you prove the AP version of the pth, qth and rth term identity?
There is a parallel problem for arithmetic progressions (AP). If the p-th, q-th and r-th terms of an AP are a, b and c, the identity to prove is:
a(q - r) + b(r - p) + c(p - q) = 0
The n-th term of an arithmetic progression is a_n = a_1 + (n-1)d 2. Let the first term be A and the common difference be D. The three terms are:
a = A + (p - 1)D
b = A + (q - 1)D
c = A + (r - 1)D
Substitute these into the left-hand side of the identity: [A + (p - 1)D](q - r) + [A + (q - 1)D](r - p) + [A + (r - 1)D](p - q)
Expand the expression and group the A terms and D terms: A(q - r + r - p + p - q) + D[(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)]
Just as in the geometric progression proof, the coefficient of A sums to 0. The coefficient of D expands identically to the exponent of R in the GP proof: (pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q) = 0
Both parts reduce to zero, leaving A(0) + D(0) = 0.
How do you overcome algebra blocks?
Proofs that rely on large algebraic expansions can look intimidating, but they all follow a predictable structure: state the definitions, substitute them in, and sort by variables. When tackling questions about inserting arithmetic and geometric means or selecting three numbers in AP or GP, relying on the fundamental definitions is always the safest starting point.
If you find yourself stuck not knowing where to begin with sequence proofs, the AI tutor in the Fermi Problem Library builds a plan for your session and explains the reasoning step by step rather than just printing an answer. Writing out the expansion is a mechanical process once you have set up the first substitution correctly.
