G.P. of Increasing Positive Terms, a1a5=28 and a2+a4=29, Find a6

JEE Main 2025 — 22 January, Shift 1. Previous Year Question.

Problem

Let $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive terms. If $a_1a_5 = 28$ and $a_2 + a_4 = 29$, then $a_6$ is equal to:

Key insight. In any G.P., the product of two terms equidistant from a third is always the same: $a_1a_5 = a_2a_4$, because both equal $a^2r^4$. That turns the messy pair of conditions into a simple “sum and product” problem — the classic setup for a quadratic.

Watch this explained step by step →  ·  More on the Shiwam’s Classes channel

Approach

Writing $a_1 = a$ and common ratio $r$ turns $a_1a_5$ into $a^2r^4$ — but so does $a_2a_4 = (ar)(ar^3)$. Since both products equal $a^2r^4$, they must be equal to each other, and both equal 28. Combined with $a_2 + a_4 = 29$, this pins down $a_2$ and $a_4$ as the two roots of a quadratic whose sum is 29 and product is 28 — no need to solve for $a$ and $r$ individually at all.

Solution

Step 1 — Use the equidistant-terms property

In a G.P., $a_1a_5 = a_2a_4$, since

$$a_1a_5 = a \cdot ar^4 = a^2r^4, \qquad a_2a_4 = ar \cdot ar^3 = a^2r^4$$

So $a_2a_4 = 28$ as well, alongside $a_2 + a_4 = 29$.

Step 2 — Form and solve the quadratic

$a_2$ and $a_4$ are the two roots of $x^2 – (\text{sum})x + (\text{product}) = 0$:

$$x^2 – 29x + 28 = 0$$

Factorising:

$$x^2 – x – 28x + 28 = 0 \implies x(x-1) – 28(x-1) = 0 \implies (x-1)(x-28) = 0$$

So $x = 1$ or $x = 28$ — meaning $\{a_2, a_4\} = \{1, 28\}$ in some order.

Step 3 — Use “increasing” to fix the order

Since the terms are increasing, $a_2 < a_4$, so:

$$a_2 = 1, \qquad a_4 = 28$$

Step 4 — Find the common ratio, then $a_6$

Since $a_4 = a_2 \cdot r^2$:

$$r^2 = \frac{a_4}{a_2} = \frac{28}{1} = 28$$

And since $a_6 = a_4 \cdot r^2$:

$$a_6 = 28 \times 28 = 784$$

Answer

Common mistakes

• Trying to solve for $a$ and $r$ separately from the start. It’s doable, but turns a two-line quadratic into a much longer slog. Spotting $a_1a_5 = a_2a_4$ first collapses the problem immediately.
• Not using “increasing” to break the symmetry. Both $(a_2,a_4)=(1,28)$ and $(28,1)$ satisfy the sum-and-product conditions; only checking which one keeps the sequence increasing (rather than decreasing partway through) picks the valid one.

Practise next

• A G.P. of increasing positive terms has $a_1a_4 = 32$ and $a_2 + a_3 = 12$; find $a_5$, using the same equidistant-terms trick.