Let a_{1} , a_{2} , a_{3} ..... be a G.P. of increasing positive terms. If a_{1} a_{5} = 28 and a_{2} + a_{4} = 29, then a_{6} is equal to (1) 628 (2) 526 (3) 784 (4) 812

1️⃣ Key ideas you must know

1. Definition of a GP
   If $a_1$ is the first term and $r$ is the common ratio, then
   $a_n = a_1 r^{\,n-1}$
2. Handling given conditions
   You convert the verbal statements into algebraic equations using the above formula.
3. Quadratic trick
   Often, products like $a_1 a_5$ and sums like $a_2 + a_4$ reduce to a neat quadratic in one variable.

2️⃣ Walk-through of the logical steps

1. Write each required term in GP form
   • $a_1$ stays $a_1$
   • $a_2 = a_1 r$
   • $a_4 = a_1 r^3$
   • $a_5 = a_1 r^4$
2. Translate the clues
   • Product clue:
   $a_1 \cdot a_5 = a_1^2 r^4 = 28$
   • Sum clue:
   $a_2 + a_4 = a_1 r + a_1 r^3 = a_1 r (1 + r^2) = 29$
3. Introduce a simpler variable
   Let $x = a_1 r$ (this is $a_2$). Immediately the sum clue turns into
   $x \bigl(1 + r^2\bigr) = 29$
4. Eliminate $a_1$
   The product clue becomes, after substituting $x = a_1 r$,
   $x^2 r^2 = 28$
5. Solve the resulting system
   You now have two equations with two unknowns ($x$ and $r$). Manipulate them (usually by expressing $r^2$ from one equation and inserting in the other) to get a quadratic in $x$.
6. Choose the physically meaningful root
   Because the GP is increasing, we want $r > 1$. One root will violate this and must be discarded.
7. Compute the desired term
   Finally, use $a_6 = a_1 r^5$. Notice that $a_1 r = x$ is already known, so $a_6 = x r^4$—quick calculation!

🎲 What is the question?

We have a number pattern (called a Geometric Progression, or GP) where every term is found by multiplying the previous term by the same number (the “common ratio”). The pattern is getting bigger and bigger and all terms are positive.

We are told two clues:

1. If you multiply the 1st number and the 5th number, you get 28.
2. If you add the 2nd number and the 4th number, you get 29.

Using only those clues, we have to find the 6th number in the pattern.

🛤️ How can we think about it like a 10-year-old?

1. Call the first number “start” and the jump-factor “step”.
2. The list looks like: start, start × step, start × step × step, and so on.
3. Build simple equations from the clues.
4. Solve the little puzzle to discover step, then plug it back to get the 6th number.

That’s all there is—just careful substitution and solving a small quadratic!

Step-by-step solution

Let the first term be $a_1$ and the common ratio be $r\;(>1)$.
So $a_n = a_1 r^{\,n-1}$

1. Translate the two conditions $a_1 a_5 = a_1 \bigl(a_1 r^4\bigr) = a_1^{\,2} r^{4} = 28 \quad\text{(i)}$ $a_2 + a_4 = a_1 r + a_1 r^{3} = a_1 r\bigl(1 + r^{2}\bigr) = 29 \quad\text{(ii)}$

2. Substitute $x = a_1 r$ (i.e. $x$ equals the 2nd term):

   From (ii): $x\bigl(1 + r^{2}\bigr) = 29 \quad\text{(iii)}$

   From (i): $a_1^{\,2} r^{4} = \bigl(a_1 r\bigr)^2 r^{2} = x^{2} r^{2} = 28 \quad\text{(iv)}$

3. Express $r^{2}$ from (iv) $r^{2} = \frac{28}{x^{2}}$

4. Insert $r^{2}$ into (iii) $x \Bigl(1 + \frac{28}{x^{2}}\Bigr) = 29$ $x + \frac{28}{x} = 29$ $x^{2} - 29x + 28 = 0$

5. Solve the quadratic Discriminant: $D = 29^{2} - 4 \times 28 = 841 - 112 = 729 = 27^{2}$

$x = \frac{29 \pm 27}{2}$ So $x_1 = 28$ or $x_2 = 1$.

6. Choose the correct root If $x = 28$: $r^{2} = \frac{28}{28^{2}} = \frac{1}{28} \;\Rightarrow\; r < 1$ Rejected because the GP is increasing.

Hence $x = 1$ is valid, giving $r^{2} = 28 \;\Rightarrow\; r = \sqrt{28} = 2\sqrt{7}$

7. Find $a_6$ $a_6 = a_1 r^{5} = \bigl(a_1 r\bigr) r^{4} = x r^{4}$ But $x = 1$ and $r^{4} = \bigl(r^{2}\bigr)^{2} = 28^{2} = 784$.

$a_6 = 1 \times 784 = 784$

[\boxed{784}]

Therefore, the correct option is (3) 784.