Key Concepts and Formulas

1. Geometric Progression (G.P.): 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 ($r$). The terms are $a, ar, ar^2, \ldots$.
2. Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant (common difference, $d$). If $X, Y, Z$ are in A.P., then $2Y = X + Z$.
3. Quadratic Equation: A second-degree polynomial equation of the form $Ax^2 + Bx + C = 0$. Its roots can be found by factorization or the quadratic formula.

---

Step-by-Step Solution

1. Expressing G.P. terms in terms of 'a' and 'r' We are given that $a, b, c$ are in G.P. with common ratio $r$. This means:

• $b = ar$
• $c = ar^2$

2. Using the A.P. property to form an equation in 'r' We are given that $3a, 7b, 15c$ are the first three terms of an A.P. Using the property that for an A.P., the middle term is the average of the other two, we have: $2(7b) = 3a + 15c$ $14b = 3a + 15c$

Now, substitute the expressions for $b$ and $c$ from Step 1 into this equation: $14(ar) = 3a + 15(ar^2)$

Since $a \ne 0$, we can divide the entire equation by $a$: $14r = 3 + 15r^2$

Rearrange this into a quadratic equation: $15r^2 - 14r + 3 = 0$

3. Solving the quadratic equation for 'r' We can solve the quadratic equation $15r^2 - 14r + 3 = 0$ by factoring. We look for two numbers that multiply to $15 \times 3 = 45$ and add up to $-14$. These numbers are $-9$ and $-5$. $15r^2 - 9r - 5r + 3 = 0$ $3r(5r - 3) - 1(5r - 3) = 0$ $(3r - 1)(5r - 3) = 0$

This gives two possible values for $r$:

• $3r - 1 = 0 \implies r = \frac{1}{3}$
• $5r - 3 = 0 \implies r = \frac{3}{5}$

4. Applying the given condition on 'r' The problem states that $0 < r \le \frac{1}{2}$. We check our two possible values for $r$:

• If $r = \frac{1}{3}$: This satisfies $0 < \frac{1}{3} \le \frac{1}{2}$. So, $r = \frac{1}{3}$ is a valid common ratio.
• If $r = \frac{3}{5}$: This does not satisfy $r \le \frac{1}{2}$ (since $\frac{3}{5} = 0.6$ and $\frac{1}{2} = 0.5$). So, $r = \frac{3}{5}$ is rejected.

Thus, the common ratio of the G.P. is $r = \frac{1}{3}$.

5. Calculating the common difference of the A.P. Let the first three terms of the A.P. be $T_1 = 3a$, $T_2 = 7b$, and $T_3 = 15c$. The common difference ($d$) of the A.P. is $T_2 - T_1$. $d = 7b - 3a$ Substitute $b = ar$ and $r = \frac{1}{3}$: $d = 7a\left(\frac{1}{3}\right) - 3a$ $d = \frac{7a}{3} - 3a$ $d = \frac{7a - 9a}{3}$ $d = -\frac{2a}{3}$

6. Calculating the 4th term of the A.P. The 4th term of the A.P. ($T_4$) can be found by adding the common difference to the 3rd term ($T_3$). $T_4 = T_3 + d$ We know $T_3 = 15c$. Substitute $c = ar^2$ and $r = \frac{1}{3}$: $T_3 = 15a\left(\frac{1}{3}\right)^2 = 15a\left(\frac{1}{9}\right) = \frac{15a}{9} = \frac{5a}{3}$

Now, substitute $T_3$ and $d$ into the equation for $T_4$: $T_4 = \frac{5a}{3} + \left(-\frac{2a}{3}\right)$ $T_4 = \frac{5a - 2a}{3}$ $T_4 = \frac{3a}{3}$ $T_4 = a$

Alternatively, we can use the formula $T_4 = T_1 + 3d$: $T_4 = 3a + 3\left(-\frac{2a}{3}\right)$ $T_4 = 3a - 2a$ $T_4 = a$

Both methods confirm that the 4th term of the A.P. is $a$.

---

Common Mistakes & Tips

• Ignoring the constraint on 'r': The condition $0 < r \le \frac{1}{2}$ is crucial for selecting the correct value of $r$. Failing to apply it would lead to an ambiguous answer.
• Algebraic errors: Be meticulous with algebraic manipulations, especially when solving quadratic equations and working with fractions.
• Confusing G.P. and A.P. terms: Clearly distinguish between the terms of the G.P. ($a, b, c$) and the terms of the A.P. ($3a, 7b, 15c$).

---

Summary The problem requires us to use the definitions of Geometric Progression and Arithmetic Progression. By expressing the terms of the G.P. in terms of the first term and common ratio, and then applying the property of A.P. to the given terms, we formed a quadratic equation for the common ratio $r$. After solving for $r$ and using the given constraint to select the valid value, we calculated the common difference of the A.P. and subsequently found the 4th term.

The final answer is $\boxed{a}$.