Sum of Specific Terms of Arithmetic and Geometric Progressions

To find the value of $a_{10} + g_5$, we analyze the given conditions step-by-step:

Let the common difference of the Arithmetic Progression (A.P.) $a_1, a_2, a_3, \dots$ be $d$, and let the common ratio of the Geometric Progression (G.P.) $g_1, g_2, g_3, \dots$ be $r$.

We are given that: $g_1 = a_1 = 1$

Since $g_1 = 1$ and the G.P. is increasing, the common ratio must satisfy $r > 1$.

Step 1: Express $a_2$, $g_2$, $a_3$, and $g_3$ in terms of $d$ and $r$ The terms of the A.P. are given by: $a_2 = a_1 + d = 1 + d$ $a_3 = a_1 + 2d = 1 + 2d$

The terms of the G.P. are given by: $g_2 = g_1 r = r$ $g_3 = g_1 r^2 = r^2$

Step 2: Use the condition $a_2 + g_2 = 1$ Given that $a_2 + g_2 = 1$, we substitute $a_2 = 1 + d$ and $g_2 = r$: $(1 + d) + r = 1$ $d + r = 0 \implies d = -r \quad \text{--- (Equation 1)}$

Step 3: Use the condition $a_3 + g_3 = 4$ Given that $a_3 + g_3 = 4$, we substitute $a_3 = 1 + 2d$ and $g_3 = r^2$: $(1 + 2d) + r^2 = 4$

Substituting $d = -r$ into this equation: $1 + 2(-r) + r^2 = 4$ $r^2 - 2r + 1 = 4$ $(r - 1)^2 = 4$

Taking the square root on both sides: $r - 1 = \pm 2$

Since the G.P. is an increasing sequence with $g_1 = 1 > 0$, we must have $r > 1$. Therefore: $r - 1 = 2 \implies r = 3$

Using Equation 1, the common difference $d$ is: $d = -r = -3$

Step 4: Calculate $a_{10}$ and $g_5$ Using the values $a_1 = 1$, $d = -3$, $g_1 = 1$, and $r = 3$:

$a_{10} = a_1 + 9d = 1 + 9(-3) = 1 - 27 = -26$ $g_5 = g_1 r^4 = 1 \cdot (3)^4 = 81$

Step 5: Compute $a_{10} + g_5$ $a_{10} + g_5 = -26 + 81 = 55$

Thus, the value of $a_{10} + g_5$ is $55$.