The problem involves an Arithmetic Progression (AP) and a Geometric Progression (GP) with specific conditions. We need to find the value of $a_{10} + g_5$.
AP and GP Definitions
- AP: $a_n = a_1 + (n-1)d$, where $d$ is the common difference.
- GP: $g_n = g_1 \cdot r^{n-1}$, where $r$ is the common ratio.
Given Conditions
- $a_1 = 1$.
- $g_1 = a_1$, so $g_1 = 1$.
- $a_2 + g_2 = 1$.
- $a_3 + g_3 = 4$.
- The GP ($g_n$) is increasing.
Deriving AP and GP Parameters
- From the given conditions, we can express terms in $a_1$ and $d$ for AP, and $g_1$ and $r$ for GP:
- $a_1 = 1$
- $a_2 = a_1 + d = 1 + d$
- $a_3 = a_1 + 2d = 1 + 2d$
- $g_1 = 1$
- $g_2 = g_1 \cdot r = 1 \cdot r = r$
- $g_3 = g_1 \cdot r^2 = 1 \cdot r^2 = r^2$
- Substitute these into the given equations:
- $a_2 + g_2 = 1 \implies (1 + d) + r = 1 \implies d + r = 0 \implies d = -r$.
- $a_3 + g_3 = 4 \implies (1 + 2d) + r^2 = 4$.
- Substitute $d = -r$ into the second equation: $(1 + 2(-r)) + r^2 = 4$ $1 - 2r + r^2 = 4$ $r^2 - 2r - 3 = 0$
- Factor the quadratic equation for $r$: $(r - 3)(r + 1) = 0$ This gives two possible values for $r$: $r = 3$ or $r = -1$.
- Consider the condition that the GP is increasing. Since $g_1 = 1$ (positive), the GP is increasing only if the common ratio $r > 1$. Therefore, we must choose $r = 3$.
- Calculate the common difference $d$: $d = -r = -3$
- Summary of parameters: $a_1 = 1, d = -3$ and $g_1 = 1, r = 3$.
Calculating the Required Terms
Now we calculate $a_{10}$ and $g_5$ using the derived parameters:
- $a_{10} = a_1 + (10-1)d = 1 + 9(-3) = 1 - 27 = -26$.
- $g_5 = g_1 \cdot r^{5-1} = 1 \cdot 3^4 = 1 \cdot 81 = 81$.
Final Calculation
Calculate the sum $a_{10} + g_5$:
$a_{10} + g_5 = -26 + 81 = 55$The value of $a_{10} + g_5$ is 55.
Similar Questions
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Important Questions from Algebra
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
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