Question
Let $a_1, a_2, a_3, \dots$ be an A.P. and $g_1 = a_1, g_2, g_3, \dots$ be an increasing G.P. If $a_1 = a_2 + g_2 = 1$ and $a_3 + g_3 = 4$, then $a_{10} + g_5$ is equal to :
The correct answer is
55
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.
Was this answer helpful?
Similar Questions
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
- The number of elements in the relation $R = \{(x, y) : 4x^2 + y^2 < 52, x, y \in \mathbb{Z}\}$ is
- Let $S = \{z \in \mathbb{C} : 4z^2 + \bar{z} = 0\}$. Then $\sum_{z \in S} |z|^2$ is equal to :
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
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 ________.
- Let $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ up to 13 terms. If $13S = \frac{2^k}{n!}, k \in \mathbb{N}$, then $n + k$ is equal to
Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
Need Expert Advice?
More Questions from JEE Main
