Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
129
Let the first term of the Geometric Progression (G.P.) be $a$ and the common ratio be $r$. Since the numbers are positive and increasing, we have $a > 0$ and $r > 1$. The terms of the G.P. are $a, ar, ar^2, ar^3, ar^4, \dots$
G.P. Conditions Analysis
- The term $a_n$ of a G.P. is given by $a_n = ar^{n-1}$.
- Given $a_3a_5 = 729$. Substituting the formula: $(ar^{3-1})(ar^{5-1}) = 729$ $(ar^2)(ar^4) = 729$ $a^2r^6 = 729$ $(ar^3)^2 = 729$ Taking the square root (since terms are positive): $ar^3 = \sqrt{729} = 27$ This means the 4th term, $a_4$, is 27.
- Given $a_2 + a_4 = \frac{111}{4}$. Substituting $a_4 = 27$: $a_2 + 27 = \frac{111}{4}$ $a_2 = \frac{111}{4} - 27$ $a_2 = \frac{111 - 108}{4} = \frac{3}{4}$ So, the 2nd term, $a_2$, is $\frac{3}{4}$.
Finding Common Ratio and First Term
- We have $a_2 = ar = \frac{3}{4}$ and $a_4 = ar^3 = 27$.
- Divide $a_4$ by $a_2$: $\frac{ar^3}{ar} = \frac{27}{3/4}$ $r^2 = 27 \times \frac{4}{3}$ $r^2 = 9 \times 4 = 36$ Taking the square root (since $r > 1$): $r = 6$
- Now, find the first term $a$ using $a_2 = ar = \frac{3}{4}$: $a(6) = \frac{3}{4}$ $a = \frac{3}{4 \times 6} = \frac{3}{24} = \frac{1}{8}$
- Check conditions: $a = \frac{1}{8}$ (positive) and $r = 6$ (greater than 1). This confirms the G.P. consists of increasing positive numbers.
Calculating the Required Value
- Calculate the first three terms ($a_1, a_2, a_3$): $a_1 = a = \frac{1}{8}$ $a_2 = ar = \frac{1}{8} \times 6 = \frac{6}{8} = \frac{3}{4}$ $a_3 = ar^2 = \frac{1}{8} \times 6^2 = \frac{36}{8} = \frac{9}{2}$
- Calculate the sum $a_1 + a_2 + a_3$: $a_1 + a_2 + a_3 = \frac{1}{8} + \frac{3}{4} + \frac{9}{2}$ Find a common denominator (8): $= \frac{1}{8} + \frac{3 \times 2}{4 \times 2} + \frac{9 \times 4}{2 \times 4}$ $= \frac{1}{8} + \frac{6}{8} + \frac{36}{8}$ $= \frac{1 + 6 + 36}{8} = \frac{43}{8}$
- Finally, calculate $24 (a_1 + a_2 + a_3)$: $24 \times \left(\frac{43}{8}\right)$ $= 3 \times 43$ $= 129$
Similar Questions
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability $P(W_n)$ of choosing the word $W_n$ satisfy $P(W_n) = 2P(W_{n-1})$, $n > 1$.
If $P(CDBEA) = \frac{2^\alpha}{2^\beta-1}$, $\alpha, \beta\in N$, then $\alpha + \beta$ is equal to :
- The largest $n \in N$ such that $3^n$ divides $50!$ is :
- The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to :
- Let A be the set of all functions $f: Z \to Z$ and R be a relation on A such that $R = \{(f, g) : f(0) = g(1) \text{ and } f(1) = g(0)\}$. Then R is :
Let $a \in \mathbf{R}$ and $A$ be a matrix of order $3 \times 3$ such that $\det (A) = -4$ and $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $I$ is the identity matrix of order $3 \times 3$. If $\det ((a+1)\text{adj}((a-1)A))$ is $2^m 3^n$, $m, n \in \{0, 1, 2, \dots, 20\}$, then $m+n$ is equal to :
The term independent of x in the expansion of $\left( \frac{(x+1)}{(x^{2/3}+1-x^{1/3})} - \frac{(x-1)}{(x-x^{1/2})} \right)^{10}$, $x > 1$, is:
Important Questions from Algebra
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability $P(W_n)$ of choosing the word $W_n$ satisfy $P(W_n) = 2P(W_{n-1})$, $n > 1$.
If $P(CDBEA) = \frac{2^\alpha}{2^\beta-1}$, $\alpha, \beta\in N$, then $\alpha + \beta$ is equal to :
