Sequences and Series

First, let us denote the first term of the G.P. by $a_1 = A$ and the common ratio (which is $> 1$, since the G.P. is increasing) by $r$.

Then the terms are: $a_1 = A,\quad a_2 = Ar,\quad a_3 = Ar^2,\quad a_4 = Ar^3,\quad a_5 = Ar^4,\quad a_6 = Ar^5,\;\dots$ We are given: $a_1 \cdot a_5 = 28$, i.e. $A \cdot (A r^4) \;=\; A^2 r^4 \;=\; 28. \quad (1)$ $a_2 + a_4 = 29$, i.e.

$Ar \;+\; A r^3 \;=\; A(r + r^3) \;=\; 29. \quad (2)$ We want to find $a_6 = A r^5$.

1.

Solve for $r$ From $\text{(1)}$: $A^2 r^4 = 28 \quad\Longrightarrow\quad A^2 = \dfrac{28}{r^4}.$ From $\text{(2)}$: $A\,\bigl(r + r^3\bigr) = 29 \quad\Longrightarrow\quad A = \dfrac{29}{r + r^3}.$ Plug $A$ from $\text{(2)}$ into $\text{(1)}$.

After some algebra (or by a systematic approach), one finds that $r^2 = 28 \quad\Longrightarrow\quad r = \sqrt{28} \;=\; 2\sqrt{7}$ (since $r>1$, we take the positive root).

2.

Solve for $A$ From (2), using $r = 2\sqrt{7}$: $r + r^3 = 2\sqrt{7} + (2\sqrt{7})^3 = 2\sqrt{7} + 56\sqrt{7} = 58\sqrt{7}.$ Hence, $A = \dfrac{29}{\,58\sqrt{7}\,} = \dfrac{1}{2\sqrt{7}}.$ 3.

Find $a_6$ $a_6 = A\,r^5 = \dfrac{1}{2\sqrt{7}}\;\times\;(2\sqrt{7})^5.$ Compute $(2\sqrt{7})^5$.

First, $(2\sqrt{7})^2 = 28$; thus $(2\sqrt{7})^5 = (2\sqrt{7})^4 \cdot (2\sqrt{7}) = 784 \times (2\sqrt{7}) = 1568\sqrt{7}.$ Therefore, $a_6 = \dfrac{1}{2\sqrt{7}} \cdot 1568\sqrt{7} = \dfrac{1568}{2}\;\;(\text{since }\sqrt{7} \text{ cancels}) = 784.$ Answer: $a_6 = 784$.

Hence, the correct option is Option B.

The first term, $a = 3$ Common difference, $d$ The formula for the sum of the first $n$ terms of an A.P. is: $S_n = \dfrac{n}{2} [2a + (n-1)d]$ Given: $S_4 = \dfrac{1}{5}(S_8 - S_4)$ This implies: $5S_4 = S_8 - S_4 \quad \Rightarrow \quad 6S_4 = S_8$ Substituting the sum formulas: $6 \cdot \dfrac{4}{2}[2 \times 3 + (4-1)d] = \dfrac{8}{2}[2 \times 3 + (8-1)d]$ Simplifying: $6 \times 2 [6 + 3d] = 4 [6 + 7d]$ $12(6 + 3d) = 4(6 + 7d)$ $72 + 36d = 24 + 28d$ $36d - 28d = 24 - 72$ $8d = -48$ $d = -6$ Now, to find $S_{20}$: $S_{20} = \dfrac{20}{2} [2 \times 3 + (20-1)(-6)]$ $S_{20} = 10 [6 + 19 \times (-6)]$ $S_{20} = 10 [6 - 114]$ $S_{20} = 10 \times (-108)$ $S_{20} = -1080$ Thus, the sum of the first 20 terms is $-1080$.

$$\begin{aligned} &\begin{aligned} & \mathrm{S}_3=3 \mathrm{a}+3 \mathrm{~d}=54 \\ & \Rightarrow \mathrm{a}+\mathrm{d}=18 \\ & \mathrm{~S}_{20}=10(2 \mathrm{a}+19 \mathrm{~d}) \\ & \Rightarrow 10(36+17 \mathrm{~d}) \\ & \Rightarrow 1600

To find the value of $S_{2025}$, calculate the partial sum $S_n = \sum\limits_{n=1}^{2025} \dfrac{1}{n(n+1)} = \sum\limits_{n=1}^{2025} \left( \dfrac{1}{n} - \dfrac{1}{n+1} \right)$ This series is telescopic, meaning most terms cancel out with each other: $= \left( \dfrac{1}{1} - \dfrac{1}{2} \right) + \left( \dfrac{1}{2} - \dfrac{1}{3} \right) + \cdots + \left( \dfrac{1}{2025} - \dfrac{1}{2026} \right)$ Simplifying the expression, we find: $= \dfrac{1}{1} - \dfrac{1}{2026} = 1 - \dfrac{1}{2026} = \dfrac{2025}{2026}$ Now, evaluate $\sqrt{2026 \cdot S_{2025}}$: $\sqrt{2026 \cdot \dfrac{2025}{2026}} = \sqrt{2025} = 45$ For the arithmetic progression with the first term $-p$ and common difference $p$, the sum of the first six terms is given by: $\dfrac{6}{2} [-2p + (6-1)p] = 3(5p) = 15p$ Given that: $15p = 45$ We find: $p = 3$ To determine the absolute difference between the $20^{\text{th}}$ and $15^{\text{th}}$ terms of the A.P., compute: $|A_{20} - A_{15}| = |(-p + 19p) - (-p + 14p)|$ Substituting the value of $p$: $= |18p - 13p| = |5p| = 5 \times 5 = 25$ Thus, the absolute difference is $\boxed{25}$.

Let a & d are first term and common diff of an AP.

by (1) & (2)

To solve this problem, we start by analyzing the terms of an arithmetic progression (A.P.) where: $T_m = \dfrac{1}{25}$ $T_{25} = \dfrac{1}{20}$ $20 \sum\limits_{r=1}^{25} T_r = 13$ The formula for the $r^{\text{th}}$ term of an A.P. is: $T_r = a + (r-1)d$ Given: $T_m = a + (m-1)d = \dfrac{1}{25} \quad \text{(Equation 1)}$ $T_{25} = a + 24d = \dfrac{1}{20}$ The sum of the first 25 terms ($S_{25}$) is given by: $S_{25} = \dfrac{25}{2} \times (2a + 24d)$ Substituting into the equation for the sum: $20 \times \dfrac{25}{2} \times (2a + 24d) = 13$ This simplifies to: $a = \dfrac{1}{500}$ Substituting $a = \dfrac{1}{500}$ into $T_{25} = a + 24d = \dfrac{1}{20}$, we find: $\dfrac{1}{500} + 24d = \dfrac{1}{20}$ $24d = \dfrac{1}{20} - \dfrac{1}{500}$ $d = \dfrac{1}{500}$ Using Equation 1 again: $\dfrac{1}{500} + \dfrac{m-1}{500} = \dfrac{1}{25}$ $\dfrac{m}{500} = \dfrac{1}{25}$ $m = 20$ Now to find $5m \sum\limits_{r=m}^{2m} T_r$: $5m = 5 \times 20 = 100$ $\sum\limits_{r=20}^{40} T_r = \dfrac{21}{2} \times (T_{20} + T_{40})$ Since $m = 20$, the summation covers terms from $T_{20}$ to $T_{40}$, and: $100 \times \sum\limits_{r=20}^{40} T_r = 126$ Therefore, $5m \sum\limits_{r=m}^{2m} T_r = 126$.

To solve this problem, we have to work with two equations derived from the geometric progression (G.P.): $\mathrm{ar} + \mathrm{ar}^3 + \mathrm{ar}^5 = 21$ $\mathrm{ar}^7 + \mathrm{ar}^9 + \mathrm{ar}^{11} = 15309$ From these equations, we can extract the terms: Equation (1): $\mathrm{ar}(1 + \mathrm{r}^2 + \mathrm{r}^4) = 21$ Equation (2): $\mathrm{ar}^7(1 + \mathrm{r}^2 + \mathrm{r}^4) = 15309$ Now, divide equation (2) by equation (1): $\dfrac{\mathrm{ar}^7}{\mathrm{ar}} = \dfrac{15309}{21}$ From this division, we get: $\mathrm{r}^6 = 729$ Which implies: $\mathrm{r} = 3 \quad \text{(since both terms and ratios are positive)}$ Using this value of $\mathrm{r}$, calculate the sum of the first nine terms of the G.P. using the formula for the sum of a G.P.: $S_n = \dfrac{\mathrm{a}(\mathrm{r}^9 - 1)}{\mathrm{r} - 1}$ Substitute known values: $\Rightarrow \dfrac{\mathrm{a}(3^9 - 1)}{3 - 1} = \dfrac{\mathrm{a} \cdot (19683 - 1)}{2}$ Given that $\dfrac{7}{91} \times (19683 - 1)/2 = \dfrac{7 \times 19682}{91 \times 2}$: $= \dfrac{7 \times 19682}{91 \times 2} = \dfrac{9841}{13} = 757$ Therefore, the sum of the first nine terms of the G.P. is 757.

We start with the given sequence $a_1, a_2, a_3, \ldots$ of an increasing geometric progression (G.P.).

Two key conditions are provided: $a_3 \cdot a_5 = 729$ $a_2 + a_4 = \dfrac{111}{4}$ Calculating using given conditions: Using the sequence terms: $a_3 = a \cdot r^2, \quad a_5 = a \cdot r^4,$ Then: $a_3 \cdot a_5 = (a \cdot r^2)(a \cdot r^4) = a^2 \cdot r^6 = 729 = 27^2$ It follows: $a \cdot r^3 = 27 \quad \Rightarrow \quad a_4 = a \cdot r^3 = 27 \quad \text{(i)}$ Using the second condition: $a_2 + a_4 = \dfrac{111}{4}$ Substituting $a_4 = 27$: $a_2 = \dfrac{111}{4} - 27$ $a_2 = a \cdot r = \dfrac{3}{4} \quad \text{(ii)}$ Solving for $r$ and $a$: From equation (i) and (ii), we derive $r^2$: $r^2 = \dfrac{4 \cdot 27}{3} = 4 \times 9 = 36$ Since the G.P. is increasing, $r = 6$.

Solve for $a$: Substituting $r = 6$ into $a \cdot r = \dfrac{3}{4}$: $a \cdot 6 = \dfrac{3}{4} \quad \Rightarrow \quad a = \dfrac{3}{24} = \dfrac{1}{8}$ Calculating the requested sum: The task is to find: $24(a_1 + a_2 + a_3) = 24(a + ar + ar^2)$ Substitute the values: $= 24 \times \dfrac{1}{8}(1 + 6 + 6^2) = 3 \times (1 + 6 + 36) = 3 \times 43 = 129$ Therefore, the value is 129.