Sequences and Series
If $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$ and $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=\mathrm{n}$, then the point $(\mathrm{m}, \mathrm{n})$ lies on the line
The value of $\frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots .+100^2 \times 101}$ is
Let three real numbers $a, b, c$ be in arithmetic progression and $a+1, b, c+3$ be in geometric progression. If $a>10$ and the arithmetic mean of $a, b$ and $c$ is 8, then the cube of the geometric mean of $a, b$ and $c$ is
Let the first three terms 2, p and q, with $q \neq 2$, of a G.P. be respectively the $7^{\text {th }}, 8^{\text {th }}$ and $13^{\text {th }}$ terms of an A.P. If the $5^{\text {th }}$ term of the G.P. is the $n^{\text {th }}$ term of the A.P., then $n$ is equal to:
Let $2^{\text {nd }}, 8^{\text {th }}$ and $44^{\text {th }}$ terms of a non-constant A. P. be respectively the $1^{\text {st }}, 2^{\text {nd }}$ and $3^{\text {rd }}$ terms of a G. P. If the first term of the A. P. is 1, then the sum of its first 20 terms is equal to -
For $0 < c < b < a$, let $(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b)=0$ and $\alpha \neq 1$ be one of its root. Then, among the two statements
(I) If $\alpha \in(-1,0)$, then $b$ cannot be the geometric mean of $a$ and $c$
(II) If $\alpha \in(0,1)$, then $b$ may be the geometric mean of $a$ and $c$
The sum of the series $\frac{1}{1-3 \cdot 1^2+1^4}+\frac{2}{1-3 \cdot 2^2+2^4}+\frac{3}{1-3 \cdot 3^2+3^4}+\ldots$ up to 10 -terms is
Let $a$ and $b$ be be two distinct positive real numbers. Let $11^{\text {th }}$ term of a GP, whose first term is $a$ and third term is $b$, is equal to $p^{\text {th }}$ term of another GP, whose first term is $a$ and fifth term is $b$. Then $p$ is equal to
Let $S_n$ denote the sum of first $n$ terms of an arithmetic progression. If $S_{20}=790$ and $S_{10}=145$, then $\mathrm{S}_{15}-\mathrm{S}_5$ is :
If $\log _e \mathrm{a}, \log _e \mathrm{~b}, \log _e \mathrm{c}$ are in an A.P. and $\log _e \mathrm{a}-\log _e 2 \mathrm{~b}, \log _e 2 \mathrm{~b}-\log _e 3 \mathrm{c}, \log _e 3 \mathrm{c} -\log _e$ a are also in an A.P, then $a: b: c$ is equal to
If each term of a geometric progression $a_1, a_2, a_3, \ldots$ with $a_1=\frac{1}{8}$ and $a_2 \neq a_1$, is the arithmetic mean of the next two terms and $S_n=a_1+a_2+\ldots . .+a_n$, then $S_{20}-S_{18}$ is equal to
If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to
In an A.P., the sixth term $a_6=2$. If the product $a_1 a_4 a_5$ is the greatest, then the common difference of the A.P. is equal to
$\text { The } 20^{\text {th }} \text { term from the end of the progression } 20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots,-129 \frac{1}{4} \text { is : }$
$4,9,14,19, \ldots \ldots$, up to $25^{\text {th }}$ term and
$3,6,9,12, \ldots \ldots$, up to $37^{\text {th }}$ term is :
If $\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots . .+\frac{1}{\alpha+1012}\right)-\left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots \ldots+\frac{1}{2024 \cdot 2023}\right)=\frac{1}{2024}$, then $\alpha$ is equal to ___________.
Explanation:
$\begin{aligned} & \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}- \\ & \left(\frac{1}{2 \times 21}+\frac{1}{4 \times 3}+\ldots+\frac{1}{2024} \cdot \frac{1}{2023}\right)=\frac{1}{2024} \\ & \quad \sum_{r=1}^{1012} \frac{1}{2 r(2 r-1)}=\sum_{r=1}^{1012}\left(\frac{1}{2 r-1}-\frac{1}{2 r}\right) \\ & \quad=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}\right) \\ & \quad=\left(1+\frac{1}{3}+\frac{1}{5}+\ldots .+\frac{1}{2023}\right) \\ & \quad-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{2024}\right) \\ & \quad=\left(1+\frac{1}{3}+\ldots+\frac{1}{2023}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{1012}\right) \end{aligned}$
$\begin{aligned} & =\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2023}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\ldots+\frac{1}{1011}\right) \\ & \frac{-1}{2}\left(1+\frac{1}{2}+\ldots+\frac{1}{1012}\right) \\ & =\frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023}-\frac{1}{2024} \\ & \Rightarrow \alpha+1012=2023 \\ & \Rightarrow \alpha=1011 \end{aligned}$
An arithmetic progression is written in the following way
The sum of all the terms of the 10th row is _________.
Explanation:
First term is each row form pattern
$\begin{aligned} & \Rightarrow T_n=a n^2+b n+c \\ & \Rightarrow T_1=a+b+c=2 \\ & \Rightarrow T_2=4 a+2 b+c=5 \\ & \Rightarrow T_3=9 a+3 b+c=11 \\ & \Rightarrow 3 a+b=3 \\ & 5 a+b=6 \\ & \Rightarrow 2 a=3 \Rightarrow a=\frac{3}{2}, \quad b=\frac{-3}{2} \Rightarrow c=2 \\ & \Rightarrow T_n=\frac{3}{2} n^2-\frac{3(n)}{2}+2 \Rightarrow \frac{3 n^2-3 n+4}{2} \\ & T_{10}=\frac{3 \times 100-3 \times 10+4}{2}=\frac{274}{2}=137 \end{aligned}$
Terms in $10^{\text {th }}$ row is 10 with 3 differences
$\begin{aligned} & \Rightarrow 137,140,143 \ldots \\ & \Rightarrow \quad S_{10}=\frac{10}{2}(2 \times 137+(10-1) \times 3) \\ & =5(274+27)=5 \times 301=1505 \end{aligned}$
Let the positive integers be written in the form :
If the $k^{\text {th }}$ row contains exactly $k$ numbers for every natural number $k$, then the row in which the number 5310 will be, is __________.
Explanation:
To solve this problem, we need to determine in which row the number $5310$ appears when positive integers are arranged in rows such that the $k^\text{th}$ row contains exactly $k$ numbers.
Understanding the Pattern
First row ($k = 1$): Contains 1 number.
Second row ($k = 2$): Contains 2 numbers.
Third row ($k = 3$): Contains 3 numbers.
…
$k^\text{th}$ row: Contains $k$ numbers.
Total Numbers Up to the $k^\text{th}$ Row
The total number of integers up to the $k^\text{th}$ row is given by the sum of the first $k$ natural numbers:
$ S(k) = 1 + 2 + 3 + \dots + k = \frac{k(k+1)}{2} $
Finding the Row Containing 5310
We need to find the smallest integer $k$ such that $S(k-1) < 5310 \leq S(k)$. This means:
$ \frac{(k-1)k}{2} < 5310 \leq \frac{k(k+1)}{2} $
Step 1: Estimate $k$
Let's approximate $k$ by solving the inequality:
$ \frac{k(k+1)}{2} \geq 5310 $
Multiply both sides by 2:
$ k(k+1) \geq 10620 $
This is a quadratic inequality. We can approximate $k$ by taking the square root:
$ k \approx \sqrt{10620} \approx 103 $
Step 2: Calculate $S(k)$ for $k = 102$ and $k = 103$
For $k = 102$:
$ S(102) = \frac{102 \times 103}{2} = \frac{10506}{2} = 5253 $
For $k = 103$:
$ S(103) = \frac{103 \times 104}{2} = \frac{10712}{2} = 5356 $
Step 3: Determine the Correct Row
Since $S(102) = 5253 < 5310 \leq 5356 = S(103)$, the number $5310$ lies between $5254$ and $5356$, inclusive. Therefore, it is in the $103^\text{rd}$ row.
Conclusion
The number $5310$ will be in the $103^\text{rd}$ row.
Let $\alpha=\sum_\limits{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r$ and $\beta=\left(\sum_\limits{r=0}^n \frac{{ }^n C_r}{r+1}\right)+\frac{1}{n+1}$. If $140<\frac{2 \alpha}{\beta}<281$, then the value of $n$ is _________.
Explanation:
$\begin{aligned} \alpha= & \sum_{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r \\ & =4 \sum_{r=0}^n r^2{ }^n C_r+2 \sum_{r=0}^n r \cdot{ }^n C_r+\sum_{r=0}^n{ }^n C_r \\ & =4 n(n+1) 2^{n-2}+2 \cdot n \cdot 2^{n-1}+2^n \\ & =2^n(n(n+1)+n+1)=2^n(n+1)^2 \\ & \beta=\sum_{r=0}^n\left(\frac{{ }^n C_r}{r+1}\right)+\left(\frac{1}{n+1}\right) \\ & (1+x)^n=\sum_{r=0}^n{ }^n C_r x^r \end{aligned}$
$\begin{aligned} & \int_\limits0^1(1+x)^n d x=\left.\sum_{r=0}^n \frac{{ }^n C_r x^{r+1}}{r+1}\right|_0 ^1=\sum_\limits{r=0} \frac{{ }^n C}{r+1} \\ & \left.\frac{(1+x)^{+1}}{n+1}\right|_0 ^1=\frac{2^n-1}{n+1} \\ \Rightarrow & \beta=\frac{2^{n+1}-1+1}{(n+1)}=\frac{2}{n+1} \\ \Rightarrow & \frac{2 \alpha}{\beta}=\frac{2^{n+1}(n \quad 1)}{\left(\frac{2^{n+1}}{n+1}\right)}=(n+1)^3 \in(140,281) \\ \Rightarrow & (n+1)^3=216 \\ \Rightarrow & n+1=6 \Rightarrow n=5 \end{aligned}$
If $\mathrm{S}(x)=(1+x)+2(1+x)^2+3(1+x)^3+\cdots+60(1+x)^{60}, x \neq 0$, and $(60)^2 \mathrm{~S}(60)=\mathrm{a}(\mathrm{b})^{\mathrm{b}}+\mathrm{b}$, where $a, b \in N$, then $(a+b)$ equal to _________.
Explanation:
$\begin{aligned} & (-x)(S(x))=\frac{(1+x)\left[(1+x)^{60}-1\right]}{(1+x-1)}-60(1+x)^{61} \\ & (-x) S(x)=\frac{(1+x)\left[(1+x)^{60}-1\right]}{x}-60(1+x)^{61} \\ & x S(x)=60(1+x)^{61}-\frac{(1+x)\left[(1+x)^{60}-1\right]}{x} \end{aligned}$
Multiplying $x$ on both side,
$x^2 S(x)=60 x(1+x)^{61}-(1+x)\left[(1+x)^{60}-1\right]$
Putting $x=60$
$\begin{aligned} & (60)^2 S(60)=60 \times 60(61)^{61}-(61)\left[61^{60}-1\right] \\ & =60 \times 60(61)^{61}-(61) \cdot 61^{60}+61 \\ & =(61)^{61}[60 \times 60-1]+61 \\ & =(3600-1) \cdot 61^{61}+61 \\ & a=3600-1, \quad b=61 \Rightarrow a+b=3660 \end{aligned}$
Let the first term of a series be $T_1=6$ and its $r^{\text {th }}$ term $T_r=3 T_{r-1}+6^r, r=2,3$, ............ $n$. If the sum of the first $n$ terms of this series is $\frac{1}{5}\left(n^2-12 n+39\right)\left(4 \cdot 6^n-5 \cdot 3^n+1\right)$, then $n$ is equal to ___________.
Explanation:
$\begin{aligned} & T_r=3 T_{r-1}+6^r \\ & \Rightarrow \text { solving homogenous part } \\ & T_r=3 T_{r-1} \\ & \Rightarrow x=3 \text { is the root } \end{aligned}$
$\therefore T_r=a .3^r$
Solving for particular part
$\begin{aligned} & T_r=b .6^r \\ & b .6^r=3 b 6^{r-1}+6^r \\ & \Rightarrow 6 b=3 b+6 \\ & \Rightarrow 3 b=6 \\ & \Rightarrow b=2 \\ & T_r=a^n+a^p \\ & T_r=a 3^{b r}+2.6^r \quad \text{.... (i)} \\ & T_r=3 T_{r-1}+6^r \end{aligned}$
Putting $r=2$
$T_2=18+36=54 \quad \text{.... (ii)}$
Using equation (i) and (ii)
$\begin{aligned} & 54=9 a+72 \Rightarrow-18=9 a \Rightarrow a=-2 \\ & \therefore T_r=2 \cdot 6^r-2 \cdot 3^r=2\left(6^r-3^r\right) \\ & \sum_{r=1}^n T_r=2 \sum 6^r-2 \sum 3^r \\ & =2 \cdot 6 \frac{\left(6^n-1\right)}{5}-2 \cdot 3 \frac{\left(3^n-1\right)}{2} \\ & =\frac{3}{5}\left(4 \cdot 6^n-5 \cdot 3^n+1\right) \\ & \therefore n^2-12 n+39=3 \\ & n^2-12 n+36=0 \\ & (n-6)^2=0 \\ & \therefore n=6 \end{aligned}$
If $1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\frac{49-20 \sqrt{6}}{180}+\ldots$ upto $\infty=2+\left(\sqrt{\frac{b}{a}}+1\right) \log _e\left(\frac{a}{b}\right)$, where a and b are integers with $\operatorname{gcd}(a, b)=1$, then $\mathrm{11 a+18 b}$ is equal to __________.
Explanation:
$\begin{aligned} & S=1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\ldots \infty \\\\ & =1+\frac{(1-\sqrt{2} / \sqrt{3})}{2}+\frac{(1-\sqrt{2} / \sqrt{3})^2}{6}+\frac{(1-\sqrt{2} / \sqrt{3})^3}{12}+\ldots \infty \end{aligned}$
$\text { let } 1-\frac{\sqrt{2}}{\sqrt{3}}=a$
$\begin{aligned} & S=1+\frac{a}{2}+\frac{a^2}{6}+\frac{a^3}{12}+\ldots \\\\ & =1+\left(1-\frac{1}{2}\right) a+\left(\frac{1}{2}-\frac{1}{3}\right) a^2+\left(\frac{1}{3}-\frac{1}{4}\right) a^3+\ldots \\\\ & =1+\left(a+\frac{a^2}{2}+\frac{a^3}{3} \ldots \infty\right)+\frac{1}{a}\left(\frac{-a^2}{2}-\frac{a^3}{3}-\frac{a^4}{4} \ldots \infty\right) \\\\ & =-\ln (1-a)+\frac{1}{a}\left(-a-\frac{a^2}{2}-\frac{a^3}{3} \ldots \infty\right)+2 \\\\ & =-\ln (1-a)+\frac{1}{a} \ln (1-a)+2 \\\\ & =2+\left(\frac{1}{a}-1\right) \ln (1-a) \\\\ & =2+\left(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}-1\right) \ln \left(1-1+\sqrt{\frac{2}{3}}\right) \\\\ & =2+\frac{\sqrt{2}}{\sqrt{3}}-\sqrt{2} \ln \sqrt{\frac{2}{3}} \\\\ & =2+\left(\frac{\sqrt{6}+2}{1} \cdot \frac{1}{2} \ln \frac{2}{3}\right) \\\\ & \therefore 2+\left(\sqrt{\frac{3}{2}}+1\right) \ln \frac{2}{3} \\\\ & \therefore 11 a+18 b=76 \end{aligned}$
Let $a_1, a_2, a_3, \ldots$ be in an arithmetic progression of positive terms.
Let $A_k=a_1^2-a_2^2+a_3^2-a_4^2+\ldots+a_{2 k-1}^2-a_{2 k}^2$.
If $\mathrm{A}_3=-153, \mathrm{~A}_5=-435$ and $\mathrm{a}_1^2+\mathrm{a}_2^2+\mathrm{a}_3^2=66$, then $\mathrm{a}_{17}-\mathrm{A}_7$ is equal to ________.
Explanation:
Let $a_n=a+(n-1) d \forall n \in N$
$\begin{aligned} A_k & =\left(a_1^2-a_2^2\right)+\left(a_3^2-a_4^2\right)+\ldots a_{2 k-1}^2-a_{2 k}^2 \\ & =(-d)\left(a_1+a_2+\ldots+a_{2 k}\right) \\ & A_k=(-d k)(2 a+(2 k-1) d) \\ \Rightarrow & A_3=(-3 d)(2 a+5 d)=-153 \\ \Rightarrow & d(2 a+5 d)=51 \quad \text{... (i)}\\ & A_5=(-5 d)(2 a+9 d)=-435 \end{aligned}$
$\begin{aligned} \Rightarrow & d(2 a+9 d)=87 \\ \Rightarrow & 4 d^2=36 \Rightarrow d= \pm 3(d=3 \text { positive terms }) \\ \Rightarrow & 3(2 a+27)=87 \\ \Rightarrow & 2 a=29-27 \\ \Rightarrow & a=1 \\ & a_{17}-A_7=(a+16 d)-(-7 d)(2+13 d) \\ & =49+7 \times 3(2+39) \\ & =49+21 \times 41=910 \end{aligned}$
Explanation:
To solve this problem, let's first denote the three successive terms of a geometric progression (G.P.) with common ratio $r$ as $a$, $ar$, and $ar^2$, where $a$ is the first term and $r > 1$. These three terms represent the lengths of the sides of a triangle.
According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Therefore, for the three terms to form a triangle, the following inequalities must hold:
$
1) \ a + ar > ar^2 \\\\
$
$2) \ a + ar^2 > ar $
$
3) \ ar + ar^2 > a
$
Given that $r > 1$, inequalities 2 and 3 will always hold because:
$ ar < ar^2 \ \text{and} \ a < ar, $
indicating that both $a + ar^2$ and $ar + ar^2$ will be greater than $ar$ and $a$ respectively. Therefore, we only need to check the first inequality to ensure that the three terms can form a triangle:
$ a + ar > ar^2 $
Simplifying this, we get:
$ a(1 + r) > a r^2 $
Since $a$ is positive (as it represents the length of a side of a triangle), we can divide both sides of the inequality by $a$ without changing the direction of the inequality:
$ 1 + r > r^2 $
We can then subtract $r$ from both sides:
$ 1 > r^2 - r $
Simplifying the right side by factoring $r$:
$ 1 > r(r - 1) $
Given that $r > 1$, the quantity $(r - 1)$ is positive; hence, $r(r - 1)$ is also positive. This means the actual value for $r$ to satisfy the inequality is within the interval $(1, \sqrt{2})$ because $r(r - 1)$ increases with increasing $r$, and it would be 1 when $r = \sqrt{2}$. It should be greater than 1, and less than $\sqrt{2}$ such that $r^2 - r$ stays below 1.
Now let’s consider the expressions $[r]$ and $[-r]$. The symbol $[x]$ denotes the greatest integer less than or equal to $x$ (also known as the floor function).
Since $1 < r < \sqrt{2}$, $[r] = 1$, because 1 is the greatest integer less than $r$ within that interval.
For $[-r]$, we need the greatest integer less than or equal to $-r$. Since $-r$ is negative and less than $-1$ (because $r > 1$), $[-r] = -2$, as this is the greatest integer that does not exceed the negative value of $r$ (which lies between $-\sqrt{2}$ and $-1$).
Now we can substitute these values into the expression:
$ 3[r] + [-r] = 3 \cdot 1 + (-2) = 3 - 2 = 1 $
Therefore $3[r] + [-r]$ is equal to 1.
Explanation:
To find the common terms in the two given arithmetic progressions (AP), we need to first identify the common difference for each sequence and then find the sequence that represents their overlap by employing the concept of least common multiple (LCM).
The first AP is:
$3, 7, 11, 15, \ldots, 403$
The common difference ($d_1$) for the first AP can be calculated by subtracting the first term from the second term:
$d_1 = 7 - 3 = 4$
The second AP is:
$2, 5, 8, 11, \ldots, 404$
The common difference ($d_2$) for the second AP is:
$d_2 = 5 - 2 = 3$
To find the terms common to both sequences, we need to find a term that appears in both sequences. Any common term must be of the form $3 + 4k$ and $2 + 3l$ for some integers $k$ and $l$. We want to find when these two forms will give us the same number, so we set them equal to each other:
$3 + 4k = 2 + 3l$
Rearranging the terms gives us:
$4k - 3l = 2 - 3$
This simplifies to:
$4k - 3l = -1 ....... (1)$
The solutions to equation $(1)$ will give us the common terms. Notice this is a Diophantine equation (A Diophantine equation is a polynomial equation, usually with two or more variables,) and has an infinite number of solutions. Let's find one such solution. We can see that:
$k = 1 \quad \text{yields} \quad 4(1) - 3l = -1 \implies 4 - 3l = -1 \implies 3l = 5 \implies l = 1\frac{2}{3}$
This is not an integer solution for $l$, so $k = 1$ does not work. Trying $k = 2$ gives:
$4(2) - 3l = -1 \implies 8 - 3l = -1 \implies 3l = 9 \implies l = 3$
Now we've found integers $k = 2$ and $l = 3$ that satisfy the equation. The corresponding term in both sequences would be:
$3 + 4(2) = 3 + 8 = 11 \quad \text{and} \quad 2 + 3(3) = 2 + 9 = 11$
Since $11$ is a common term, we can assert that every common term in both APs will be of the form $11 + m(4 \times 3)$, where $m$ is a non-negative integer, and $4 \times 3 = 12$ is the LCM of the common differences of the two APs. Thus, the general form for the common terms would be:
$11 + 12m$
Now we are to find all terms that are common up to $403$ in the first sequence and up to $404$ in the second sequence. Because the first sequence doesn't exceed $403$, we'll use this as our limit:
$11 + 12m \leq 403$
To find the largest possible integer value for $m$, we solve the inequality:
$12m \leq 403 - 11$
$12m \leq 392$
$m \leq 32\frac{2}{3}$
Since $m$ has to be an integer, the largest possible value for $m$ is $32$. Therefore, the common terms are generated by $m = 0, 1, 2, \ldots, 32$. There are $32 + 1 = 33$ terms in total.
We will now sum these up. The sum of an AP is given by the formula:
$S = \frac{n}{2}(a_1 + a_n)$
Where $S$ is the sum, $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the last term. Using the formula:
$S = \frac{33}{2}(11 + (11 + 12 \times 32))$
$S = \frac{33}{2}(11 + 11 + 384)$
$S = \frac{33}{2}(11 + 11 + 384)$
$S = \frac{33}{2}(406)$
$S = 33 \times 203$
$S = 6699$
Therefore, the sum of the common terms in the two arithmetic progressions is 6699.
Let $S_n$ be the sum to $n$-terms of an arithmetic progression $3,7,11$, If $40<\left(\frac{6}{n(n+1)} \sum_\limits{k=1}^n S_k\right)<42$, then $n$ equals ________.
Explanation:
$\begin{aligned} & \mathrm{S}_{\mathrm{n}}= 3+7+11+\ldots \ldots \mathrm{n} \text { terms } \\ &=\frac{\mathrm{n}}{2}(6+(\mathrm{n}-1) 4)=3 \mathrm{n}+2 \mathrm{n}^2-2 \mathrm{n} \\ &=2 \mathrm{n}^2+\mathrm{n} \\ & \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}=2 \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{K}^2+\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{K} \\ &=2 \cdot \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2} \\ &=\mathrm{n}(\mathrm{n}+1)\left[\frac{2 \mathrm{n}+1}{3}+\frac{1}{2}\right] \\ &=\frac{\mathrm{n}(\mathrm{n}+1)(4 \mathrm{n}+5)}{6} \\ & \Rightarrow 40<\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{k=1}^n S_{\mathrm{k}}<42 \\ & 40<4 \mathrm{n}+5<42 \\ & 35<4 n<37 \\ & \mathrm{n}=9 \end{aligned}$
Let $\alpha=1^2+4^2+8^2+13^2+19^2+26^2+\ldots$ upto 10 terms and $\beta=\sum_\limits{n=1}^{10} n^4$. If $4 \alpha-\beta=55 k+40$, then $\mathrm{k}$ is equal to __________.
Explanation:
$\begin{gathered} \alpha=1^2+4^2+8^2 \ldots . \\ t_n=a^2+b n+c \end{gathered}$
$\begin{aligned} & 1=a+b+c \\ & 4=4 a+2 b+c \\ & 8=9 a+3 b+c \end{aligned}$
On solving we get, $\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{3}{2}, \mathrm{c}=-1$
$\begin{aligned} & \alpha=\sum_{n=1}^{10}\left(\frac{n^2}{2}+\frac{3 n}{2}-1\right)^2 \\ & 4 \alpha=\sum_{n=1}^{10}\left(n^2+3 n-2\right)^2, \beta=\sum_{n=1}^{10} n^4 \\ & 4 \alpha-\beta=\sum_{n=1}^{10}\left(6 n^3+5 n^2-12 n+4\right)=55(353)+40 \end{aligned}$
Explanation:
$8=\frac{3}{1-\frac{1}{4}}+\frac{p \cdot \frac{1}{4}}{\left(1-\frac{1}{4}\right)^2}$
$\text { (sum of infinite terms of A.G.P }=\frac{a}{1-r}+\frac{d r}{(1-r)^2} \text { ) }$
$\Rightarrow \frac{4 p}{9}=4 \Rightarrow p=9$
means of two distinct positive numbers. Then $G_{1}^{4}+G_{2}^{4}+G_{3}^{4}+G_{1}^{2} G_{3}^{2}$ is equal to :
Let a$_1$, a$_2$, a$_3$, .... be a G.P. of increasing positive numbers. Let the sum of its 6th and 8th terms be 2 and the product of its 3rd and 5th terms be $\frac{1}{9}$. Then $6(a_2+a_4)(a_4+a_6)$ is equal to
Let $s_{1}, s_{2}, s_{3}, \ldots, s_{10}$ respectively be the sum to 12 terms of 10 A.P. s whose first terms are $1,2,3, \ldots .10$ and the common differences are $1,3,5, \ldots \ldots, 19$ respectively. Then $\sum_\limits{i=1}^{10} s_{i}$ is equal to :
Let $< a_{\mathrm{n}} > $ be a sequence such that $a_{1}+a_{2}+\ldots+a_{n}=\frac{n^{2}+3 n}{(n+1)(n+2)}$. If $28 \sum_\limits{k=1}^{10} \frac{1}{a_{k}}=p_{1} p_{2} p_{3} \ldots p_{m}$, where $\mathrm{p}_{1}, \mathrm{p}_{2}, \ldots ., \mathrm{p}_{\mathrm{m}}$ are the first $\mathrm{m}$ prime numbers, then $\mathrm{m}$ is equal to
Let $a, b, c$ and $d$ be positive real numbers such that $a+b+c+d=11$. If the maximum value of $a^{5} b^{3} c^{2} d$ is $3750 \beta$, then the value of $\beta$ is
Let $x_{1}, x_{2}, \ldots, x_{100}$ be in an arithmetic progression, with $x_{1}=2$ and their mean equal to 200 . If $y_{i}=i\left(x_{i}-i\right), 1 \leq i \leq 100$, then the mean of $y_{1}, y_{2}, \ldots, y_{100}$ is :
If $\mathrm{S}_{n}=4+11+21+34+50+\ldots$ to $n$ terms, then $\frac{1}{60}\left(\mathrm{~S}_{29}-\mathrm{S}_{9}\right)$ is equal to :
Let the first term $\alpha$ and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to
Let $\mathrm{a}_{\mathrm{n}}$ be the $\mathrm{n}^{\text {th }}$ term of the series $5+8+14+23+35+50+\ldots$ and $\mathrm{S}_{\mathrm{n}}=\sum_\limits{k=1}^{n} a_{k}$. Then $\mathrm{S}_{30}-a_{40}$ is equal to :
Let $S_{K}=\frac{1+2+\ldots+K}{K}$ and $\sum_\limits{j=1}^{n} S_{j}^{2}=\frac{n}{A}\left(B n^{2}+C n+D\right)$, where $A, B, C, D \in \mathbb{N}$ and $A$ has least value. Then
If $\operatorname{gcd}~(\mathrm{m}, \mathrm{n})=1$ and $1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+(2021)^{2}-(2022)^{2}+(2023)^{2}=1012 ~m^{2} n$ then $m^{2}-n^{2}$ is equal to :
The sum of the first $20$ terms of the series $5+11+19+29+41+\ldots$ is :
The sum $\sum\limits_{n = 1}^\infty {{{2{n^2} + 3n + 4} \over {(2n)!}}} $ is equal to :
The sum of 10 terms of the series
${1 \over {1 + {1^2} + {1^4}}} + {2 \over {1 + {2^2} + {2^4}}} + {3 \over {1 + {3^2} + {3^4}}}\, + \,....$ is
If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , respectively, then the sum of common ratios of all such GPs is
If ${a_n} = {{ - 2} \over {4{n^2} - 16n + 15}}$, then ${a_1} + {a_2}\, + \,....\, + \,{a_{25}}$ is equal to :
For three positive integers p, q, r, ${x^{p{q^2}}} = {y^{qr}} = {z^{{p^2}r}}$ and r = pq + 1 such that 3, 3 log$_yx$, 3 log$_zy$, 7 log$_xz$ are in A.P. with common difference $\frac{1}{2}$. Then r-p-q is equal to
$\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^{2}}-\frac{1}{2 \cdot 3}+\frac{1}{3^{2}}\right)+\left(\frac{1}{2^{3}}-\frac{1}{2^{2} \cdot 3}+\frac{1}{2 \cdot 3^{2}}-\frac{1}{3^{3}}\right)+$
$\left(\frac{1}{2^{4}}-\frac{1}{2^{3} \cdot 3}+\frac{1}{2^{2} \cdot 3^{2}}-\frac{1}{2 \cdot 3^{3}}+\frac{1}{3^{4}}\right)+\ldots$
is $\frac{\alpha}{\beta}$, where $\alpha$ and $\beta$ are co-prime, then $\alpha+3 \beta$ is equal to __________.
Explanation:
$S = \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{4}-\frac{1}{6}+\frac{1}{9}\right) + \left(\frac{1}{8}-\frac{1}{12}+\frac{1}{18}-\frac{1}{27}\right) + \left(\frac{1}{16}-\frac{1}{24}+\frac{1}{36}-\frac{1}{54}+\frac{1}{81}\right) + \ldots$
The first few terms of the series are :
$S = \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots\right) - \left(\frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \ldots\right) + \left(\frac{1}{9} + \frac{1}{18} + \frac{1}{36} + \ldots\right) - \left(\frac{1}{27} + \frac{1}{54} + \ldots\right) + \ldots$
We can now see that each group of terms forms a geometric series with a common ratio of $\frac{1}{2}$:
$S = \frac{\frac{1}{2}}{1-\frac{1}{2}} - \frac{\frac{1}{3}}{1-\frac{1}{2}} + \frac{\frac{1}{9}}{1-\frac{1}{2}} - \frac{\frac{1}{27}}{1-\frac{1}{2}} + \ldots$
The series can be rewritten as :
$S = 2 \left(\frac{1}{{2}} - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots\right)$
Now, we can simplify and rewrite the series inside the parentheses as:
$S = 2 \left[\frac{1}{{2}} + (- \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots\right)]$
The series inside the parentheses is an infinite geometric series with the first term $a = - \frac{1}{3}$ and the common ratio $r = -\frac{1}{3}$:
$S = 2 \left(\frac{1}{{2}} +\frac{a}{1-r}\right) = 2 \left(\frac{1}{{2}} + \frac{- \frac{1}{3}}{1+\frac{1}{3}}\right) = 2 \left(\frac{1}{{2}} + \frac{- \frac{1}{3}}{\frac{4}{3}}\right) $
= $2\left( {{1 \over 2} - {1 \over 3} \times {3 \over 4}} \right)$
$ = 2\left( {{1 \over 2} - {1 \over 4}} \right) = 2\left( {{1 \over 4}} \right) = {1 \over 2}$
Thus, the sum of the series is $\frac{1}{2}$, and $\alpha = 1$ and $\beta = 2$ are co-prime.
Therefore, $\alpha + 3\beta = 1 + 6 = 7$