In an arithmetic progression, if ${\mathrm{S}}_{40}=1030$ and ${\mathrm{S}}_{12}=57$, then ${\mathrm{S}}_{30}-{\mathrm{S}}_{10}$ is equal to :

If $7=5+\frac{1}{7}(5+\alpha )+\frac{1}{7^2}(5+2\alpha )+\frac{1}{7^3}(5+3\alpha )+\dots \dots \dots \dots \infty$, then the value of $\alpha$ is :

Let $⟨a_{\mathrm{n}}⟩$ be a sequence such that $a_0=0,a_1=\frac{1}{2}$ and $2a_{\mathrm{n}+2}=5a_{\mathrm{n}+1}-3a_{\mathrm{n}},\mathrm{n}=0,1,2,3,\dots$. Then $\sum _{k=1}^{100}{a}_k$ is equal to

Let ${\mathrm{T}}_{\mathrm{r}}$ be the ${\mathrm{r}}^{\text{th }}$ term of an A.P. If for some $\mathrm{m},{\mathrm{T}}_{\mathrm{m}}=\frac{1}{25},{\mathrm{T}}_{25}=\frac{1}{20}$, and $20\sum _{\mathrm{r}=1}^{25}{\mathrm{T}}_{\mathrm{r}}=13$, then $5\mathrm{m}\sum _{\mathrm{r}=\mathrm{m}}^{2\mathrm{m}}{\mathrm{T}}_{\mathrm{r}}$ is equal to

If $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\dots \infty =\frac{{\pi }^4}{90}$, $\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\dots \infty =\alpha$, $\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\dots \infty =\beta$, then $\frac{\alpha }{\beta }$ is equal to :

Let $a_1,a_2,a_3,\dots$ be in an A.P. such that \sum_\limits{k=1}^{12} a_{2 k-1}=-\frac{72}{5} a_1, a_1 \neq 0. If \sum_\limits{k=1}^n a_k=0, then $n$ is :

Let $a_n$ be the $n^{th}$ term of an A.P. If $S_n=a_1+a_2+a_3+\dots +a_n=700$, $a_6=7$ and $S_7=7$, then $a_n$ is equal to :

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :

Let $x_1,x_2,x_3,x_4$ be in a geometric progression. If $2,7,9,5$ are subtracted respectively from $x_1,x_2,x_3,x_4$, then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24}\left(x_1{x}_2{x}_3{x}_4\right)$ is:

The number of terms of an A.P. is even; the sum of all the odd terms is 24 , the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$. Then the number of terms which are integers in the A.P. is :

Let $a_1,a_2,a_3,\dots$. be a G.P. of increasing positive numbers. If $a_3{a}_5=729$ and $a_2+a_4=\frac{111}{4}$, then $24\left(a_1+a_2+a_3\right)$ is equal to

The sum $1+3+11+25+45+71+\dots$ upto 20 terms, is equal to

If the sum of the first 20 terms of the series $\frac{4\cdot 1}{4+3\cdot 1^2+1^4}+\frac{4\cdot 2}{4+3\cdot 2^2+2^4}+\frac{4\cdot 3}{4+3\cdot 3^2+3^4}+\frac{4\cdot 4}{4+3\cdot 4^2+4^4}+\dots \cdot$ is $\frac{\mathrm{m}}{\mathrm{n}}$, where m and n are coprime, then $\mathrm{m}+\mathrm{n}$ is equal to :

Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and $q$ respectively. Let d and D be the common differences of ${\mathrm{AP}}^{\prime }\mathrm{s}$ in $A$ and $B$ respectively such that $D=d+3, d>0$. If $\frac{p+q}{p-q}=\frac{19}{5}$, then $\mathrm{p}-\mathrm{q}$ is equal to

Let $A=\{1,6,11,16,\dots \}$ and $B=\{9,16,23,30,\dots \}$ be the sets consisting of the first 2025 terms of two arithmetic progressions. Then $n(A\cup B)$ is

$1+3+5^2+7+9^2+\dots$ upto 40 terms is equal to

The sum $1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\dots$ upto $\infty$ terms, is equal to

Let $f(x)={\int }_0^{x^2}\frac{{\mathrm{t}}^2-8\mathrm{t}+15}{{\mathrm{e}}^{\mathrm{t}}}\mathrm{dt},x\in \mathbf{R}$. Then the numbers of local maximum and local minimum points of $f$, respectively, are :

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm , the ice-cream melts at the rate of $81{\mathrm{cm}}^3/\min$ and the thickness of the ice-cream layer decreases at the rate of $\frac{1}{4\pi }\mathrm{cm}/\min$. The surface area (in ${\mathrm{cm}}^2$ ) of the chocolate ball (without the ice-cream layer) is :

Consider the region $R=\left\{(x,y):x\le y\le 9-\frac{11}{3}{x}^2,x\ge 0\right\}$. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R , is: