Let a_n, n 1, be an arithmetic progression with first term 2 and … - Vidyadip

MCQ

Let $a_n, n \geq 1$, be an arithmetic progression with first term $2$ and common difference $4$ . Let $M_n$ be the average of the first $n$ terms. Then the sum $\sum \limits_{n=1}^{10} M_n$ is

✓
$110$
B
$335$
C
$770$
D
$1100$

✓

Answer

Correct option: A.

$110$

a
(a)

The sum of first $n, n \geq 1$ terms of arithmetic progression with first term $2$ and common difference $4$, is

$S_n=\frac{n}{2}[4+(n-1) 4]=2 n^2$

So, the average of the first $n$ terms

$M_n=\frac{S_n}{n}=2 n$

Now, $\sum\limits_{n=1}^{10} M_n=2 \sum\limits_{n=1}^{10} n$

$=2 \times\left(\frac{10 \times 11}{2}\right)=110$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If $y = a\log |x| + b{x^2} + x$ has its extremum values at $x = - 1$ and $x = 2,$ then

If the curve $y=y(x)$ is the solution of the differential equation $2\left(x^{2}+x^{5 / 4}\right) d y-y\left(x+x^{1 / 4}\right) d x=2 x^{9 / 4} d x, x > 0$ which passes through the point $\left(1,1-\frac{4}{3} \log _{e} 2\right),$ then the value of $y(16)$ is equal to :

What will be the equation of that chord of ellipse $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{9} = 1$ which passes from the point $(2,1)$ and bisected on the point

$I = \int\limits_0^1 {\sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx} $ is equal to

The equation $\sqrt 3 \sin x + \cos x = 4$ has

The function $f(x) = \frac{{1 - \sin x + \cos x}}{{1 + \sin x + \cos x}}$ is not defined at $x = \pi .$ The value of $f(\pi ),$ so that $f(x)$ is continuous at $x = \pi $, is

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors such that $\overrightarrow{ a } \times \overrightarrow{ b }=4 \overrightarrow{ c }, \overrightarrow{ b } \times \overrightarrow{ c }=9 \overrightarrow{ a }$ and $\overrightarrow{ c } \times \overrightarrow{ a }=\alpha \overrightarrow{ b }, \alpha>0$ . If $|\vec{a}|+|\vec{b}|+|\vec{c}|=36$, then $\alpha$ is equal to$....$

The sum to $(n + 1)$ terms of the following series $\frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - \frac{{{C_3}}}{5} + $..... is

Let $f:[0,1] \rightarrow[0,1]$ be the function defined by $f(x)=\frac{x^5}{3}-x^2+\frac{5}{9} x+\frac{17}{36}$. Consider the square region $S=[0,1] \times[0,1]$. Let $G=\{(x, y) \in S: y>f(x)\}$ be called the green region and $R=\{(x, y) \in S$ : $\mathrm{y}<\mathrm{f}(\mathrm{x})\}$ be called the red region. Let $\mathrm{L}_{\mathrm{h}}=\{(\mathrm{x}, \mathrm{h}) \in \mathrm{S}: \mathrm{x} \in[0,1]\}$ be the horizontal line drawn at a height $\mathrm{h} \in[0,1]$. Then which of the following statements is(are) ture?

($A$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the green region below the line $\mathrm{L}_{\mathrm{h}}$

($B$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the red region below the line $\mathrm{L}_h$

($C$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the red region below the line $L_h$

($D$) There exists an $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $\mathrm{L}_{\mathrm{h}}$ equals the area of the green region below the line $\mathrm{L}_{\mathrm{h}}$

Let $ABCD$ is a parallelogram where $\overrightarrow {AB} = \overrightarrow a ,\,\overrightarrow {AD} = \overrightarrow b ,\,\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = 2$ and $\left| {\overrightarrow a \times \overrightarrow b } \right| + \overrightarrow a \cdot \overrightarrow b = \sqrt 2 \,\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\,\left( {\overrightarrow a \cdot \overrightarrow b \, > \,0} \right),$ then area of this parallelogram, is (in square units)-