Sequence of series practice · Easy

The series of natural numbers is divided into groups $(1), (2,3,4),(5,6,7,8,9)$, and so on. The sum of first and last term of the $n$th group is

$\left(\frac{a-b}{a}\right)+\frac{1}{2}\left(\frac{a-b}{a}\right)^2+\frac{1}{3}\left(\frac{a-b}{a}\right)^3+\ldots=$ ?

If $a, b$, c are in H.P. then $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ will be in-

If the $n^{\text {th}}$ term of a series be $3+n(n-1)$, then the sum of $n$ terms of the series is

If $f(x)$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in N$ such that $f(1)=3$ and $\sum\limits_{x=1}^n f(x)=120$. Then, the value of $n$ is

Suppose A.M. of two positive numbers be 7 and G.M. between them be 5. The A.M. between their squares is

If H.M. : G.M. $=4: 5$ for the two positive numbers, then the ratio of the numbers is

The solution of the equation ${\log _\mathrm{\pi }}\left( {{{\log }_2}\left( {{{\log }_7}x} \right)} \right) = 0,$ is

If $\frac{{5 + 9 + 13 + ... + n\;\mathrm{terms}}}{{7 + 9 + 11 + ... + 12\;\mathrm{terms}}} = \frac{5}{{12}}$, then $n$ is equal to

$\log _{10} 2, \log _{10}\left(2^x-1\right)$ and $\log _{10}\left(2^x+3\right)$ are three consecutive terms of an A.P. for

If ${\log _3}x \times {\log _x}2x \times {\log _{2x}}y = {\log _x}{x^2},$ then $y$ equals

If ${a^2} + 4{b^2} = 12ab,$ then $\log \left( {a + 2b} \right) =$

I. If each term of a G.P. be raised to the same power, the resulting sequence also forms a G.P.

II. $25^{\text {th}}$ term of the sequence 4, 9, 14, 19,.... is 124.

Choose the correct option.

$1 + \frac{{{2^3}}}{{2!}} + \frac{{{3^3}}}{{3!}} + \frac{{{4^3}}}{{4!}} + \ldots \infty$ equals

The largest interval for which the series $1+(x-1)+(x-1)^{2}$ $+\cdots \infty$ may be summed is

Three numbers are in A.P, such that their sum is 18 and sum of their squares is 158. The greatest among them is

If ${\log _x}\left( {4{x^{{{\log }_5}x}} + 5} \right) = 2{\log _5}x,$ then $x$ equals to

The set of real values of $x$ satisfying ||$x-1|-1| \mid \leq 1$ is

Let $a, b, c, d$ and $e$ be distinct positive numbers. If $a, b$, and $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ both are in A.P. and $b, c, d$ are in G.P., then

The sum to $n$ terms of the series $1+2\left(1+\frac{1}{n}\right)+3\left(1+\frac{1}{n}\right)^2+\ldots$, is given by

The sum of the series ${1^3} + {2^3} + {3^3} + ... + {15^3}$ is

Let $S_n$ denote the sum of the cubes of the first $n$ natural numbers and $s_n$ denote the sum of the first $n$ natural numbers. Then, $\sum\limits_{r=1}^n \frac{S_r}{s_r}$ is equal to

The value of $\frac{1}{{2!}} + \frac{2}{{3!}} + ... + \frac{{999}}{{1000!}}$ is equal to

The greatest positive integer k, for which ${49^k} + 1$ is a factor of the sum ${49^{125}} + {49^{124}} + ... + {49^2} + 49 + 1$ is

The sum of the infinite series $\frac{5}{17}+\frac{55}{(17)^2}+\frac{555}{(17)^3}+\ldots$ is

Let $a,b,c$ be in AP. If $0 < a,\;b,c < 1,\;x = \sum \limits_{n = 0}^\infty {a^n},\;y = \sum \limits_{n = 0}^\infty {b^n}$ and $z = \sum \limits_{n = 0}^\infty {c^n}$ , then

The sum to the series $\frac{1}{{1.2}} + \frac{{1.3}}{{1.2.3.4}} + \frac{{1.3.5}}{{1.2.3.4.5.6}} + ...$ is

Three numbers form a GP. If the 3rd term is decreased by 64, then the three numbers thus obtained will constitute an AP. If the second term of this AP is decreased by 8, a GP will be formed again, then the numbers will be

The sum of ${\left( {x + 2} \right)^{n - 1}} + {\left( {x + 2} \right)^{n - 2}}\left( {x + 1} \right) + {\left( {x + 2} \right)^{n - 3}}{\left( {x + 1} \right)^2} + ... + {\left( {x + 1} \right)^{n - 1}}$ is equal to

The sum to $n$ terms of the series $\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots$, is