Given that $n$ AM’s are inserted between two sets of numbers $a,\;2b\;{\rm{and}}\;2a,\;b\;{\rm{where}}\;a,\;b \in R.$. Suppose further that $m$th mean between these sets of numbers is same, then the ratio $a\;:b$ equals

$n\;:\left( {n - m + 1} \right)$

The harmonic mean of $ \frac{a}{1-a b} $ and $ \frac{a}{1+a b} $ is

$ \frac{a}{\sqrt{1-a^{2} b^{2}}} $

$\begin{array}{*{20}{c}} {{{\left( {666 \ldots .6} \right)}^2}}\\ {n - {\rm{digits}}} \end{array}\begin{array}{*{20}{c}} {\; + \;\left( {888 \ldots .8} \right)}\\ {n - {\rm{digits}}} \end{array}$ is equal to

$\frac{4}{9}\left( {{{10}^n} - 1} \right)$

$\frac{4}{9}{\left( {{{10}^n} - 1} \right)^2}$

Back to SEQUENCE OF SERIES

Your Progress

0 / 564 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Medium

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →

Showing 421-450 of 564 questions

Preparing practice and saved progress…

421 Medium

421 of 564

Analyzing +4 | -1

Observe that

1^3=1,2^3=3+5

3^3=7+9+11,4^3=13+15+17+19

. Then,

n^3

as a similar series is

View answer and explanation

Correct option: C

\left(n^2-n+1\right)+\left(n^2-n+3\right)+\left(n^2-n+5\right)

+\ldots+\left(n^2+n-1\right)

Option C (

\left(n^2-n+1\right)+\left(n^2-n+3\right)+\left(n^2-n+5\right) +\ldots+\left(n^2+n-1\right)

) is correct because it represents the sum of n consecutive odd numbers starting from

n^2 - n + 1

, which equals

n^3

422 Medium

422 of 564

Analyzing +4 | -1

p^{\text {th }}, q^{\text {th }}, r^{\text {th }}

, terms of a G.P. are the positive numbers

a, b, c

respectively then angle between the vectors

\log a^3 \hat{i}+\log b^3 \hat{j}+\log c^3 \hat{k}

and

(q-r) \hat{i}+(r-p) \hat{j}+(p-q) \hat{k}

is:

View answer and explanation

Correct option: A

\frac{\pi}{2}

Option A (

\frac{\pi}{2}

) is the correct answer because the dot product of the two vectors is zero, indicating they are orthogonal.

423 Medium

423 of 564

Application +4 | -1

{2^{\frac{3}{{{{\log }_3}x}}}} = \frac{1}{{64}},

then

x =

View answer and explanation

Correct option: C

\frac{1}{{\sqrt 3 }}

Option C (

\frac{1}{{\sqrt 3 }}

) is correct because by simplifying the exponential equation using logarithm properties, the value of

x

is found to be

\frac{1}{{\sqrt 3 }}

424 Medium

424 of 564

Analyzing +4 | -1

\log _3 2, \log _6 2

and

\log _{12} 2

are in

View answer and explanation

Correct option: C

\mathrm{HP}

Option C (HP) is correct because the reciprocals of the given logarithmic terms are in Arithmetic Progression (AP).

425 Medium

425 of 564

Analyzing +4 | -1

a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right)

and

c\left(\frac{1}{a}+\frac{1}{b}\right)

are in

\mathrm{AP}

, then

View answer and explanation

Correct option: D

a, b

and

c

are in

\mathbf{A P}

Option D (

a, b

and

c

are in AP) is correct because the given condition leads to the conclusion that

a, b,

and

c

have a constant difference.

426 Medium

426 of 564

Application +4 | -1

Let

a=\log 25

and

b=\log 225

, then

\log \left(\frac{1}{81}\right)+\log \left(\frac{1}{2250}\right)

is equal to:

View answer and explanation

Correct option: C

2 a-3 b-1

Option C (

2a-3b-1

) is correct because by expressing the given logarithms in terms of a and b and simplifying, the expression

\log (\frac{{1}}{{81}})+\log (\frac{{1}}{{2250}})

simplifies to

2a-3b-1

427 Medium

427 of 564

Analyzing +4 | -1

The next term of the sequence

1,5,14,30,55, \ldots

View answer and explanation

Correct option: A

Option A (91) is correct because the next term in the sequence is found by adding the square of the term number to the previous term, resulting in 91.

428 Medium

428 of 564

Application +4 | -1

The five geometric means between 576 and 9 are

View answer and explanation

Correct option: B

288,144,72,36,18

Option B (288, 144, 72, 36, 18) is correct because these are the five geometric means between 576 and 9, calculated using the common ratio of 1/2.

429 Medium

429 of 564

Application +4 | -1

\sum\limits_{{n=1}}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}

, then

\sum\limits_{{n=1}}^{\infty} \frac{1}{(2 n-1)^2}

View answer and explanation

Correct option: A

\frac{\pi^2}{8}

Option A (

\frac{\pi^2}{8}

) is correct because it is derived by subtracting the sum of the reciprocals of even squares from the sum of the reciprocals of all squares.

430 Medium

430 of 564

Application +4 | -1

\frac{1^3+2^3+3^3+4^3+\ldots \ldots . .12^3}{1^2+2^2+3^2+4^2+\ldots \ldots . .12^2}=

View answer and explanation

Correct option: A

\frac{234}{25}

Option A (234/25) is correct because it is the result of simplifying the ratio of the sum of cubes to the sum of squares for the first 12 natural numbers.

431 Medium

431 of 564

Application +4 | -1

|2 x-3| < |x+5|

, then

x

belongs to

View answer and explanation

Correct option: C

\left(-\frac{2}{3}, 8\right)

Option C (

\left(-\frac{{2}}{{3}}, 8\right)

) is correct because it represents the solution set for the inequality

|2x-3| < |x+5|

after considering different intervals.

432 Medium

432 of 564

Analyzing +4 | -1

The sum of the series

\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\ldots

upto 16 terms is

View answer and explanation

Correct option: C

446

Option C (446) is correct because it is the sum of the series up to 16 terms after simplification.

433 Medium

433 of 564

Application +4 | -1

\frac{{x - y}}{x} + \frac{1}{2}{\left( {\frac{{x - y}}{x}} \right)^2} + \frac{1}{3}{\left( {\frac{{x - y}}{x}} \right)^3} + ...

is equal to

View answer and explanation

Correct option: C

{\log _e}\left( {\frac{x}{y}} \right)\;

Option C (

{\log _e}\left( {\frac{x}{y}} \right)\;

) is correct because the given series is the expansion of

-\log_e(1 - \frac{{x-y}}{x})

which simplifies to

{\log _e}\left( {\frac{x}{y}} \right)

434 Medium

434 of 564

Application +4 | -1

Given that

n

AM’s are inserted between two sets of numbers

a,\;2b\;\mathrm{and}\;2a,\;b\;\mathrm{where}\;a,\;b \in R.

. Suppose further that

m

th mean between these sets of numbers is same, then the ratio

a\;:b

equals

View answer and explanation

Correct option: D

m\;:\left( {n - m + 1} \right)

Option D (

m : (n - m + 1)

) is correct because it represents the ratio

a:b

derived from equating the

m

th means of the two sets of numbers.

435 Medium

435 of 564

Application +4 | -1

The next term of the sequence

1,3,6,10, \ldots

View answer and explanation

Correct option: C

Option C (15) is correct because the sequence follows a pattern where each term is the sum of the previous term and the next consecutive integer.

Explore Other Chapters in MATHEMATICS

View All →

DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs

436 Medium

436 of 564

Analyzing +4 | -1

a = 1 + {\log _x}y\;z,\;b = 1 + {\log _y}zx,

c = 1 + {\log _z}xy,

then

ab + bc + ca =

View answer and explanation

Correct option: C

abc

Option C (

abc

) is correct because

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1

, which implies

ab + bc + ca = abc

437 Medium

437 of 564

Analyzing +4 | -1

The sum of the infinite series

{\left( {\frac{1}{3}} \right)^2} + \frac{1}{3}{\left( {\frac{1}{3}} \right)^4} + \frac{1}{5}{\left( {\frac{1}{3}} \right)^6} + \ldots

View answer and explanation

Correct option: C

\frac{1}{6}{\log _e}2

Option C (

\frac{1}{6}{\log _e}2

) is correct because the series can be related to the logarithmic series expansion.

438 Medium

438 of 564

Analyzing +4 | -1

The sum

\sum\limits_{{n=1}}^{\infty} \frac{2 n^{2}+3 n+4}{(2 n) !}

is equal to:

View answer and explanation

Correct option: A

\frac{13 e}{4}+\frac{5}{4 e}

The correct answer is A (

\frac{13 e}{4}+\frac{5}{4 e}

) because it is obtained by manipulating the series and expressing it in terms of exponential series.

439 Medium

439 of 564

Application +4 | -1

\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\ldots=

View answer and explanation

Correct option: B

\frac{1}{e}

Option B (

\frac{{1}}{{e}}

) is correct because the given series is a direct result of substituting x = -1 into the Taylor series expansion of

e^x

and rearranging the terms.

440 Medium

440 of 564

Application +4 | -1

y = 1 + x + {x^2} + {x^3} + ...

then

x

is equal to

View answer and explanation

Correct option: A

\frac{{y - 1}}{y}

Option A is correct because x can be expressed as (y-1)/y by recognizing the geometric series and solving for x.

441 Medium

441 of 564

Application +4 | -1

Let

{S_k} = \frac{{1 + 2 + 3 + ... + k}}{k}.\mathrm{ }\,\,\mathrm{If}

S_1^2 + S_2^2 + ... + S_{10}^2 = \frac{5}{{12}}A,

then A is equal to

View answer and explanation

Correct option: D

303

Option C (

2\int_0^a {f\left( x \right)dx}

) is correct because using the properties of definite integrals and the given conditions, the integral simplifies to this expression.

442 Medium

442 of 564

Application +4 | -1

The harmonic mean of

\frac{a}{1-a b}

and

\frac{a}{1+a b}

View answer and explanation

Correct option: C

a

Option C (

a

) is correct because when the harmonic mean is calculated between the two given expressions, the result simplifies to 'a'.

443 Medium

443 of 564

Application +4 | -1

The sum of integers from 1 to 100 that are divisible by 2 or 5 is

View answer and explanation

Correct option: B

3050

Option B (3050) is correct because it's the sum of numbers divisible by 2 or 5, calculated using the principle of inclusion-exclusion.

444 Medium

444 of 564

Application +4 | -1

Sum of

n

terms of the series

\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ldots

View answer and explanation

Correct option: C

\frac{n(n+1)}{\sqrt{2}}

Option C (

\frac{n(n+1)}{\sqrt{2}}

) is correct because it is the sum of the given arithmetic series.

445 Medium

445 of 564

Application +4 | -1

The sum of the series

1.3 .5+.2 .5 .8+3.7 .11+\ldots

upto '

n

' terms is

View answer and explanation

Correct option: A

\frac{n(n+1)\left(9 n^2+23 n+13\right)}{6}

Option A (

\frac{n(n+1)(9n^2+23n+13)}{6}

) is correct because it is the sum of the series calculated by finding the general term and then summing it using standard summation formulas.

446 Medium

446 of 564

Analyzing +4 | -1

\begin{array}{cccccccccccccccccccc} {{{\left( {666 \ldots .6} \right)}^2}}\\ {n - \mathrm{digits}} \end{array}\begin{array}{cccccccccccccccccccc} {\; + \;\left( {888 \ldots .8} \right)}\\ {n - \mathrm{digits}} \end{array}

is equal to

View answer and explanation

Correct option: B

\frac{4}{9}\left( {{{10}^{2n}} - 1} \right)

Option B (

\frac{4}{9}\left( {{{10}^{2n}} - 1} \right)

) is correct because the expression simplifies to this result after converting the repeating digits to a geometric series and applying algebraic manipulations.

447 Medium

447 of 564

Application +4 | -1

a_1, a_2, a_3, \ldots

are in A.P., and

a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=240

, then the sum of the first 24 terms will be

View answer and explanation

Correct option: D

960

Option D (960) is correct because the sum of the first 24 terms can be found by recognizing that the sum of terms equidistant from the beginning and end is constant.

448 Medium

448 of 564

Analyzing +4 | -1

\log (a+c), \log (c-a)

and

\log (a-2 b+c)

are in AP, then

View answer and explanation

Correct option: D

a, b

and

c

are in HP

Option C (2n) is correct because the summation telescopes, and the final result simplifies to 2n.

449 Medium

449 of 564

Understanding +4 | -1

The sum

1(1 !)+2(2 !)+3(3 !)+\ldots+n(n !)

equals

View answer and explanation

Correct option: C

(n+1) !-1

Option C (

(n+1) !-1

) is correct because by rewriting each term as ((k+1) - 1)k! and telescoping the sum, we arrive at (n+1)! - 1.

450 Medium

450 of 564

Analyzing +4 | -1

If three real numbers

a,b,c

are in harmonic progression, then which of the following is true?

View answer and explanation

Correct option: B

\frac{1}{{bc}},\frac{1}{{ca}},\frac{1}{{ab}}

are in HP

Option C (

ab,bc,ca

are in HP) is incorrect. Option D (

\frac{a}{b},\frac{b}{c},\frac{c}{a}

are in HP) is incorrect. Option A (

\frac{1}{a},b,\frac{1}{c}

are in AP) is incorrect. Option B (

\frac{1}{{bc}},\frac{1}{{ca}},\frac{1}{{ab}}

are in HP) is correct because if

a, b, c

are in HP, then

bc, ca, ab

are in AP, and thus

\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}

are in HP.

Revision notes and key takeaways

Review SEQUENCE OF SERIES in your textbook or class notes, then use this set to check your understanding.

Showing 421-450 of 564 questions

Your Progress

Sequence of series practice · Medium

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice