If ${\log _2}7 = x,$ then $x$ is:

A rational number such that $0 < x < 2$

A rational number such that $2 < x < 3$

A prime number of the form $7x + 2$

If the sum of the series $20 + 19\frac{3}{5} + 19\frac{1}{5}\, + \,18\frac{4}{5} + \,...\,\,{\rm{upto}}\,\,\,n{\rm{th}}\,$ term is 488 and the $n\,{\rm{th}}$ term is negative, then

$\,n{\rm{th}}\,\,{\rm{term}}\,{\rm{is}}\, - 4$

$n{\rm{th}}\,\,{\rm{term}}\,\,{\rm{is}}\, - 4\frac{2}{5}$

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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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451 Medium

451 of 998

Understanding +4 | -1

{\log _2}7 = x,

then

x

is:

View answer and explanation

Correct option: B

An irrational number such that

2 < x < 3

Option B (An irrational number such that

2 < x < 3

) is correct because 7 is not a power of 2, and it lies between 2² and 2³.

452 Medium

452 of 998

Application +4 | -1

\frac{{\log 3}}{{x - y}} = \frac{{\log 5}}{{y - z}} = \frac{{\log 7}}{{z - x}},

then

{3^{x + y}}\;{5^{y + z}}\;{7^{z + x}} =

View answer and explanation

Correct option: C

Option C is correct because by setting each ratio equal to a constant and using logarithmic properties, the expression simplifies to 1.

453 Easy

453 of 998

Application +4 | -1

a, a_1, a_2, \ldots ., a_{2 n-1}, b

are in A.P.,

a, b_1, b_2, \ldots . b_{2 n-1}

b

are in G.P. and

a, c_1, c_2, \ldots, c_{2 n-1}, b

, are in H.P. where

a

b

are positive, then

View answer and explanation

Correct option: B

b_n^2=c_n a_n

Option B (

b_n^2=c_n a_n

) is correct because it is derived from the relationship between the arithmetic, geometric, and harmonic means of two positive numbers a and b.

454 Easy

454 of 998

Application +4 | -1

The second, third and sixth terms of an A.P. are consecutive terms of a G.P. The common ratio of the G.P. is

View answer and explanation

Correct option: C

Option C (3) is correct because by setting up the equation for consecutive terms in a G.P., the common ratio is found to be 3.

455 Easy

455 of 998

Application +4 | -1

The sum to n terms of the series

\frac{1}{{\sqrt 1 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 5 }} + \;...

View answer and explanation

Correct option: D

\frac{1}{2}\left( {\sqrt {2n + 1} - 1} \right)

Option D (

\frac{1}{2}\left( {\sqrt {2n + 1} - 1} \right)

) is correct because the sum to n terms of the given series is found by rationalizing the denominator and simplifying the resulting telescoping series.

456 Medium

456 of 998

Application +4 | -1

If three numbers are in H.P., then the numbers obtained by subtracting half of the middle number from each of them are in

View answer and explanation

Correct option: B

G.P.

Option B (G.P.) is correct because subtracting half of the middle term from each term in a H.P. results in a G.P.

457 Easy

457 of 998

Application +4 | -1

\frac{a^n+b^n}{a^{n-1}+b^{n-1}}

is the AM between

a

and

b

, then the value of

n

, is

View answer and explanation

Correct option: B

Option B (1) is correct because when n=1, the given expression simplifies to (a+b)/2, which is the arithmetic mean of a and b.

458 Medium

458 of 998

Application +4 | -1

Let

a_n

be the

n

th term of an AP. If

\sum\limits_{{r=1}}^{10^{99}} a_{2 r}=10^{100}

and

\sum\limits_{{r=1}}^{10^{99}} a_{2 r-1}=10^{99}

, then the common difference of an AP is

View answer and explanation

Correct option: B

Option B (9) is correct because the common difference of the AP is calculated to be 9 based on the given summations.

459 Easy

459 of 998

Application +4 | -1

The common difference of the A.P.

b_1, b_2 \ldots, b_m

is 2 more than the common difference of the A.P.

a_1, a_2, \ldots, a_n

. If

a_{40}=-159, a_{100}=-399

and

b_{100}=a_{70}

, then

b_1

is equal to

View answer and explanation

Correct option: D

-81

Option D (-81) is correct because it calculates the first term b_1 of the second A.P. using the given information about both A.P.s.

460 Easy

460 of 998

Application +4 | -1

The 1025th term in the sequence 1, 22,

4444,88888888, \ldots

View answer and explanation

Correct option: B

2^{10}

Option B (

2^{10}

) is correct because the 1025th term falls within the range defined by

2^{10}

based on the geometric progression of digits.

461 Easy

461 of 998

Application +4 | -1

The interior angles of a polygon are in AP. If the smallest angle be 120

^\circ

and the common difference be 5, then the number of side is

View answer and explanation

Correct option: C

Option C (9) is correct because it is the number of sides of the polygon derived from the given conditions about its interior angles being in AP.

462 Medium

462 of 998

Application +4 | -1

S_k=\lim \limits_{{n \rightarrow \infty}} \sum\limits_{{i=0}}^n \frac{1}{(k+1)^i}

, then

\sum\limits_{{k=1}}^n k S_k

is equal to

View answer and explanation

Correct option: D

\frac{n(n+3)}{2}

Option D (n(n+3)/2) is correct because it is the result of simplifying the summation after finding that S_k = (k+1)/k.

463 Easy

463 of 998

Application +4 | -1

If the sum of an infinite decreasing G.P. is 3 and sum of the cubes of its terms is

108 / 13

, then common ratio is given by

View answer and explanation

Correct option: C

\frac{1}{3}

Option C (

\frac{1}{3}

) is correct because it's the common ratio derived from the given conditions on the sum of an infinite GP and the sum of the cubes of its terms.

464 Medium

464 of 998

Application +4 | -1

The solution of the equation

\left( {x + 1} \right) + \left( {x + 4} \right) + \left( {x + 7} \right) + ... + \left( {x + 28} \right) = 155

View answer and explanation

Correct option: A

Option A (1) is the correct solution because substituting x=1 into the given equation satisfies the condition after summing the arithmetic progression.

465 Easy

465 of 998

Application +4 | -1

If the sum of the series

20 + 19\frac{3}{5} + 19\frac{1}{5}\, + \,18\frac{4}{5} + \,...\,\,\mathrm{upto}\,\,\,n\mathrm{th}\,

term is 488 and the

n\,\mathrm{th}

term is negative, then

View answer and explanation

Correct option: A

\,n\mathrm{th}\,\,\mathrm{term}\,\mathrm{is}\, - 4

Option A (

\n\mathrm{th}\,\mathrm{term}\,\mathrm{is}\, - 4

) is correct because the nth term is -4 when the sum of the series is 488 and the nth term is negative.

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466 Easy

466 of 998

Application +4 | -1

Let there be a G.P. whose first term is

a

and the common ratio is

r

. If

A

and

H

are the arithmetic mean and the harmonic mean respectively for the first

n

terms of the G.P. Then,

A H

is equal to

View answer and explanation

Correct option: A

a^2 r^{n-1}

The correct answer is A (

a^2 r^{n-1}

) because it is the product of the arithmetic mean (A) and harmonic mean (H) of the first n terms of the given geometric progression.

467 Medium

467 of 998

Application +4 | -1

The harmonic mean between two numbers is

14\frac{2}{5}

and the geometric mean is 24. The greater number between them is

View answer and explanation

Correct option: A

Option A (72) is correct because it is the larger of the two numbers calculated using the given harmonic and geometric means.

468 Medium

468 of 998

Application +4 | -1

The sum of an infinite geometric series is 3 . A series which is formed by squares of its terms also have sum equal to 3 . Then, the sequence corresponding to the series will be

View answer and explanation

Correct option: A

\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{3}{16}, \ldots

Option A (

\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{3}{16}, \ldots

) is correct because solving the system of equations derived from the sum of the infinite GP and the sum of the squares of its terms gives a = 3/2 and r = 1/2.

469 Easy

469 of 998

Application +4 | -1

If the sum of

n

terms of the series

704 + \frac{1}{2}\left( {704} \right) + \frac{1}{4}\left( {704} \right) + \ldots \mathrm{and},1984 - \frac{1}{2}\left( {1984} \right) + \frac{1}{4}\left( {1984} \right) \ldots

are equal, then

n =

View answer and explanation

Correct option: A

Option A (5) is correct because the sum of the two series are equal when n=5, which is found by solving the equation formed by equating the sums.

470 Medium

470 of 998

Application +4 | -1

The sum of 10 terms of the sequence

1,6,27,108, \ldots

View answer and explanation

Correct option: D

280483

Option D (280483) is correct because the series is an arithmetico-geometric progression, and the sum of the first 10 terms is calculated using the appropriate formula.

471 Easy

471 of 998

Application +4 | -1

If the AM and GM between two numbers are in the ratio

m:n

, then the numbers are in the ratio

View answer and explanation

Correct option: C

m + \sqrt {{m^2} - {n^2}} :m - \sqrt {{m^2} - {n^2}}

Option C (

m + \sqrt {{m^2} - {n^2}} :m - \sqrt {{m^2} - {n^2}}

) is correct because it represents the ratio of two numbers when the ratio of their AM and GM is m:n.

472 Easy

472 of 998

Analyzing +4 | -1

The sum of the series

{\log _4}2 - {\log _8}2 + {\log _{16}}2 - {\log _{32}}2 + ...

View answer and explanation

Correct option: D

1 - {\log _e}2

Option D (

1 - {\log _e}2

) is correct because the series can be simplified to

1 - {\log _e}2

using logarithm properties and series expansion.

473 Easy

473 of 998

Analyzing +4 | -1

The sum of the infinite series

\frac{{{2^2}}}{{2!}} + \frac{{{2^4}}}{{4!}} + \frac{{{2^6}}}{{6!}} + \ldots

is equal to

View answer and explanation

Correct option: C

\frac{{{{\left( {{e^2} - 1} \right)}^2}}}{{2{e^2}}}

Option C (

\frac{{{{\left( {{e^2} - 1} \right)}^2}}}{{2{e^2}}}

) is correct because it is derived from the power series expansion of

e^x

and

e^{-x}

, and then subtracting 1 to match the given series.

474 Easy

474 of 998

Application +4 | -1

{\log _2}x + {\log _4}x + {\log _{16}}x = \frac{{21}}{4},

then

x

equals to

View answer and explanation

Correct option: A

Option A (8) is correct because by changing the base of the logarithms to 2 and solving the resulting equation, we find that x = 8.

475 Easy

475 of 998

Understanding +4 | -1

\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots+\ldots \frac{1}{n .(n+1)}

equals

View answer and explanation

Correct option: B

\frac{n}{n+1}

Option B (

\frac{n}{n+1}

) is correct because the series can be expressed as a telescoping sum, where each term is of the form

\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}

, leading to the result

\frac{n}{n+1}

476 Easy

476 of 998

Application +4 | -1

An A.P. consists of

2 n

terms. If two middle terms are

\alpha-\beta

and

\alpha+\beta

, then sum of the

2 n

terms of the A.P. is

View answer and explanation

Correct option: A

2 n \alpha

Option A (

2n \alpha

) is correct because the sum of the 2n terms of the A.P. is

2n \alpha

, given that the two middle terms are

\alpha - \beta

and

\alpha + \beta

477 Easy

477 of 998

Application +4 | -1

The value of

2.\overline {357}

View answer and explanation

Correct option: B

\frac{{2355}}{{999}}

Option B (

\frac{{2355}}{{999}}

) is correct because it represents the fractional equivalent of the repeating decimal

2.\overline {357}

478 Easy

478 of 998

Application +4 | -1

{\log _{10}}2 = 0.3010,

then

{\log _5}64 =

View answer and explanation

Correct option: A

\frac{{602}}{{233}}

Option A (

\frac{{602}}{{233}}

) is correct because it is the result of converting the logarithm to base 10 and using the given value of

{\log _{10}}2

479 Medium

479 of 998

Understanding +4 | -1

The coefficient of

{x^n}

in the expansion of

{\log _a}\left( {1 + x} \right)

View answer and explanation

Correct option: B

\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}{\log _a}e

Option B is correct because the coefficient of x^n in the expansion of log_a(1+x) is given by ((-1)^(n-1) / n) * log_a(e).

480 Medium

480 of 998

Application +4 | -1

1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+\ldots

upto 22nd terms is

View answer and explanation

Correct option: A

23276

Option A (23276) is correct. By finding the general term and summing the series, we get the result 23276.

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