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

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

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

If ${\rm{1/(b + c),1/(c + a),1/(a + b)}}$ are in A.P., then

$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.

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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Easy

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451 Easy

451 of 998

Application +4 | -1

x = 1 + a + {a^2} + \ldots \infty

and

y = 1 + b + {b^2} + \ldots \infty

where

a

and

b

are proper fractions, then

1 + ab + {a^2}{b^2} + \ldots \infty

equals

View answer and explanation

Correct option: A

\frac{{xy}}{{y + x - 1}}

Option A (

\frac{{xy}}{{y + x - 1}}

) is correct because, given the infinite geometric series for x and y, the expression

1 + ab + {a^2}{b^2} + \ldots \infty

simplifies to

\frac{{xy}}{{x + y - 1}}

452 Easy

452 of 998

Application +4 | -1

y = {2^{1/{{\log }_x}8}},

, then

x

equal to

View answer and explanation

Correct option: C

{y^3}

Option C is correct because by using the properties of logarithms and simplifying the expression, we can find that x = y³.

453 Easy

453 of 998

Application +4 | -1

If the sum of first

n

terms of an AP is

c n^2

, then the sum of squares of these

n

terms is

View answer and explanation

Correct option: C

\frac{n\left(4 n^2-1\right) c^2}{3}

Option C (

\frac{n(4n^2-1)c^2}{3}

) is correct because it is the sum of the squares of the first n terms of the AP, derived from the given sum of the first n terms.

454 Easy

454 of 998

Application +4 | -1

The minimum value of

{4^x} + {4^{1 - x}},x \in R,

View answer and explanation

Correct option: C

Option C (4) is correct because the minimum value is found using the AM-GM inequality, resulting in a minimum value of 4.

455 Easy

455 of 998

Analyzing +4 | -1

The product of

(32)-(32)^{1 / 6} \cdot(32)^{1 / 36}+\ldots

is equal to

View answer and explanation

Correct option: B

Option B (64) is correct because the given expression is an infinite geometric progression in the exponent, which simplifies to 64.

456 Easy

456 of 998

Application +4 | -1

n

arithmetic means are inserted between

a

and 100 such that the ratio of the first mean to the last mean is 1: 7 and

a + n = 33,

then the value of

n

is:

View answer and explanation

Correct option: C

Option C (23) is correct because it satisfies the given conditions for the arithmetic means inserted between a and 100.

457 Easy

457 of 998

Analyzing +4 | -1

A=1+r^a+r^{2 a}+r^{3 a}+\ldots \ldots

\infty

and

B=1+r^b+r^{2 b}+\ldots \ldots

\infty

, then

\frac{a}{b}

is equal to

View answer and explanation

Correct option: B

{\log _{\left( {\frac{{B - 1}}{B}} \right)}}\left( {\frac{{A - 1}}{A}} \right)

Option B is correct because it correctly expresses the relationship between a, b, A, and B derived from the given infinite geometric series.

458 Easy

458 of 998

Analyzing +4 | -1

\frac{1}{2} - \frac{1}{{{{2.2}^2}}} + \frac{1}{{{{3.2}^3}}} - \frac{1}{{{{4.2}^4}}} + \ldots

is equal to

View answer and explanation

Correct option: C

{\log _e}\left( {\frac{3}{2}} \right)

Option C (

{\log _e}(\frac{3}{2})

) is correct because the given series is a direct application of the Maclaurin series expansion of

\log_e(1+x)

with

x = \frac{1}{2}

459 Easy

459 of 998

Application +4 | -1

The value of

n

for which

704+\frac{1}{2}(704)+

\frac{1}{4}(704)+\ldots

upto

n

terms

=1984-\frac{1}{2}(1984)+\frac{1}{4}(1984)

-\ldots

upto

n

terms is

View answer and explanation

Correct option: A

Option A (5) is correct because solving the equation derived from the given geometric series leads to n=5 as the only valid solution.

460 Easy

460 of 998

Application +4 | -1

Which of the following is the sum of 20 terms of an

A P 1,4,7,10, \ldots

View answer and explanation

Correct option: B

590

Option A (590) is correct because the sum of the first 20 terms of the AP 1, 4, 7, 10,... is 590.

461 Easy

461 of 998

Analyzing +4 | -1

2 \lambda, \lambda

and

\left[\lambda^2-14\right], \lambda \in R-\{0\}

and [.] denotes the greatest integer function are the first three terms of a GP in order, then the 51th term of the sequence,

1,3 \lambda, 6 \lambda, 10 \lambda, \ldots

View answer and explanation

Correct option: B

5304

Option B (5304) is correct because after finding λ = 4, the 51th term of the sequence is calculated using the sum of the first 51 natural numbers multiplied by 4.

462 Easy

462 of 998

Application +4 | -1

The difference between the fourth and the first term of a GP is 52 and the sum of the first three terms is 26 . Then the sum of the first six terms is

View answer and explanation

Correct option: C

728

Option C (728) is correct because solving the equations for the GP yields the sum of the first six terms as 728.

463 Easy

463 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.

464 Easy

464 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.

465 Easy

465 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.

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466 Easy

466 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.

467 Easy

467 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.

468 Easy

468 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.

469 Easy

469 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.

470 Easy

470 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.

471 Easy

471 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.

472 Easy

472 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.

473 Easy

473 of 998

Analyzing +4 | -1

{S_n} = \left( {{1^2} - 1 + 1} \right)\left( {1!} \right) + \left( {{2^2} - 2 + 1} \right)\left( {2!} \right) + ...... + \left( {{n^2} - n + 1} \right)\left( {n!} \right),

then

{S_{50}}

is:

View answer and explanation

Correct option: B

1 + 49 \times 51!

Option B (

1 + 49 \times 51!

) is correct because by expressing the general term and summing the series, we find the value of S_{50}.

474 Easy

474 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.

475 Easy

475 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.

476 Easy

476 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.

477 Easy

477 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.

478 Easy

478 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.

479 Easy

479 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}

480 Easy

480 of 998

Analyzing +4 | -1

\mathrm{1/(b + c),1/(c + a),1/(a + b)}

are in A.P., then

View answer and explanation

Correct option: B

{a^2},{b^2},{c^2}

are in A.P.

Option B (

{a^2},{b^2},{c^2}

are in A.P.) is correct because the given condition leads to the relationship

2b^2 = a^2 + c^2

, which defines an arithmetic progression for

a^2, b^2, c^2

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