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 $7x + 2$

The sum of the series $1^2 \cdot 2+2^2 \cdot 3+3^2 \cdot 4+\ldots$ to $n$ terms is

$\frac{n(n+1)}{6}[n(n+1)+(2 n+1)]$

$\frac{n(n+1)}{12}[6 n(n+1)+2(2 n+1)]$

If ${\log _a}x,{\log _b}x,{\log _c}x$ are in A.P., where $x \ne 1,$ then ${c^2} = $

${\left( {ab} \right)^{{{\log }_a}b}}$

${\left( {ab} \right)^{{{\log }_b}a}}$

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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Medium

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241 Medium

241 of 564

Application +4 | -1

f(n)=\left[\frac{1}{2}+\frac{n}{100}\right]

, where

[x]

denotes the integral part of

x

, then the value of

\sum\limits_{{n=1}}^{100} f(n)

View answer and explanation

Correct option: B

Option B (51) is correct because the sum of the integral parts of the given function from n=1 to 100 is equal to 51.

242 Medium

242 of 564

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.

243 Medium

243 of 564

Application +4 | -1

Numbers

x_1, x_2, x_3

and

y_1, y_2, y_3

are in G.P. and the common ratio in both the sequences is the same. Then points

\mathrm{A}\left(x_1, y_1\right), \mathrm{B}\left(x_2, y_2\right), \mathrm{C}\left(x_3, y_3\right)

are

View answer and explanation

Correct option: A

Collinear

Option A (Collinear) is correct because the points lie on a straight line due to the constant ratio between their coordinates.

244 Medium

244 of 564

Analyzing +4 | -1

a_1=0

and

a_1, a_2, a_3, \ldots, a_n

are real numbers such that

\left|a_i\right|=\left|a_{i-1}+1\right|

for all

i

, then the AM of the numbers

a_1, a_2, \ldots, a_n

has value

x

, where

View answer and explanation

Correct option: B

x \geq-\frac{1}{2}

Option B (

x \geq-\frac{1}{2}

) is correct because the AM of the numbers

a_1, a_2, \ldots, a_n

is greater than or equal to

-\frac{1}{2}

as derived from the given condition.

245 Medium

245 of 564

Application +4 | -1

n=(1999)

! then

\sum\limits_{{x=1}}^{1999} \log _{n} x

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because the sum of logarithms with the same base is the logarithm of the product, which simplifies to 1.

246 Medium

246 of 564

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.

247 Medium

247 of 564

Analyzing +4 | -1

Suppose

a, b, c

are in A.P. and

a^2, b^2, c^2

are in G.P. If

a < b < c

and

a+b+c=3 / 2

, then value of

a

View answer and explanation

Correct option: D

\frac{1}{2}-\frac{1}{\sqrt{2}}

Option D (

\frac{1}{2}-\frac{1}{\sqrt{2}}

) is correct because it satisfies the A.P. and G.P. conditions along with the given constraints.

248 Medium

248 of 564

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.

249 Medium

249 of 564

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.

250 Medium

250 of 564

Application +4 | -1

\log _2 x+\log _2 y \geq 6

, then the least value of

x+y

View answer and explanation

Correct option: A

Option A (16) is correct because by using the properties of logarithms and the AM-GM inequality, we can find the least value of

x+y

251 Medium

251 of 564

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.

252 Medium

252 of 564

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.

253 Medium

253 of 564

Analyzing +4 | -1

\operatorname{gcd}(m, n)=1

and

1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+(2021)^{2} -(2022)^{2}+(2023)^{2}=1012 m^{2} n

then

m^{2}-n^{2}

is equal to

View answer and explanation

Correct option: A

240

The correct answer is A (240) because simplifying the series results in 2023 * 1012, which factors to (17² * 7) * (2² * 11 * 23). Since gcd(m,n) = 1, m = 17 and n = 7, so m² - n² = 240.

254 Medium

254 of 564

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.

255 Medium

255 of 564

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.

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256 Medium

256 of 564

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.

257 Medium

257 of 564

Analyzing +4 | -1

L=\lim \limits_{{n \rightarrow \infty}} \frac{1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\ldots \text { upto } n \text { terms }}{n(1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots \text { upto } n \text { terms })}

is equal to

View answer and explanation

Correct option: A

\frac{3}{4}

Option A (

\frac{3}{4}

) is correct because after simplifying the numerator and denominator using summation formulas and taking the limit, the expression evaluates to

\frac{3}{4}

258 Medium

258 of 564

Analyzing +4 | -1

The sum of infinite terms of a decreasing GP is equal to the greatest value of the function

f(x)=x^3+3 x-9

in the interval

[-2,3]

and the difference between the first two terms is

f^{\prime}(0)

. Then, the common ratio of a GP is

View answer and explanation

Correct option: C

\frac{2}{3}

Option C (

\frac{2}{3}

) is correct because by setting up the equations based on the given conditions and solving for the common ratio, we find that r =

\frac{2}{3}

259 Medium

259 of 564

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).

260 Medium

260 of 564

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.

261 Medium

261 of 564

Application +4 | -1

The sum of the series

1^2 \cdot 2+2^2 \cdot 3+3^2 \cdot 4+\ldots

n

terms is

View answer and explanation

Correct option: B

\frac{n(n+1)\left(3 n^2+7 n+2\right)}{12}

Option B is correct because the sum of the series is derived by expressing the general term and then summing the series using known summation formulas.

262 Medium

262 of 564

Application +4 | -1

x,1,z

are in A.P. and

x,2,z

are in G.P., then

x,4,z

are in

View answer and explanation

Correct option: C

H.P.

Option C (H.P.) is correct because

x, 4, z

satisfy the condition for being in Harmonic Progression, which is

4 = \frac{{2xz}}{{x + z}}

263 Medium

263 of 564

Application +4 | -1

m

is the root of the equation

(1-a b) x^2-\left(a^2+b^2\right) x-(1+a b)=0

and

m

harmonic means are inserted between

a

and

b

, then the difference between the last and the first of the means equals

View answer and explanation

Correct option: C

a b(b-a)

Option C (

ab(b-a)

) is correct because it is derived by finding the first and last Harmonic Means and then calculating their difference using the given equation.

264 Medium

264 of 564

Analyzing +4 | -1

{\log _a}x,{\log _b}x,{\log _c}x

are in A.P., where

x \ne 1,

then

{c^2} =

View answer and explanation

Correct option: B

{\left( {ac} \right)^{{{\log }_a}b}}

Option D (

{\left( {ac} \right)^{{{\log }_b}a}}

) is correct because by using the properties of logarithms and the given condition that

{\log _a}x,{\log _b}x,{\log _c}x

are in A.P., we can derive the relationship

{c^2} = {\left( {ac} \right)^{{{\log }_a}b}}

265 Medium

265 of 564

Application +4 | -1

The lengths of the sides of a triangle are

\log _{10} 12

\log _{10} 75

and

\log _{10} n

, where

n \in N

. If a and

b

are the least and greatest values of

n

respectively, the value of

b-a

is divisible by

View answer and explanation

Correct option: C

223

Option C (223) is correct because the triangle inequality gives the range of possible values for

n

, and the difference between the largest and smallest integer values of

n

is divisible by 223.

266 Medium

266 of 564

Analyzing +4 | -1

The sum

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

\frac{1^{3}+2^{3}+3^{3}+\ldots+15^{3}}{1+2+3+\ldots+15}-\frac{1}{2}(1+2+3+\ldots+15

is equal to :

View answer and explanation

Correct option: A

620

Option A (620) is correct because after simplifying the series and subtracting the given term, the result is 620.

267 Medium

267 of 564

Application +4 | -1

Sum of

n

terms of the following series

{1^3} + {3^3} + {5^3} + {7^3} + ...

View answer and explanation

Correct option: A

{n^2}\left( {2{n^2} - 1} \right)

Option A (

{n^2}\left( {2{n^2} - 1} \right)

) is correct because it represents the sum of the given series

{1^3} + {3^3} + {5^3} + {7^3} + ...

after simplification.

268 Medium

268 of 564

Analyzing +4 | -1

a,b

and

c

are in AP, then which one of the following is not true?

View answer and explanation

Correct option: D

{a^2},{b^2}

and

{c^2}

are in AP

Option D (

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

are in AP) is correct because if a, b, and c are in AP, their squares are generally not in AP unless a=c.

269 Medium

269 of 564

Application +4 | -1

\mathrm{n}

- arithmetic means are inserted between 1 and 31 such that the

7^{\text {th }}

mean : the

(n-1)^{\text {th }}

mean

=5: 9

, then '

n

' is equal to

View answer and explanation

Correct option: C

Option C (14) is correct because solving the ratio of the 7th and (n-1)th means gives n=14.

270 Medium

270 of 564

Application +4 | -1

\log _{\sqrt{2}} x+\log _{\sqrt[4]{2}} x+\log _{\sqrt[6]{2}} x+\ldots+\log _{\sqrt[32]{2}} x=136

, then

x=

View answer and explanation

Correct option: B

\sqrt{2}

Option B (

\sqrt{{2}}

) is correct because it is the value of x that satisfies the given logarithmic equation, found by simplifying the equation using logarithm properties.

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