${\log _e}3 - \frac{{{{\log }_e}9}}{{{2^2}}} + \frac{{{{\log }_e}27}}{{{3^2}}} - \frac{{{{\log }_e}81}}{{{4^2}}} + ...$ is

$\frac{{{{\log }_e}5}}{{{{\log }_e}3}}$

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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Easy

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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.

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151 Easy

151 of 998

Analyzing +4 | -1

{\log _e}3 - \frac{{{{\log }_e}9}}{{{2^2}}} + \frac{{{{\log }_e}27}}{{{3^2}}} - \frac{{{{\log }_e}81}}{{{4^2}}} + ...

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the series can be simplified to a known logarithmic series.

152 Easy

152 of 998

Application +4 | -1

Let

{a_1},{a_2},{a_3}, \ldots

be terms of an AP. If

\frac{{{a_1} + {a_2} + \ldots + {a_p}}}{{{a_1} + {a_2} + \ldots + {a_q}}} = \frac{{{p^2}}}{{{q^2}}},\;p \ne q

, then

\frac{{{a_6}}}{{{a_{21}}}}

equals

View answer and explanation

Correct option: C

\frac{{11}}{{41}}

Option C is correct because by using the given ratio of sums and properties of arithmetic progressions, we can find the ratio of the 6th term to the 21st term.

153 Easy

153 of 998

Application +4 | -1

x > 0

, and

\log _2 x+\log _2(\sqrt{x})+

\log _2(\sqrt[4]{x})+\log _2(\sqrt[8]{x})+\log _2(\sqrt[16]{x})+\ldots=4

then

x

equals

View answer and explanation

Correct option: C

Option C (4) is correct because simplifying the logarithmic expression using properties of logarithms and geometric series leads to x=4.

154 Easy

154 of 998

Analyzing +4 | -1

\frac{1}{{{3^2} - 1}} + \frac{1}{{{5^2} - 1}} + \frac{1}{{{7^2} - 1}} + .... + \frac{1}{{{{(201)}^2} - 1}}

is equal to

View answer and explanation

Correct option: B

\frac{{25}}{{101}}

Option B (25/101) is correct because the sum can be expressed as a telescoping series and simplified.

155 Easy

155 of 998

Application +4 | -1

Let

\frac{1}{x_{1}}, \frac{1}{x_{2}}, \frac{1}{x_{3}}, \ldots .,\left(x_{i} \neq 0\right.

for

\left.i=1,2, \ldots, n\right)

be in A.P. such that

x_{1}=4

and

x_{21}=20

. If

n

is the least positive integer for which

x_{n}>50

, then

\sum_{i=1}^{n}\left(\frac{1}{x_{i}}\right)

is equal to.

View answer and explanation

Correct option: C

\frac{13}{4}

Option C (13/4) is correct because it is the sum of the reciprocals of the first n terms of the A.P., where n is the least positive integer for which x_n > 50.

156 Easy

156 of 998

Analyzing +4 | -1

a,b,c

are in G.P. and

{\log _c}a,{\log _b}c,{\log _a}b

are in A.P., then the common difference of the A.P. is

View answer and explanation

Correct option: B

3/2

Option B (3/2) is correct because, given the conditions, the common difference of the arithmetic progression formed by the logarithms is calculated to be 3/2.

157 Easy

157 of 998

Application +4 | -1

a, b, c

are in AP and

(a+2 b-c)(2 b+c-a)(c+a-b)

=\lambda a b c

, then

\lambda

View answer and explanation

Correct option: C

Option C (16) is correct because the minimum value of the expression is 16, obtained by applying AM-GM inequality.

158 Easy

158 of 998

Analyzing +4 | -1

The series expansion of

\log \left\{ {{{\left( {1 + x} \right)}^{1 + x}}{{\left( {1 - x} \right)}^{1 - x}}} \right\},

View answer and explanation

Correct option: A

2\left\{ {\frac{{{x^2}}}{{1 \cdot 2}} + \frac{{{x^4}}}{{3 \cdot 4}} + \frac{{{x^6}}}{{5 \cdot 6}} + \ldots } \right\}

Option A (

2\left\{ {\frac{{{x^2}}}{{1 \cdot 2}} + \frac{{{x^4}}}{{3 \cdot 4}} + \frac{{{x^6}}}{{5 \cdot 6}} + \ldots } \right\}

) is correct because it is the series expansion of

\log \left\{ {\left( {1 + x} \right)^{1 + x} {\left( {1 - x} \right)^{1 - x}}} \right\}

after simplification.

159 Easy

159 of 998

Application +4 | -1

{\log _7}\left\{ {{{\log }_5}\left( {\sqrt {x + 5} + \sqrt x } \right)} \right\} = 0

then

x =

View answer and explanation

Correct option: B

Option B (4) is correct because solving the logarithmic equation leads to x = 4 after simplification and squaring.

160 Easy

160 of 998

Application +4 | -1

2, a, 2^3, b

and

2^5

are in

G P

, then

a b

View answer and explanation

Correct option: B

Option B (64) is correct because in a GP, the product of terms equidistant from the beginning and end is constant, so

ab = 2 \times 2^5 = 64

161 Easy

161 of 998

Application +4 | -1

Fifth term of a G.P. is 2, then the product of its 9 terms is

View answer and explanation

Correct option: B

512

Option B (512) is correct because the product of 9 terms in a GP with the fifth term being 2 is equal to 2 raised to the power of 9, which is 512.

162 Easy

162 of 998

Application +4 | -1

a, b, c

, are in A.P. and

p, p^{\prime}

are respectively A.M. and GM. between

a

and

b

while

q, q^{\prime}

are respectively AM and

\mathrm{GM}

, between

\mathrm{b}

and

\mathrm{c}

, then

View answer and explanation

Correct option: C

p^2-q^2=p^{\prime 2}-q^{\prime 2}

Option A (1) is correct because the sum of the solutions for x divided by pi equals 1, after solving the given equation involving cosine.

163 Easy

163 of 998

Analyzing +4 | -1

Let

\sum\limits_{{r=1}}^n r^6=f(n)

, then

\sum\limits_{{r=1}}^n(2 r-1)^6

is equal to

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because the sum of (2r-1)⁶ from r=1 to n simplifies to f(2n) - 64f(n), which is not represented by any of the given options.

164 Easy

164 of 998

Understanding +4 | -1

The sum of

n

arithmetic means between

a

and

b

View answer and explanation

Correct option: A

\frac{n(a+b)}{2}

Option A (

\frac{n(a+b)}{2}

) is correct because the sum of

n

arithmetic means between

a

and

b

is given by

n

times the average of

a

and

b

, which is

\frac{n(a+b)}{2}

165 Easy

165 of 998

Application +4 | -1

Let

a_1, a_2, \ldots, a_{10}

be in AP and

h_1, h_2, \ldots, h_{10}

be in HP. If

a_1=h_1=2

and

a_{10}=h_{10}=3

, then

a_4 h_7

View answer and explanation

Correct option: D

Option D (6) is correct because it is the product of the 4th term of the AP and the 7th term of the HP, calculated using the given conditions.

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166 Easy

166 of 998

Application +4 | -1

a_1, a_2, a_3, \ldots

are in GP with first term

a

and common ratio

r

, then

\frac{a_1 a_2}{a_1^2-a_2^2}+\frac{a_2 a_3}{a_2^2-a_3^2}+\frac{a_3 a_4}{a_3^2-a_4^2}+\ldots+\frac{a_{n-1} a_n}{a_{n-1}^2-a_n^2}

is equal to

View answer and explanation

Correct option: B

\frac{(n-1) r}{1-r^2}

Option C (4) is correct because the given condition

a, b, c

are in AP leads to the simplification of the expression to

4abc

, thus

\lambda = 4

167 Easy

167 of 998

Application +4 | -1

The first 3 terms of the sequence which is defined by

a_n=\frac{n}{n^2+1}

, are

View answer and explanation

Correct option: B

\frac{1}{2}, \frac{2}{5}, \frac{3}{10}

Option B (

\frac{1}{2}, \frac{2}{5}, \frac{3}{10}

) is correct because these are the first three terms obtained by substituting

n=1, 2, 3

into the given sequence formula.

168 Easy

168 of 998

Analyzing +4 | -1

Let

a_n

be the

n

th term of an AP. If

\sum\limits_{r = 1}^{100} {{a_{2r}}} = \alpha {\text{ and }}\sum\limits_{r = 1}^{100} {{a_{2r - 1}}} = \beta ,

the common difference of the AP is

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct. The common difference of the AP is calculated as

\frac{{\alpha - \beta}}{{100}}

, which is not among the given options.

169 Easy

169 of 998

Application +4 | -1

The value of

\sum_{\mathrm{r}=16}^{30}(\mathrm{r}+2)(\mathrm{r}-3)

is equal to :

View answer and explanation

Correct option: A

7770

Option A (7770) is correct because it is the result of correctly summing the given series.

170 Easy

170 of 998

Understanding +4 | -1

a

and

b

are two different positive real numbers, then which of the following statements is true?

View answer and explanation

Correct option: B

2\sqrt {ab} < a + b

Option B is correct because for two different positive real numbers, the Arithmetic Mean is always greater than the Geometric Mean.

171 Easy

171 of 998

Analyzing +4 | -1

a_n

be the nth term of an A.P and if

a_7=15

, then the value of the common difference that would make

a_2 a_7 a_{12}

greatest, is

View answer and explanation

Correct option: C

Option C (0) is correct because the product

a_2 a_7 a_{12}

is maximized when the common difference is 0, making all terms equal to

a_7 = 15

172 Easy

172 of 998

Analyzing +4 | -1

a_r > 0, r \in N

and

a_1, a_2, a_3, \ldots . a_{2 n}

are in A.P., then

\frac{a_1+a_{2 n}}{\sqrt{a_1}+\sqrt{a_2}}+\frac{a_2+a_{2 n-1}}{\sqrt{a_2}+\sqrt{a_3}}+\frac{a_3+a_{2 n-2}}{\sqrt{a_3}+\sqrt{a_4}}+\ldots .+\frac{a_n+a_{n+1}}{\sqrt{a_n}+\sqrt{a_{n+1}}}

is equal to

View answer and explanation

Correct option: B

\frac{n\left(a_1+a_{2 n}\right)}{\sqrt{a_1}+\sqrt{a_{n+1}}}

Option B is correct because the given expression can be simplified using the properties of arithmetic progressions and then summed to obtain the final result.

173 Easy

173 of 998

Application +4 | -1

The first and second terms of a G.P. are

x^{-4}

and

x^n

respectively. If

x^{52}

is the eight th term of the same progression, then

n

is equal to

View answer and explanation

Correct option: B

Option B (4) is correct because solving the G.P. equation yields n=4.

174 Easy

174 of 998

Application +4 | -1

The value of

\sqrt {\log _{0.5}^24} ,

View answer and explanation

Correct option: A

- 2

Option A (-2) is correct because the expression simplifies to the logarithm of 4 to the base 0.5, which equals -2.

175 Easy

175 of 998

Analyzing +4 | -1

The sum

\frac{3 \times 1^3}{1^2}+\frac{5 \times\left(1^3+2^3\right)}{1^2+2^2}+\frac{7 \times\left(1^3+2^3+3^3\right)}{1^2+2^2+3^2}+\ldots

View answer and explanation

Correct option: A

660

The sum of the series is 660. This is found by determining the general term of the series and then summing the first 10 terms.

176 Easy

176 of 998

Application +4 | -1

The value of 0.234 is

View answer and explanation

Correct option: A

\frac{{232}}{{990}}

Option A (

\frac{{232}}{{990}}

) is correct because the repeating decimal 0.234 (with 34 repeating) can be expressed as the fraction

\frac{{232}}{{990}}

177 Easy

177 of 998

Analyzing +4 | -1

The sum

\sum\limits_{k = 1}^{20} {k\frac{1}{{{2^k}}}}

is equal to

View answer and explanation

Correct option: A

2 - \frac{{11}}{{{2^{19}}}}

Option A (

2 - \frac{{11}}{{{2^{19}}}}

) is correct because the sum of the series is found by manipulating the series and using the formula for the sum of an infinite geometric progression.

178 Easy

178 of 998

Application +4 | -1

If 100 times the 100 th term of an

\mathrm{AP}

with non-zero common difference equals the 50 times its 50th term, then the 150th term of this AP is

View answer and explanation

Correct option: C

zero

Option C (zero) is correct because the 150th term of the AP is zero, derived from the given relationship between the 100th and 50th terms.

179 Easy

179 of 998

Application +4 | -1

Given,

\log 2=0.301

and

\log 3=0.477,

then the number of digits before decimal in

3^{12} \times 2^8

View answer and explanation

Correct option: C

Option C (9) is correct because the number of digits is found by calculating the logarithm and adding 1 to the integer part.

180 Easy

180 of 998

Application +4 | -1

The product of three geometric means between

4

and

\frac{1}{4}

will be

View answer and explanation

Correct option: D

Option D (1) is correct because the product of geometric means between two numbers is equal to the geometric mean of the two numbers raised to the power of the number of means.

Revision notes and key takeaways

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