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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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Showing 661-690 of 998 questions

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661 Easy
661 of 998
Understanding +4 | -1
Non zero equal numbers are in
View answer and explanation
Correct option: C
A.P., G.P. and H.P.
Option C (A.P., G.P. and H.P.) is correct because non-zero equal numbers satisfy the conditions for arithmetic, geometric, and harmonic progressions.
662 Easy
662 of 998
Application +4 | -1
Three positive numbers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then, the common ratio of the GP is
View answer and explanation
Correct option: D
2+32 + \sqrt 3
Option D (2+32 + \sqrt 3) is correct because solving the quadratic equation derived from the AP condition yields this value for the common ratio, considering the GP is increasing.
663 Easy
663 of 998
Understanding +4 | -1
The value of ∑i=1n∑j=1i∑k=1j1\sum\limits_{i=1}^n \sum\limits_{j=1}^i \sum\limits_{k=1}^j 1, is
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the correct result is n(n+1)(n+2)6\frac{{n(n+1)(n+2)}}{{6}}, which doesn't match any of the given summations.
664 Easy
664 of 998
Understanding +4 | -1
The sum of the series 11.2+12.3+13.4+…….+1n(n+1)\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots \ldots .+\frac{1}{n(n+1)}, is
View answer and explanation
Correct option: B
1−1n+11-\frac{1}{n+1}
Option B (1−1n+11-\frac{{1}}{{n+1}}) is correct because the series telescopes, leaving only the first and last terms after cancellation.
665 Easy
665 of 998
Application +4 | -1
Sum of the series ∑r=1nr(r+1)!\sum\limits_{{r=1}}^n \frac{r}{(r+1) !} is
View answer and explanation
Correct option: B
1−1(n+1)!1-\frac{1}{(n+1) !}
Option B (1−1(n+1)!1-\frac{1}{(n+1) !}) is correct because the sum of the series can be simplified using partial fractions.
666 Medium
666 of 998
Application +4 | -1
If log⁡32,log⁡3(2x−5)\log _3 2, \log _3\left(2^x-5\right) and log⁡3(2x−7/2)\log _3\left(2^x-7 / 2\right) are in AP, then xx is equal to
View answer and explanation
Correct option: B
3
Option B (3) is correct because it satisfies the condition for the given logarithmic terms to be in arithmetic progression, after checking for valid logarithmic arguments.
667 Easy
667 of 998
Analyzing +4 | -1
If log⁡2,log⁡(2x−1)\log 2,\log \left( {{2^x} - 1} \right) and log⁡(2x+3)\log \left( {{2^x} + 3} \right) are in A.P., then 2,  2x−1,  2x+32,\;{2^x} - 1,\;{2^x} + 3 are in
View answer and explanation
Correct option: C
G.P.
Option C (G.P.) is correct because if the logarithms are in A.P., then the original terms are in G.P.
668 Medium
668 of 998
Application +4 | -1
If a1=a2=2,an=an−1−1(n>2)a_1=a_2=2, a_n=a_{n-1}-1(n>2), then a5a_5 is
View answer and explanation
Correct option: C
-1
Option C (-1) is correct because by recursively applying the formula an=an−1−1a_n = a_{n-1} - 1 with the given initial values, we find that a5=−1a_5 = -1.
669 Medium
669 of 998
Application +4 | -1
Given that a,ba, b and cc are in AP. The determinant ∣x+1x+2x+ax+2x+3x+bx+3x+4x+c∣\left|\begin{array}{lll}x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c\end{array}\right| in its simplest form is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because the determinant simplifies to 0 when the given condition that a, b, and c are in AP is applied after performing column and row operations.
670 Easy
670 of 998
Application +4 | -1
If (1−y)(1+2x+4x2+8x3+16x4+32x5)=1−y6,(y≠1)(1-y)\left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5\right)=1-y^6,(y \neq 1), then a value y/xy / x is
View answer and explanation
Correct option: B
2
Option A (2) is correct because, from the given equation, a possible value for y/x is 2.
671 Easy
671 of 998
Application +4 | -1
The value of log⁡ba×log⁡cb×log⁡ac,{\log _b}a \times {\log _c}b \times {\log _a}c, is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the expression log⁡ba×log⁡cb×log⁡ac{\log _b}a \times {\log _c}b \times {\log _a}c simplifies to 1 using the change of base formula for logarithms.
672 Medium
672 of 998
Application +4 | -1
The sum of nn terms of the series 41⋅2⋅3+52⋅3⋅4+63⋅4⋅5+…\frac{4}{1 \cdot 2 \cdot 3}+\frac{5}{2 \cdot 3 \cdot 4}+\frac{6}{3 \cdot 4 \cdot 5}+\ldots is
View answer and explanation
Correct option: C
54−2n+52(n+1)(n+2)\frac{5}{4}-\frac{2 n+5}{2(n+1)(n+2)}
Option D (54−2n+52(n+1)(n+2)\frac{5}{4}-\frac{2 n+5}{2(n+1)(n+2)}) is correct because the sum of n terms of the series is calculated to be in this form by partial fraction decomposition.
673 Medium
673 of 998
Application +4 | -1
Three arithmetic means between 3 and 19 are
View answer and explanation
Correct option: A
7,11,157,11,15
Option A (7,11,157, 11, 15) is correct because these are the three arithmetic means between 3 and 19 with a common difference of 4.
674 Easy
674 of 998
Application +4 | -1
Let a1, a2, .... , a10{a_1},\,{a_2},\,....\,,\,{a_{10}} be a GP. If a3a1=25,\frac{{{a_3}}}{{{a_1}}} = 25, then a9a5\frac{{{a_9}}}{{{a_5}}} equals
View answer and explanation
Correct option: D
54{5^4}
Option D (54{5^4}) is correct because the ratio a9a5\frac{{{a_9}}}{{{a_5}}} simplifies to r4{r^4}, and given a3a1=25\frac{{{a_3}}}{{{a_1}}} = 25, we find r4=54{r^4} = {5^4}.
675 Easy
675 of 998
Application +4 | -1
If a,b,ca, b, c are in H.P., then
View answer and explanation
Correct option: B
an+cn>2bna^n+c^n>2 b^n
Option B (an+cn>2bna^n+c^n>2 b^n) is correct because it is derived from the relationship between arithmetic and harmonic means, where A.M. > H.M., and using the inequality an+cn2>(a+c2)n\frac{a^n+c^n}{2}>\left(\frac{a+c}{2}\right)^n.

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676 Medium
676 of 998
Application +4 | -1
Let S(n)S(n) denotes the sum of first nn terms of an AP. Then, the value of S(3n)S(2n)−S(n)\frac{S(3 n)}{S(2 n)-S(n)} is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the value of the expression simplifies to 3 based on the properties of arithmetic progressions.
677 Easy
677 of 998
Application +4 | -1
The first, second and middle terms of an AP are a,b,ca, b, c respectively. Their sum is
View answer and explanation
Correct option: B
2c(c−a)b−a+c\frac{2 c(c-a)}{b-a}+c
Option D (2c(c−a)b−a+c\frac{{2 c(c-a)}}{{b-a}}+c) is correct because it represents the sum of an A.P. with an odd number of terms, where the middle term is known.
678 Medium
678 of 998
Application +4 | -1
If a1,  a2,  a3,…,  a4001{a_1},\;{a_2},\;{a_3}, \ldots ,\;{a_{4001}} are terms of an AP such that 1a1a2+1a2a3+...+1a4000a4001=10  \frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + ... + \frac{1}{{{a_{4000}}{a_{4001}}}} = 10\; and a2+a4000=50,{a_2} + {a_{4000}} = 50,, then ∣a1−a4001∣\left| {{a_1} - {a_{4001}}} \right| is equal to
View answer and explanation
Correct option: B
30
Option B (30) is correct because the given conditions are used to find a relationship between the first and last terms of the AP, and then the difference between them is calculated.
679 Easy
679 of 998
Understanding +4 | -1
The sum of 1+3+5+7+⋯1+3+5+7+\cdots up to nn terms is
View answer and explanation
Correct option: C
n2n^{2}
Option C (n2n^{2}) is correct because the sum of the first n odd numbers is n2n^{2}.
680 Easy
680 of 998
Understanding +4 | -1
If a,ba, b and cc are in APA P, then which of the following will also be in AP?
View answer and explanation
Correct option: A
3+a,3+b,3+c\sqrt{3}+a, \sqrt{3}+b, \sqrt{3}+c
Option A (3+a,3+b,3+c\sqrt{3}+a, \sqrt{3}+b, \sqrt{3}+c) is correct because adding a constant to each term of an arithmetic progression results in another arithmetic progression.
681 Easy
681 of 998
Application +4 | -1
The sum of first 9 terms of the series 131+13+231+3+13+23+331+3+5+…\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\ldots, is
View answer and explanation
Correct option: D
96
Option D (96) is correct because the sum of the first 9 terms of the series is calculated by simplifying the general term and then summing the series.
682 Easy
682 of 998
Application +4 | -1
In the sequence 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8,…1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, \ldots, where nn consecutive terms have the value nn, the 1025th term is
View answer and explanation
Correct option: B
2102^{10}
Option B (2102^{10}) is correct because the 1025th term falls within the block of terms where the value is 2102^{10} in the given sequence.
683 Medium
683 of 998
Application +4 | -1
In the expansion of e7x+e3xe5x\frac{{{e^{7x}} + {e^{3x}}}}{{{e^{5x}}}} , the constant term is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the constant term in the expansion is obtained by simplifying the expression to e2x+e−2xe^{2x} + e^{-2x} and then observing that the constant term is 1 + 1 = 2.
684 Medium
684 of 998
Understanding +4 | -1
The coefficient of xn{x^n} in the expansion of (1+x22!+x44!+…)2,{\left( {1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} + \ldots } \right)^2},, when nn is odd, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the expression simplifies to cosh2(x), which only contains even powers of x. Therefore, the coefficient of x^n when n is odd is 0.
685 Easy
685 of 998
Application +4 | -1
The sum to nn terms of the infinite series 1.32+2.52+3.72+...∞{1.3^2} + {2.5^2} + {3.7^2} + ...\infty
View answer and explanation
Correct option: A
n6(n+1)(6n2+14n+7)\frac{n}{6}\left( {n + 1} \right)\left( {6{n^2} + 14n + 7} \right)
Option A is correct because the sum to n terms of the given series is derived by expressing the general term, summing it, and simplifying the expression.
686 Easy
686 of 998
Application +4 | -1
99th term of the series 2+7+14+23+34…2 + 7 + 14 + 23 + 34 \ldots is
View answer and explanation
Correct option: A
9998
Option A (9998) is correct because the 99th term of the given series is calculated by finding the general term and substituting n=99.
687 Easy
687 of 998
Analyzing +4 | -1
If log⁡x{log⁡4(log⁡x(5x2+4x3))}=0,{\log _x}\left\{ {{{\log }_4}\left( {{{\log }_x}\left( {5{x^2} + 4{x^3}} \right)} \right)} \right\} = 0, then
View answer and explanation
Correct option: D
5
Option D (5) is correct because solving the logarithmic equation step-by-step leads to the solution x = 5.
688 Easy
688 of 998
Application +4 | -1
If a,b,ca, b, c be respectively the pthp^{t h}, qth q^{\text {th }} and rthr^{th} terms of a G.P., then

Δ=∣log⁡alog⁡blog⁡cpqr111∣ equals \Delta=\left|\begin{array}{ccc} \log a & \log b & \log c \\ p & q & r \\ 1 & 1 & 1 \end{array}\right| \text { equals }
View answer and explanation
Correct option: B
0
Option B (0) is correct because the determinant simplifies to zero due to the logarithmic properties of terms in a G.P.
689 Medium
689 of 998
Application +4 | -1
The arithmetic mean of 7 consecutive integers starting with aa is mm. Then, the arithmetic mean of 11 consecutive integers starting with a+2a + 2 is
View answer and explanation
Correct option: D
m+4m + 4
Option D (m+4m + 4) is correct because the arithmetic mean of 11 consecutive integers starting with a+2a + 2 is a+7a + 7, and since a+3=ma + 3 = m, the result is m+4m + 4.
690 Easy
690 of 998
Application +4 | -1
If AM of two numbers is twice of their GM, then the ratio of greatest number to smallest number is
View answer and explanation
Correct option: B
7+437 + 4\sqrt 3
Option B (7 + 4\sqrt 3) is correct because it is the ratio of the greatest to the smallest number when the AM is twice the GM.
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