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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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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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541 Easy
541 of 998
Application +4 | -1
In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then, the common ratio of this progression equals
View answer and explanation
Correct option: D
12(5−1)\frac{1}{2}\left( {\sqrt 5 - 1} \right)
Option D (12(5−1)\frac{1}{2}\left( {\sqrt 5 - 1} \right)) is correct because it is the positive root of the quadratic equation derived from the given condition of the geometric progression.
542 Easy
542 of 998
Application +4 | -1
The product of x1/2⋅x1/4⋅x1/8x^{1 / 2} \cdot x^{1 / 4} \cdot x^{1 / 8}.. Equals
View answer and explanation
Correct option: C
xx
Option C (xx) is correct because the product of the terms can be simplified using exponent rules and the sum of an infinite geometric series.
543 Easy
543 of 998
Analyzing +4 | -1
If three positive real numbers a,  b,  ca,\;b,\;c are in AP and abc=4abc = 4, then the minimum possible value of bb is
View answer and explanation
Correct option: B
22/3{2^{2/3}}
Option B (22/3{2^{2/3}}) is correct because the minimum value of b is found using the AM-GM inequality, given that a, b, c are in AP and abc = 4.
544 Easy
544 of 998
Application +4 | -1
Consider an infinite geometric series with first term aa and common ratio rr. If its sum is 4 and the second term is 3/43 / 4, then
View answer and explanation
Correct option: D
a=3,r=14a=3, r=\frac{1}{4}
Option D (a=3,r=14a=3, r=\frac{1}{4}) is correct because it satisfies both the sum to infinity formula and the given second term condition for the geometric series.
545 Medium
545 of 998
Application +4 | -1
If a1,a2…,ana_1, a_2 \ldots, a_n are positive real numbers whose product is a fixed number c\mathrm{c}, then the minimum value of a1+a2+……+an−1+2ana_1+a_2+\ldots \ldots+a_{n-1}+2 a_n is
View answer and explanation
Correct option: A
n(2c)1/nn(2 c)^{1 / n}
Option A (n(2c)1/nn(2 c)^{1 / n}) is correct because it applies the AM-GM inequality to find the minimum value of the given expression.
546 Easy
546 of 998
Application +4 | -1
If S is the sum to infinity of a G.P., whose first term is a,a, then the sum of the first nn terms is
View answer and explanation
Correct option: B
S{1−(1−as)n}S\left\{1-\left(1-\frac{a}{s}\right)^n\right\}
Option B (S{1−(1−as)n}S\{1-(1-\frac{a}{s})^n\}) is correct because it correctly expresses the sum of the first n terms of the G.P. in terms of S, a, and n.
547 Easy
547 of 998
Application +4 | -1
For a sequence <an>,a1=2< {a_n} > ,{a_1} = 2 and an+1an=13.\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{1}{3}. Then, ∑r=120ar\sum \limits_{r = 1}^{20} {a_r} is
View answer and explanation
Correct option: B
3(1−1320)3\left( {1 - \frac{1}{{{3^{20}}}}} \right)
Option B (3(1−1320)3\left( {1 - \frac{1}{{{3^{20}}}}} \right)) is correct because it is the sum of the first 20 terms of the geometric progression.
548 Medium
548 of 998
Application +4 | -1
If cos⁡(x−y),cos⁡x\cos (x-y), \cos x and cos⁡(x+y)\cos (x+y) are in HP, then cos⁡xsec⁡(y2)\cos x \sec \left(\frac{y}{2}\right) is equal to
View answer and explanation
Correct option: A
±2\pm \sqrt{2}
Option A (±2\pm \sqrt{{2}}) is correct because by using the Harmonic Progression (HP) condition and trigonometric identities, we can derive that cos⁡xsec⁡(y2)=±2\cos x \sec \left(\frac{{y}}{{2}}\right) = \pm \sqrt{{2}}.
549 Easy
549 of 998
Application +4 | -1
Every term of a G.P. is positive and also every term is the sum of two preceding terms. Then, the common ratio of the G.P. is
View answer and explanation
Correct option: B
5+12\frac{\sqrt{5}+1}{2}
Option B (5+12\frac{{\sqrt{{5}}+1}}{{2}}) is correct because it is the positive root of the quadratic equation derived from the given condition.
550 Medium
550 of 998
Application +4 | -1
sin⁡−1(a−a23+a39+….)+cos⁡−1(1+b+b2+…)=π2\sin ^{-1}\left(a-\frac{a^2}{3}+\frac{a^3}{9}+\ldots.\right)+\cos ^{-1}\left(1+b+b^2+\ldots\right)=\frac{\pi}{2} when
View answer and explanation
Correct option: B
a=1  &  b=−13a=1 \,\, \& \,\, b=-\frac{1}{3}
Option B (a=1&b=−13a=1 \& b=-\frac{1}{3}) is correct because it satisfies the condition that the two series are equal and converge within their respective domains.
551 Medium
551 of 998
Application +4 | -1
If 32x+1.4x−1=36,{3^{2x + 1}}{.4^{x - 1}} = 36, then x=x =
View answer and explanation
Correct option: A
log⁡3648{\log _{36}}48
Option B (log⁡4836{\log _{48}}36) is correct because the equation can be simplified and solved for x using logarithms.
552 Easy
552 of 998
Application +4 | -1
The GM of roots of the equation x2−18x+9=0{x^2} - 18x + 9 = 0 is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the geometric mean (GM) of the roots of the quadratic equation is the square root of the product of the roots, which equals 3.
553 Easy
553 of 998
Application +4 | -1
Let x∈(1,∞)x \in \left( {1,\infty } \right) and nn be a positive integer greater than 1. If fn(x)=n1log⁡2x+1log⁡3x+…+1log⁡nx{f_n}\left( x \right) = \frac{n}{{\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_3}x}} + \ldots + \frac{1}{{{{\log }_n}x}}}}, then (n!)fn(x){\left( {n!} \right)^{{f_n}\left( x \right)}} equals to
View answer and explanation
Correct option: B
xn{x^n}
Option B (xn{x^n}) is correct because it is the simplified form of the expression after applying logarithm properties.
554 Medium
554 of 998
Application +4 | -1
The sum of the series 1+3x+6x2+10x3+...∞1 + 3x + 6{x^2} + 10{x^3} + ...\infty will be
View answer and explanation
Correct option: D
1(1−x)3\frac{1}{{{{\left( {1 - x} \right)}^3}}}
Option D (1(1−x)3\frac{1}{{{{\left( {1 - x} \right)}^3}}}) is correct because it is derived by using the method of difference to simplify the series.
555 Medium
555 of 998
Application +4 | -1
If log⁡10(12x+x−1)=x(log⁡105−1)\log _{10}\left(\frac{1}{2^x+x-1}\right)=x\left(\log _{10} 5-1\right) then xx is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because substituting x=1 into the equation satisfies the given logarithmic equation.
556 Easy
556 of 998
Application +4 | -1
The sum of nn terms of the following series 1+(1+x)+(1+x+x2)+…1 + \left( {1 + x} \right) + \left( {1 + x + {x^2}} \right) + \ldots is
View answer and explanation
Correct option: C
n(1−x)−x(1−xn)(1−x)2\frac{{n\left( {1 - x} \right) - x\left( {1 - {x^n}} \right)}}{{{{\left( {1 - x} \right)}^2}}}
Option C (n(1−x)−x(1−xn)(1−x)2\frac{{n\left( {1 - x} \right) - x\left( {1 - {x^n}} \right)}}{{{{\left( {1 - x} \right)}^2}}}) is correct because it is derived by summing the series and using the formula for the sum of a geometric progression.
557 Easy
557 of 998
Analyzing +4 | -1
If a+b1−ab,b,b+c1−bc\frac{{a + b}}{{1 - ab}},b,\frac{{b + c}}{{1 - bc}} are in A.P., then a,1b,ca,\frac{1}{b},c are in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because the given condition leads to the relationship 2b=1a+1c2b = \frac{1}{a} + \frac{1}{c}, which defines a harmonic progression for a,1b,ca, \frac{1}{b}, c.
558 Easy
558 of 998
Application +4 | -1
If 3,A\sqrt{3}, A and 2\sqrt{2} are in APA P, then 3+22\frac{\sqrt{3}+\sqrt{2}}{2} is greater than or equal to
View answer and explanation
Correct option: B
6\sqrt{6}
Option B (6\sqrt{6}) is correct because, since 3,A,2\sqrt{3}, A, \sqrt{2} are in AP, A is the AM of 3\sqrt{3} and 2\sqrt{2}, and AM >= GM.
559 Medium
559 of 998
Analyzing +4 | -1
The sum of series 11.2.3+13.4.5+15.6.7+…∞\frac{1}{{1.2.3}} + \frac{1}{{3.4.5}} + \frac{1}{{5.6.7}} + \ldots \infty is equal to
View answer and explanation
Correct option: A
log⁡e2−12{\log _e}2 - \frac{1}{2}
Option A (log⁡e2−12{\log _e}2 - \frac{1}{2}) is the correct answer because it is the result of summing the given infinite series using partial fraction decomposition and logarithmic series expansion.
560 Medium
560 of 998
Application +4 | -1
The sum of all two digit numbers which, when divided by 4, yield unity as a remainder is
View answer and explanation
Correct option: C
1210
Option C (1210) is correct because the sum of the arithmetic progression 13, 17, ..., 97 is calculated correctly using the formula for the sum of an AP.
561 Easy
561 of 998
Analyzing +4 | -1
The sum of the series 1+13.14+15.142+17.143+…∞1 + \frac{1}{3}.\frac{1}{4} + \frac{1}{5}.\frac{1}{{{4^2}}} + \frac{1}{7}.\frac{1}{{{4^3}}} + \ldots \infty is
View answer and explanation
Correct option: C
log⁡e3{\log _e}3
Option C (log⁡e3{\log _e}3) is correct because the given series can be related to the logarithmic series expansion after a suitable substitution.
562 Easy
562 of 998
Application +4 | -1
If a,b,ca,b,c are in H.P., then ab+c,bc+a,ca+b\frac{a}{{b + c}},\frac{b}{{c + a}},\frac{c}{{a + b}} will be in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because if a,b,ca, b, c are in H.P., then ab+c,bc+a,ca+b\frac{a}{{b + c}}, \frac{b}{{c + a}}, \frac{c}{{a + b}} are also in H.P.
563 Easy
563 of 998
Application +4 | -1
If the sum to 2n2n terms of the AP 2, 5, 8, 11,… is equal to the sum to n terms of the AP 57, 59, 61, 63,…, then nn is equal to
View answer and explanation
Correct option: B
11
Option B (11) is correct because equating the sum of 2n terms of the first AP to the sum of n terms of the second AP and solving for n yields n=11.
564 Easy
564 of 998
Application +4 | -1
If a,b,ca, b, c are in H.P., then ab+c−a,bc+a−b,ca+b−c\frac{a}{b+c-a}, \frac{b}{c+a-b}, \frac{c}{a+b-c} are in
View answer and explanation
Correct option: C
H.P.
Option C is correct because if a, b, c are in H.P., then the given expression is also in H.P.
565 Easy
565 of 998
Application +4 | -1
If x=1+a+a2+...∞x = 1 + a + {a^2} + ...\infty and y=1+b+b2+...∞y = 1 + b + {b^2} + ...\infty where aa and bb are proper fractions, then 1+ab+a2b2+...∞1 + ab + {a^2}{b^2} + ...\infty equals
View answer and explanation
Correct option: A
xyy+x−1\frac{{xy}}{{y + x - 1}}
Option A (xyy+x−1\frac{{xy}}{{y + x - 1}}) is correct because it is derived by expressing aa and bb in terms of xx and yy, and then substituting them into the infinite geometric series formula.
566 Medium
566 of 998
Application +4 | -1
The sum of all two digit odd numbers is
View answer and explanation
Correct option: A
2475
Option A (2475) is correct because it is the sum of all two-digit odd numbers, which form an arithmetic progression.
567 Easy
567 of 998
Application +4 | -1
The eighth term of a geometric progression is 128 and common ratio is 2. The product of the first five terms is
View answer and explanation
Correct option: B
454^5
Option B (454^5) is correct because, with a=1 and r=2, the product of the first five terms of the geometric progression is 2102^{10}, which equals 454^5.
568 Medium
568 of 998
Understanding +4 | -1
∑i=1n∑j=1i∑k=1j1=\sum\limits_{{i=1}}^n \sum\limits_{{j=1}}^i \sum\limits_{{k=1}}^j 1=
View answer and explanation
Correct option: C
n(n+1)(n+2)6\frac{n(n+1)(n+2)}{6}
Option C (n(n+1)(n+2)6\frac{n(n+1)(n+2)}{6}) is correct because evaluating the triple summation results in the given expression.
569 Medium
569 of 998
Application +4 | -1
If the sum to first nn terms of the AP 2, 4, 6,… is 240, then the value of nn is
View answer and explanation
Correct option: B
15
Option B (15) is correct because the sum of the first n terms of an AP is used to form a quadratic equation, which is then solved for n.
570 Medium
570 of 998
Application +4 | -1
A club consists of members whose ages are in AP, the common difference being 3 months. If the youngest member of the club is just 7 yr old and the sum of the ages of all the members is 250yr250 \mathrm{yr}, then the number of members in the club is
View answer and explanation
Correct option: B
25
Option B (25) is correct because it is the number of members in the club, calculated using the arithmetic progression properties and the given conditions.
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