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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Easy

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871 Easy
871 of 998
Analyzing +4 | -1
Consider an A.P. with first term ' aa ' . Let SnS_n denote the sum of its nn terms. If SkxSx\frac{S_{k x}}{S_x} is independent of xx, then Sn=S_n=
View answer and explanation
Correct option: A
n2an^2 a
Option A (n2an^2 a) is correct because with the condition SkxSx\frac{S_{k x}}{S_x} independent of xx implying d=2ad=2a, the sum SnS_n simplifies to n2an^2 a.
872 Easy
872 of 998
Application +4 | -1
The value of 21/4⋅41/8⋅81/16…∞2^{1 / 4} \cdot 4^{1 / 8} \cdot 8^{1 / 16} \ldots \infty is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the infinite product can be simplified to a power of 2, where the exponent is the sum of an infinite series.
873 Easy
873 of 998
Analyzing +4 | -1
If x=log⁡2a(bcd2),y=log⁡3b(acd3),x=\log _{2 a}\left(\frac{b c d}{2}\right), y=\log _{3 b}\left(\frac{a c d}{3}\right), z=log⁡4c(abd4)z=\log _{4 c}\left(\frac{a b d}{4}\right) and w=log⁡5d(abc5)w=\log _{5 d}\left(\frac{a b c}{5}\right) and 1x+1+1y+1+1z+1+1w+1=log⁡abcdN+1,\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}+\frac{1}{w+1}=\log _{a b c d} N+1, the value of NN is
View answer and explanation
Correct option: C
120
Option C (120) is correct because by simplifying the given logarithmic expressions and summing them, we find that N = 120.
874 Easy
874 of 998
Application +4 | -1
Suppose mm arithmetic means are inserted between 1 and 31. If the ratio of the second mean to the mm th mean is 1:41: 4, then mm is equal to
View answer and explanation
Correct option: B
9
Option B (9) is correct because by using the properties of arithmetic means and the given ratio, we can solve for the number of means inserted.
875 Easy
875 of 998
Analyzing +4 | -1
For what value of b,b, will the roots of the equation cos⁡x=b,−1≤b≤1\cos x = b, - 1 \le b \le 1 when arranged in ascending order of their magnitudes, form an A.P.?
View answer and explanation
Correct option: A
−1- 1
Option A (−1-1) is correct because when cos⁡x=−1\cos x = -1, the solutions for x form an arithmetic progression.
876 Easy
876 of 998
Analyzing +4 | -1
In a GP the sum of three numbers is 14, if 1 is added to first two numbers and subtracted from third number the series becomes AP, then the greatest number is
View answer and explanation
Correct option: A
8
Option A (8) is correct because the numbers 2, 4, and 8 satisfy the given conditions of being in GP initially and then forming an AP after the specified modifications.
877 Easy
877 of 998
Application +4 | -1
If ax=b,  by=c,  cz=a,{a^x} = b,\;{b^y} = c,\;{c^z} = a,, then value of xyzxyz is
View answer and explanation
Correct option: B
1
Option B (1) is correct because by substituting and simplifying the given equations, we find that xyz=1xyz = 1.
878 Easy
878 of 998
Analyzing +4 | -1
The sum of the products of 2n2 n numbers ±1,±2,±3,…,±n\pm 1, \pm 2, \pm 3, \ldots, \pm n taking two at a time is
View answer and explanation
Correct option: C
−∑r=1nr2-\sum\limits_{{r=1}}^n r^2
Option C (−∑r=1nr2-\sum\limits_{{r=1}}^n r^2) is correct because it represents the sum of the products of the given numbers taken two at a time.
879 Easy
879 of 998
Analyzing +4 | -1
The sum of the series

1⋅n+1+2⋅(n−1)+3⋅(n−2)…+n⋅11 \cdot n + 1 + 2 \cdot \left( {n - 1} \right) + 3 \cdot \left( {n - 2} \right) \ldots + n \cdot 1 is
View answer and explanation
Correct option: A
n(n+1)(n+2)6\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}
Option A (n(n+1)(n+2)6\frac{{n(n + 1)(n + 2)}}{6}) is correct because it represents the sum of the series obtained by summing the product of terms in the form r(n−r+1)r(n - r + 1).
880 Easy
880 of 998
Application +4 | -1
In an arithmetic progression, the 24th term is 100. Then, the sum of the first 47 terms of the arithmetic progression is
View answer and explanation
Correct option: D
4700
Option D (4700) is correct because the sum of the first 47 terms is calculated using the given 24th term and the arithmetic progression formula.
881 Easy
881 of 998
Application +4 | -1
The sum of the series S=∑r=1nlog⁡(ar+1br−1)S=\sum\limits_{{r=1}}^n \log \left(\frac{a^{r+1}}{b^{r-1}}\right) is
View answer and explanation
Correct option: B
n2log⁡(an+3bn−1)\frac{n}{2} \log \left(\frac{a^{n+3}}{b^{n-1}}\right)
Option B (n2log⁡(an+3bn−1)\frac{n}{2} \log \left(\frac{a^{n+3}}{b^{n-1}}\right)) is correct because it represents the simplified sum of the given logarithmic series.
882 Easy
882 of 998
Analyzing +4 | -1
Let a,b,c>1  a3,b3a, b, c>1 \,\, a^3, b^3 and c3c^3 be in A.P., and log⁡ab\log _a b, log⁡ca\log _c a and log⁡bc\log _b c be in G.P. If the sum of first 20 terms of an A.P., whose first term is a+4b+c3\frac{a+4 b+c}{3} and the common difference is a−8b+c10\frac{a-8 b+c}{10} is -444 , then abca b c is equal to:
View answer and explanation
Correct option: D
216
Option D (216) is correct because using the given conditions, it can be shown that a=b=c=6, and thus abc = 216.
883 Easy
883 of 998
Application +4 | -1
If the sum of first nn terms of an AP is cn2c{n^2}, then the sum of squares of these nn terms is
View answer and explanation
Correct option: C
n(4n2−1)c23\frac{{n\left( {4{n^2} - 1} \right){c^2}}}{3}
Option C is correct because it represents the sum of the squares of the terms of the AP, derived from the given sum of the first n terms.
884 Easy
884 of 998
Application +4 | -1
The sum of the series 6+66+666+...6 + 66 + 666 + ... upto nn term is
View answer and explanation
Correct option: B
2(10n+1−9n−10)27\frac{{2({{10}^{n + 1}} - 9n - 10)}}{{27}}
Option B is correct because the sum of the series is derived by manipulating the series and using the formula for the sum of a geometric progression.
885 Easy
885 of 998
Analyzing +4 | -1
Number of G.P's having 5,9 and 11 as its three terms is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because no such geometric progression exists with 5, 9, and 11 as its terms.

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886 Easy
886 of 998
Application +4 | -1
7th 7^{\text {th }} term of an A.P. is 40 . Then, the sum of first 13 terms is
View answer and explanation
Correct option: A
520
Option A (520) is correct because the sum of the first 13 terms can be directly calculated using the 7th term, which is the middle term.
887 Easy
887 of 998
Application +4 | -1
(0.16)log⁡2.5[(1/3)+(1/32)+⋯+∞]=(0.16)^{\log _{2.5}\left[(1 / 3)+\left(1 / 3^2\right)+\cdots+\infty\right]}=
View answer and explanation
Correct option: D
4
Option D (4) is correct because after simplifying the geometric series and using logarithm properties, the expression evaluates to 4.
888 Easy
888 of 998
Remembering +4 | -1
Consider the following statements.

I. The nthn^{\text {th}} term of a G.P. with first term 'aa' and common ratio 'rr' is given by an=a⋅rn−1a_n=a \cdot r^{n-1}.

II. Geometric mean of a{a} and b{b} is given by (ab)1/3({ab})^{1 / 3}

Choose the correct option.
View answer and explanation
Correct option: A
Only I is true
Option A (Only I is true) is correct because the formula for the nth term of a G.P. is correctly stated, while the geometric mean formula is incorrect.
889 Easy
889 of 998
Application +4 | -1
If a+c,a+b,b+ca+c, a+b, b+c are in G.P. and a,c,ba, c, b are in H.P. where a,b,c>0a, b, c>0, then the value of a+bc\frac{a+b}{c} is:
View answer and explanation
Correct option: B
2
Option B (2) is correct because by using the properties of Geometric Progression (GP) and Harmonic Progression (HP) and solving the resulting equations, we find that a+bc=2\frac{{a+b}}{{c}} = 2.
890 Easy
890 of 998
Application +4 | -1
Let x=2222… upto ∞x=\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2 \ldots \text { upto } \infty}}}} and y=2+2+2+2+… upto ∞y=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots \text { upto } \infty}}}}, then x−yx-y equals
View answer and explanation
Correct option: A
0
Option A (0) is correct because both x and y simplify to 2, making their difference zero.
891 Easy
891 of 998
Application +4 | -1
If a1,a2,…,a10a_1, a_2, \ldots, a_{10} are in G.P. such that a1a5=25\frac{a_1}{a_5}=25, then a10a8=\frac{a_{10}}{a_8}=
View answer and explanation
Correct option: A
25
Option A (25) is correct because in a G.P., the ratio of terms with a constant difference in their indices is constant.
892 Easy
892 of 998
Application +4 | -1
The minimum value of the quantity (a2+3a+1)(b2+3b+1)(c2+3c+1)abc\frac{\left(a^2+3 a+1\right)\left(b^2+3 b+1\right)\left(c^2+3 c+1\right)}{a b c}, where a,b,c∈R+a, b, c \in R^{+}, is
View answer and explanation
Correct option: B
125
Option B (125) is correct because by applying AM >= GM inequality to each factor, we find the minimum value of the expression.
893 Easy
893 of 998
Application +4 | -1
If sum of an infinite geometric series is 43\frac{4}{3} and its 1st term is 34\frac{3}{4}, then its common ration is
View answer and explanation
Correct option: A
716\frac{7}{{16}}
Option A (7/16) is correct because it is the common ratio calculated using the formula for the sum of an infinite geometric series.
894 Easy
894 of 998
Application +4 | -1
The sum of the first 15 terms of an AP is 600 and the common difference is 5 . The first term is equal to
View answer and explanation
Correct option: D
5
Option D (5) is correct because it is the first term of the arithmetic progression, calculated using the sum of the first 15 terms and the common difference.
895 Easy
895 of 998
Analyzing +4 | -1
If the sum of an infinite G.P. is equal to the maximum value of f(x)=x3+2x−8f(x)=x^3+2 x-8 in the interval [−1,4][-1,4] and the sum of first two terms is 8 . Then, the common ratio of the G.P. is
View answer and explanation
Correct option: C
78\frac{\sqrt{7}}{8}
Option C (78\frac{\sqrt{7}}{8}) is correct because it is the common ratio obtained by solving the equations derived from the sum of the infinite G.P. and the sum of the first two terms.
896 Easy
896 of 998
Application +4 | -1
If a1,a2,a3,...,an,...a_1, a_2, a_3, . . . , a_n, . . . are in A.P. such that a4−a7+a10=m,a_4 - a_7 + a_{10} = m, then sum of the first 13 terms of the A.P. is
View answer and explanation
Correct option: C
13 m13 \,m
Option C (13m13m) is correct because the given condition simplifies to relate to the 7th term, and the sum of the first 13 terms is 13 times the 7th term.
897 Easy
897 of 998
Application +4 | -1
If a,b,ca, b, c are in A.P. then the points (a,x),(b,y)(a, x),(b, y) and (c,z)(c, z) are collinear if
View answer and explanation
Correct option: D
x,y,zx, y, z are in A.P.
Option D (x,y,zx, y, z are in A.P.) is correct because the collinearity condition implies a constant slope, leading to x,y,zx, y, z being in arithmetic progression.
898 Easy
898 of 998
Analyzing +4 | -1
If 1x−1+1y−1+1z−1>0\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{y-1}}+\frac{1}{\sqrt{z-1}}>0 and x,y,zx, y, z are in G.P., then (log⁡x2)−1,(log⁡xz)−1,(log⁡z2)−1\left(\log x^2\right)^{-1},(\log x z)^{-1},\left(\log z^2\right)^{-1} are in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because if x, y, z are in GP, then log(x), log(y), log(z) are in AP, and (log x2)-1, (log xz)-1, (log z2)-1 are in HP.
899 Easy
899 of 998
Application +4 | -1
If in a geometric progression {an},a1=3,an=96\left\{a_n\right\}, a_1=3, a_n=96 and Sn=189S_n=189 then the value of nn is:
View answer and explanation
Correct option: B
6
Option B (6) is correct because the number of terms is found by using the formulas for the nth term and the sum of a geometric progression.
900 Easy
900 of 998
Application +4 | -1
If log⁡10343=2.5353,{\log _{10}}343 = 2.5353, then the least positive integer n such that 7n>105,{7^n} > {10^5}, is
View answer and explanation
Correct option: B
6
Option B (6) is correct because it is the smallest integer value of n that satisfies the inequality 7n>105{7^n} > {10^5} based on the given logarithm value.
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