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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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Showing 1-30 of 998 questions
1 Easy
1 of 998
Analyzing +4 | -1
If a,b,ca,b,c are in A.P.; a,x,ba,x,b are in G.P. and b,y,cb,y,c are in G.P., then x2,b2,y2{x^2},{b^2},{y^2} are in
View answer and explanation
Correct option: C
A.P.
Option C (A.P.) is correct because, given the conditions, it can be shown that x2,b2,y2{x^2},{b^2},{y^2} satisfy the arithmetic progression relationship.
2 Easy
2 of 998
Application +4 | -1
Suppose P(x)=1−x+x2−x3+…+x2012P(x)=1-x+x^2-x^3+\ldots+x^{2012} is expressed as a polynomial in yy, as Q(y)=a0+a1y+…+a2012y2012Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012} where y=x−2y=x-2, then ∑i=02012ai\sum\limits_{{i=0}}^{2012} a_i equals
View answer and explanation
Correct option: A
14(32013+1)\frac{1}{4}\left(3^{2013}+1\right)
Option A is correct because the sum of the coefficients of the polynomial Q(y) is equal to P(3), which can be calculated using the formula for P(x).
3 Easy
3 of 998
Application +4 | -1
If x2log⁡10x=1000x,{x^{2{{\log }_{10}}x}} = 1000x,, then xx equals to
View answer and explanation
Correct option: B
10−1,1010{10^{ - 1}},10\sqrt {10}
Option B (10−1,1010{10^{ - 1}},10\sqrt {10}) is correct because these are the two values of x that satisfy the given equation x2log⁡10x=1000x{x^{2{{\log }_{10}}x}} = 1000x.
4 Easy
4 of 998
Application +4 | -1
If p,q,rp,q,r are in A.P., then pth,qthp\mathrm{th},q\mathrm{th} and rthr\mathrm{th} terms of any G.P. are in
View answer and explanation
Correct option: B
G.P.
Option B (G.P.) is correct because if p, q, r are in A.P., then the pth, qth, and rth terms of any G.P. will also be in G.P.
5 Easy
5 of 998
Application +4 | -1
The ratio of sum of first three terms of a G.P.to the sum of first six terms is 64:9164: 91, the common ratio of G.P. is
View answer and explanation
Correct option: B
34\frac{3}{4}
Option B (34\frac{{3}}{{4}}) is correct because it is the common ratio obtained by setting up the ratio of the sums of the first three and six terms of the G.P. and solving for r.
6 Easy
6 of 998
Application +4 | -1
If log⁡(2a−3b)=log⁡a−log⁡b,\log \left( {2a - 3b} \right) = \log a - \log b, then a=a =
View answer and explanation
Correct option: A
3b22b−1\frac{{3{b^2}}}{{2b - 1}}
Option A is correct because by applying the properties of logarithms and simplifying the equation, we can isolate 'a' and express it in terms of 'b'.
7 Easy
7 of 998
Application +4 | -1
Sum of nn terms of the series 12+34+78+1516+...\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{{15}}{{16}} + ... is
View answer and explanation
Correct option: D
2−n+n−1{2^{ - n}} + n - 1
Option D (2−n+n−1{2^{ - n}} + n - 1) is correct because it is derived by summing the given series after rewriting each term.
8 Easy
8 of 998
Application +4 | -1
If a,  b,  ca,\;b,\;c are in AP then 10ax+10,  10bx+10,  10cx+10(x≠0){10^{ax + 10}},\;{10^{bx + 10}},\;{10^{cx + 10}}\left( {x \ne 0} \right) are in
View answer and explanation
Correct option: C
GP for all xx
Option C (GP for all x) is correct because the given terms form a GP regardless of the value of x, as shown by the relationship derived from the AP condition.
9 Easy
9 of 998
Application +4 | -1
The 10th common term between the series 3+7+11+…3+7+11+\ldots and 1+6+11+….1+6+11+\ldots ., is
View answer and explanation
Correct option: A
191
Option A (191) is correct because it finds the 10th common term of the two given arithmetic series.
10 Easy
10 of 998
Analyzing +4 | -1
If x>1,  y>1,  z>1x > 1,\;y > 1,\;z > 1 are in GP, then 11+log⁡x,11+log⁡y,11+log⁡z\frac{1}{{1 + \log x}},\frac{1}{{1 + \log y}},\frac{1}{{1 + \log z}} are in
View answer and explanation
Correct option: B
HP
Option B (HP) is correct because if x, y, z are in GP, then 11+log⁡x,11+log⁡y,11+log⁡z\frac{1}{{1 + \log x}},\frac{1}{{1 + \log y}},\frac{1}{{1 + \log z}} are in Harmonic Progression.
11 Medium
11 of 998
Application +4 | -1
Let S1,S2,…S_1, S_2, \ldots be squares such that for each n≥1n \geq 1, the length of a side of SnS_n equals the length of the diagonal of Sn+1S_{n+1}. If the length of a side of S1S_1 is 10 cm10 \mathrm{~cm} and the area of SnS_n less than 1sqcm1 \mathrm{sq} \mathrm{cm}, then the least value of nn is
View answer and explanation
Correct option: B
8
Option B (8) is correct because the area of the nth square must be less than 1, which leads to the inequality 2n−1>1002^{n-1} > 100, and the smallest integer n that satisfies this is 8.
12 Easy
12 of 998
Application +4 | -1
If the angles of a quadrilateral are in AP, whose common difference is 10∘10^\circ, then the angles of the quadrilateral are
View answer and explanation
Correct option: B
75∘,85∘,95∘,105∘75^\circ ,85^\circ ,95^\circ ,105^\circ
Option B (75∘,85∘,95∘,105∘75^\circ ,85^\circ ,95^\circ ,105^\circ) is correct because these angles are in AP with a common difference of 10∘10^\circ and sum up to 360∘360^\circ.
13 Easy
13 of 998
Application +4 | -1
If the sum of first nn natural numbers is 1/5 times the sum of their squares, then the value of nn is
View answer and explanation
Correct option: C
7
Option C (7) is correct because solving the equation derived from the given relationship between the sum of first n natural numbers and the sum of their squares yields n=7.
14 Easy
14 of 998
Application +4 | -1
If three positive real numbers a,b,ca, b, c are in APA P such that abc=4a b c=4, then the minimum value of bb is
View answer and explanation
Correct option: B
22/32^{2 / 3}
Option B (22/32^{2 / 3}) is correct because it is the minimum value of b, obtained by using the arithmetic progression property and minimizing the expression abc=4.
15 Easy
15 of 998
Application +4 | -1
If the sides of a right-angled triangle are in AP, then the sines of the acute angles are
View answer and explanation
Correct option: A
3/5,4/53 / 5,4 / 5
Option A (3/5,4/53 / 5,4 / 5) is correct because by assuming the sides are in A.P. and using the Pythagorean theorem, the sines of the acute angles can be determined as 3/5 and 4/5.

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16 Easy
16 of 998
Application +4 | -1
In an arithmetic progression of 16 distinct terms with a1=16a_1=16, the sum is equal to square of the last term. The common difference of the A.P. is
View answer and explanation
Correct option: C
−8/5-8 / 5
Option C (-8/5) is correct because, given the conditions, the common difference of the A.P. is -8/5.
17 Easy
17 of 998
Application +4 | -1
If AM and GM of xx and yy are in the ratio p:qp:q, then x:yx:y is
View answer and explanation
Correct option: B
p+p2−q2:p−p2−q2p + \sqrt {{p^2} - {q^2}} :p - \sqrt {{p^2} - {q^2}}
Option B (p+p2−q2:p−p2−q2p + \sqrt {{p^2} - {q^2}} :p - \sqrt {{p^2} - {q^2}}) is correct because it's derived from the given ratio of AM and GM using componendo and dividendo.
18 Medium
18 of 998
Application +4 | -1
If x,y,zx, y, z are three real numbers of the same sign then the value of xy+yz+zx\frac{x}{y}+\frac{y}{z}+\frac{z}{x} lies in the interval
View answer and explanation
Correct option: B
[3,∞)[3, \infty)
Option B ([3,∞)[3, \infty)) is correct because, by applying the AM-GM inequality, the value of the expression is always greater than or equal to 3.
19 Easy
19 of 998
Application +4 | -1
Sum to 25 terms of the series 0.5+0.55+0.555+…0.5 + 0.55+ 0.555 + \ldots is:
View answer and explanation
Correct option: A
581(224−10−25)\frac{5}{{81}}\left( {224 - {{10}^{ - 25}}} \right)
Option A (581(224−10−25)\frac{5}{{81}}\left( {224 - {{10}^{ - 25}}} \right)) is correct because by converting the decimal series into a geometric progression, we can find the sum to 25 terms.
20 Medium
20 of 998
Application +4 | -1
Let S1,S2,S3S_1, S_2, S_3 be the sums of nn terms of three series in A.P., the first term of each being 1 and the common differences 1,2,31,2,3, resp. If S1+S3=λS2S_1+S_3=\lambda S_2 then the value of λ\lambda is:
View answer and explanation
Correct option: B
2
Option B (2) is correct because substituting the given series into the equation S1+S3=λS2S_1 + S_3 = \lambda S_2 and solving for λ yields 2.
21 Medium
21 of 998
Application +4 | -1
If (2n+r)r,n∈N,r∈N(2 n+r) r, n \in N, r \in N is expressed the sum of kk consecutive odd natural numbers, then kk is equal to
View answer and explanation
Correct option: A
rr
Option A (rr) is correct because (2n+r)r(2n+r)r can be expressed as the difference of two squares, representing the sum of rr consecutive odd natural numbers.
22 Easy
22 of 998
Application +4 | -1
If the sum of the series 1+3x+9x2+27x3+…to  ∞1 + \frac{3}{x} + \frac{9}{{{x^2}}} + \frac{{27}}{{{x^3}}} + \ldots \mathrm{to\;}\infty is a finite number, then
View answer and explanation
Correct option: D
x>3x > 3
Option D (x>3x > 3) is correct because the series converges only when the absolute value of the common ratio is less than 1, which implies ∣3x∣<1|\frac{3}{x}| < 1.
23 Medium
23 of 998
Application +4 | -1
If x,∣x+1∣,∣x−1∣x,\left| {x + 1} \right|,\left| {x - 1} \right| are first three terms of an A.P., then the sum of its first 20 terms is
View answer and explanation
Correct option: B
350,180350,180
Option B (350, 180) is correct because it provides the two possible sums of the first 20 terms of the A.P., calculated based on the two possible values of x.
24 Easy
24 of 998
Application +4 | -1
Value of y=(0.36)log⁡0.25(13+132+133+… upto ∞)y=(0.36)^{\log _{0.25}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \text { upto } \infty\right)} is
View answer and explanation
Correct option: C
0.6
Option C (0.6) is correct because after simplifying the geometric series and applying logarithmic properties, the value of y is found to be 0.6.
25 Easy
25 of 998
Application +4 | -1
The sum of first 20 terms of the sequence 0.7,0.77,0.777,…0.7,0.77,0.777, \ldots, is
View answer and explanation
Correct option: B
781(179+10−20)\frac{7}{81}\left(179+10^{-20}\right)
Option D (781(179+10−20)\frac{7}{81}(179+10^{-20})) is correct because it represents the sum of the first 20 terms of the given sequence, calculated by converting the terms into a form suitable for geometric progression.
26 Medium
26 of 998
Application +4 | -1
Sum of the series (1+x)+(1+x+x2)+(1+x+x2+x3)+…n(1+x)+\left(1+x+x^2\right)+\left(1+x+x^2+x^3\right)+\ldots n terms is
View answer and explanation
Correct option: A
11−x[n−x2(1−xn)1−x]\frac{1}{1-x}\left[n-\frac{x^2\left(1-x^n\right)}{1-x}\right]
Option C (11−x[n−x(1−xn)1−x]\frac{1}{1-x}\left[n-\frac{x(1-x^n)}{1-x}\right]) is correct because the sum of the series can be found by expressing each term as a geometric series and then summing the resulting series.
27 Easy
27 of 998
Application +4 | -1
If x=1+2+42!+83!+164!+…x = 1 + 2 + \frac{4}{{2!}} + \frac{8}{{3!}} + \frac{{16}}{{4!}} + \ldots, then x−1{x^{ - 1}} is equal to
View answer and explanation
Correct option: A
e−2{e^{ - 2}}
Option A (e−2{e^{ - 2}}) is correct because the given series is the expansion of e2, and the inverse is therefore e-2.
28 Easy
28 of 998
Application +4 | -1
If the mmth term of HP be nn and nnth term be mm, then the rrth term will be
View answer and explanation
Correct option: C
mnr\frac{{mn}}{r}
Option C (mnr\frac{{mn}}{r}\) is correct because it represents the rth term of the harmonic progression given the mth and nth terms.
29 Easy
29 of 998
Application +4 | -1
If a1,a2,a3,……a_1, a_2, a_3, \ldots \ldots is an A.P such that a1+a5+a10+a15+a20+a24=225, a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225 \text {, } then a1+a2+a3+……+a23+a24a_1+a_2+a_3+\ldots \ldots+a_{23}+a_{24} is equal to
View answer and explanation
Correct option: D
900
Option D (900) is correct because the sum of the A.P. is calculated by using the given condition and the properties of an A.P.
30 Medium
30 of 998
Understanding +4 | -1
If a1,  a2,  a3...,  a20{a_1},\;{a_2},\;{a_3}...,\;{a_{20}} are AM’s between 13 and 67, then the maximum value of a1,  a2,  a3...,  a20{a_1},\;{a_2},\;{a_3}...,\;{a_{20}} is equal to
View answer and explanation
Correct option: B
(40)20{\left( {40} \right)^{20}}
Option B ((40)20{\left( {40} \right)^{20}}) is correct because the maximum value of the product occurs when all the arithmetic means are equal to their average.
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