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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Medium

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211 Medium
211 of 564
Application +4 | -1
If p,q,rp, q, r are in APA P and x,y,zx, y, z are in GP, then xq−r⋅yr−p⋅zp−qx^{q-r} \cdot y^{r-p} \cdot z^{p-q} is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because when p, q, r are in AP and x, y, z are in GP, the expression simplifies to 1.
212 Medium
212 of 564
Application +4 | -1
For a given arithmetic series the sum of first 50 terms is 200 and the sum of the next 50 terms is 2700, the first term of the series is
View answer and explanation
Correct option: B
−20.5-20.5
Option A (1y=x2−x+C\frac{1}{y}=x^2-x+C) is correct because the differential equation is separable, and the integration leads to this solution.
213 Medium
213 of 564
Application +4 | -1
Let a1,a2,a3…ana_1, a_2, a_3 \ldots a_n be in A.P. and SnS_n denotes the sum of first nn terms of this A.P. If S10=390,a10a50=157S_{10}=390, \frac{a_{10}}{a_{50}}=\frac{15}{7}, then S15−S5S_{15} - S_5 is equal to
View answer and explanation
Correct option: A
468
Option A (468) is correct because after finding the values of 'a' and 'd' using the given information about S10 and the ratio a10/a50, the value of S15 - S5 is calculated to be approximately 468.
214 Medium
214 of 564
Application +4 | -1
In a GP, if (m+n)(m+n) th term is pp and (m−n)(m-n) th term is qq, then its mm th term is
View answer and explanation
Correct option: C
pq\sqrt{p q}
Option C (pq\sqrt{p q}) is correct because the mth term of the GP is found by multiplying the (m+n)th and (m-n)th terms and taking the square root.
215 Medium
215 of 564
Analyzing +4 | -1
11⋅2⋅3⋅4+43⋅4⋅5⋅6+95⋅6⋅7⋅8+167⋅8⋅9⋅10+…to  ∞\frac{1}{{1 \cdot 2 \cdot 3 \cdot 4}} + \frac{4}{{3 \cdot 4 \cdot 5 \cdot 6}} + \frac{9}{{5 \cdot 6 \cdot 7 \cdot 8}} + \frac{{16}}{{7 \cdot 8 \cdot 9 \cdot 10}} + \ldots \mathrm{to\;}\infty is equal to
View answer and explanation
Correct option: A
16log⁡e2−124\frac{1}{6}{\log _e}2 - \frac{1}{{24}}
Option A (16log⁡e2−124\frac{1}{6}{\log _e}2 - \frac{1}{{24}}) is correct because the infinite series simplifies to 16log⁡e2−124\frac{1}{6}{\log _e}2 - \frac{1}{{24}}.
216 Medium
216 of 564
Application +4 | -1
Let 3,a,b,c3, a, b, c be in A.P. and 3,a−1,b+1,c+93, a-1, b+1, c+9 be in G.P. Then, the arithmetic mean of a,ba, b and cc is:
View answer and explanation
Correct option: C
11
Option C (11) is correct because it is the arithmetic mean of a, b, and c, calculated using the properties of arithmetic and geometric progressions.
217 Medium
217 of 564
Application +4 | -1
The sum of the series (1+2)+(1+2+22)+(1+2+22+23)+...upto  n\left( {1 + 2} \right) + \left( {1 + 2 + {2^2}} \right) + \left( {1 + 2 + {2^2} + {2^3}} \right) + ...\mathrm{upto}\;n terms is
View answer and explanation
Correct option: A
2n+2−n−4{2^{n + 2}} - n - 4
The correct answer is A (2n+2−n−4{2^{n + 2}} - n - 4) because summing the series of geometric progressions and simplifying leads to this result.
218 Medium
218 of 564
Analyzing +4 | -1
If pth,qth,rthp\mathrm{th},q\mathrm{th},r\mathrm{th} and sth terms of an A.P. are in G.P., then p−q,q−r,r−sp - q,q - r,r - s are in
View answer and explanation
Correct option: B
G.P.
Option B (G.P.) is correct because if the terms of an A.P. are in G.P., then the differences between their positions are also in G.P.
219 Medium
219 of 564
Understanding +4 | -1
The coefficients of xn{x^n} in the expansion of log⁡a(1+x){\log _a}\left( {1 + x} \right) is
View answer and explanation
Correct option: B
(−1)n−1nlog⁡ae\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}{\log _a}e
Option B ((−1)n−1nlog⁡ae\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}{\log _a}e) is correct because it is the coefficient of xnx^n in the expansion of log⁡a(1+x){\log _a}(1 + x).
220 Medium
220 of 564
Application +4 | -1
If x,y,zx,y,z are positive integers, then (x+y)(y+z)(z+x)\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right) is
View answer and explanation
Correct option: B
>8  xyz> 8\;xyz
Option B (>8  xyz> 8\;xyz) is correct because it follows from applying the AM-GM inequality to the pairs (x, y), (y, z), and (z, x).
221 Medium
221 of 564
Analyzing +4 | -1
The sixth term of an A.P. is equal to 2 the value of the common difference of the A.P. which makes the product a1a4a5{a_1}{a_4}{a_5} least is given by
View answer and explanation
Correct option: C
x=2/3x = 2/3
Option C (x=2/3x = 2/3) is correct because it minimizes the product a1a4a5{a_1}{a_4}{a_5} by finding the critical points and checking the second derivative.
222 Medium
222 of 564
Application +4 | -1
The harmonic mean of two numbers is 4 and the arithmetic and geometric means satisfy the relation 2  A+G2=27,2\;A + {G^2} = 27, the numbers are
View answer and explanation
Correct option: A
6, 3
Option A (6, 3) is correct because the harmonic mean is 4, and the arithmetic and geometric means satisfy the given relation, leading to the numbers 6 and 3.
223 Medium
223 of 564
Application +4 | -1
If p,q,rp, q, r are in one geometric progression and a,b,ca, b, c in another geometric progression, then cp,bq,arc p, b q, ar are in
View answer and explanation
Correct option: C
G.P.
The correct answer is G.P. because if p, q, r and a, b, c are in G.P., then cp, bq, ar are also in G.P.
224 Medium
224 of 564
Application +4 | -1
The sum of the series : 2+2(d)2+3(6)2+…^{2}+2(d)^{2}+3(6)^{2}+\ldots upto 10 terms is :
View answer and explanation
Correct option: C
12100
Option C (12100) is correct because the series can be simplified to a multiple of the sum of cubes of the first 10 natural numbers.
225 Medium
225 of 564
Application +4 | -1
Consider the sequence 8A+2B,6A+B,4A,2A−B8 A+2 B, 6 A+B, 4 A, 2 A-B, which term of this sequence will have a coefficient of AA which is twice the coefficient of BB ?
View answer and explanation
Correct option: D
None
Option D (None) is correct because there is no term in the sequence where the coefficient of A is twice the coefficient of B.
226 Medium
226 of 564
Analyzing +4 | -1
The value of S=∑k=0n(−1)kk3+9k2+26k+24⋅nCkS=\sum\limits_{{k=0}}^n \frac{(-1)^k}{k^3+9 k^2+26 k+24} \cdot{ }^n C_k is
View answer and explanation
Correct option: B
12(n+3)(n+4)\frac{1}{2(n+3)(n+4)}
Option B (12(n+3)(n+4)\frac{1}{2(n+3)(n+4)}) is correct because it is the simplified form of the given summation after applying partial fraction decomposition and binomial coefficient identities.
227 Medium
227 of 564
Application +4 | -1
If a,ba, b and cc are in HP, then the straight line xa+yb+1c=0\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0 always passes through a fixed point and that point is
View answer and explanation
Correct option: C
(1,−2)(1,-2)
Option C ((1,-2)) is correct because if a, b, and c are in HP, then 1/a, 1/b, and 1/c are in AP, leading to the condition that the line passes through the fixed point (1, -2).
228 Medium
228 of 564
Application +4 | -1
If α\alpha and β\beta are the roots of the equation x2−x−1=0x^2-x-1= 0 and Sn=2024αn+2024βnS_n=2024 \alpha^n+2024 \beta^n then S3S_3 is equal to
View answer and explanation
Correct option: A
8096
Option A (8096) is correct because by using the properties of roots of the quadratic equation and simplifying the expression for S_3, we arrive at the value 8096.
229 Medium
229 of 564
Analyzing +4 | -1
The consecutive numbers of a three digit number form a GP. If we subtract 792 from this number, then we get a number consisting of the same digits written in the reverse order and if we increase the second digit of the required number by 2 , then resulting number forms an AP. The number is
View answer and explanation
Correct option: C
951
Option C (951) is correct because it satisfies the conditions of being a three-digit number in GP, reversing digits after subtracting 792, and forming an AP when the middle digit is increased by 2.
230 Medium
230 of 564
Analyzing +4 | -1
Let S1=∑j=110j(j−1)10Cj,S2=∑j=110j10Cj{S}_1=\sum\limits_{{j}=1}^{10} {j}({j}-1)^{10} \mathrm{C}_{{j}}, {S}_2=\sum\limits_{{j}=1}^{10} {j}^{10} \mathrm{C}_{{j}} and S3=∑j=110j2 10Cj{S}_3=\sum\limits_{{j}=1}^{10} {j}^{2} \, {}^{10}\mathrm{C}_{{j}}. Consider

Statement -1: S3=55×29S_3=55 \times 2^9

Statement - 2: S1=90×28S_1=90 \times 2^8 and S2=10×28S_2=10 \times 2^8, then
View answer and explanation
Correct option: B
Statement -1 is true, Statement - 2 is false
Statement -1 is true and Statement -2 is false. The values of S1, S2, and S3 are calculated using binomial theorem properties, revealing the discrepancy in Statement-2.
231 Medium
231 of 564
Application +4 | -1
If f(x)f(x) is a function satisfying f(x+y)=f(x)f(y),∀x,y∈Nf(x+y)=f(x) f(y), \forall x, y \in N such that f(1)=3f(1)=3 and ∑x=1nf(x)=120\sum_{x=1}^n f(x)=120. Then, the value of nn is
View answer and explanation
Correct option: A
4
Option A (4) is correct because by using the given functional equation and the sum of the function values, we can determine that f(x) = 3^x, and solving the sum equation gives n = 4.
232 Medium
232 of 564
Analyzing +4 | -1
If Sn=∑r=0n1nCrS_n=\sum_{r=0}^n \frac{1}{{ }^n C_r} and tn=∑r=0vrrCrt_n=\sum_{r=0}^v \frac{r}{{ }^r C_r}, then tnSs\frac{t_n}{S_s} is equal to:
View answer and explanation
Correct option: D
12n\frac{1}{2} n
Option D (12n\frac{1}{2} n) is correct because by using the property nCr=nCn−r{ }^n C_r = { }^n C_{n-r} and manipulating the summation, we can show that tnSn=n2\frac{t_n}{S_n} = \frac{n}{2}.
233 Medium
233 of 564
Application +4 | -1
The sum of first nn terms of 12+34+78+1516+3132+…\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+\ldots is
View answer and explanation
Correct option: A
2−n+n−12^{-n}+n-1
Option A (2−n+n−12^{-n}+n-1) is correct because it is the sum of the first n terms of the given series.
234 Medium
234 of 564
Analyzing +4 | -1
The minimum value of f(x)=(x−2)10+(x−2)11+3(x−2)12+(x−2)15+(x−2)23+(x−2)25(x−2)15f(x)=\frac{(x-2)^{10}+(x-2)^{11}+3(x-2)^{12}+(x-2)^{15}+(x-2)^{23}+(x-2)^{25}}{(x-2)^{15}} ∀ x>2\forall \, x > 2
View answer and explanation
Correct option: D
8
Option D (8) is correct because applying AM >= GM to the terms results in a minimum value of 8.
235 Medium
235 of 564
Application +4 | -1
Compute the following

(i) 51log⁡75+1(−log⁡100.1)3\sqrt[3]{5^{\frac{1}{\log _7 5}}+\frac{1}{\left(-\log _{10} 0.1\right)}}

(ii) log⁡0.75log⁡210.125\log _{0.75} \log _2 \sqrt{\sqrt{\frac{1}{0.125}}}

(iii) (149)1+log⁡72+5−log⁡1/57\left(\frac{1}{49}\right)^{1+\log _7 2}+5^{-\log _{1 / 5} 7}

(iv) 7log⁡35+3log⁡57−5log⁡37−7log⁡537^{\log _3 5}+3^{\log _5 7}-5^{\log _3 7}-7^{\log _5 3}
View answer and explanation
Correct option: A
(i) 2, (ii) 1, (iii) 7+11967+\frac{1}{196}, (iv) 0
Option A ((i) 2, (ii) 1, (iii) 7+11967+\frac{1}{196}, (iv) 0) is correct because it accurately computes the logarithmic and exponential expressions.
236 Medium
236 of 564
Application +4 | -1
If SnS_n denotes the sum of nn terms of an AP, then Sn+3−3Sn+2+3Sn+1−SnS_{n+3}-3 S_{n+2}+3 S_{n+1}-S_n is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the expression simplifies to zero when using the properties of arithmetic progressions and their sums.
237 Medium
237 of 564
Analyzing +4 | -1
The maximum value of x2yx^2 y, subject to constraints x+y+2x2+2xy+3y2=kx+y+\sqrt{2 x^2+2 x y+3 y^2}=k (constant), x,y≥0x, y \geq 0 is
View answer and explanation
Correct option: B
4k3(3+15)3\frac{4 k^3}{(3+\sqrt{15})^3}
Option B is correct because it uses AM-GM inequality to find the maximum value of the expression under the given constraint.
238 Medium
238 of 564
Analyzing +4 | -1
Let the sum of an infinite G.P., whose first term is a and the common ratio is rr, be 5 . Let the sum of its first five terms be 9825\frac{98}{25}. Then the sum of the first 21 terms of an AP, whose first term is 10ar,nth 10 \mathrm{ar}, n^{\text {th }} term is ana_n and the common difference is 10ar210 ar^2, is equal to
View answer and explanation
Correct option: A
21a1121 a_{11}
Option A (21a1121 a_{11}) is correct because after solving for the common ratio and substituting into the sum of the AP, the result simplifies to 21a1121 a_{11}.
239 Medium
239 of 564
Application +4 | -1
The sum of the series 04+0.004+000004+…04+0.004+000004+\ldots is
View answer and explanation
Correct option: C
4099\frac{40}{99}
Option C (4099\frac{{40}}{{99}}) is correct because the sum of the infinite GP is calculated as a/(1-r).
240 Medium
240 of 564
Analyzing +4 | -1
If H1,H2,H3,…,H2n+1H_1, H_2, H_3, \ldots, H_{2 n+1} are in HP\mathrm{HP}, then ∑i=12n(−1)i(Hi+Hi+1Hi−Hi+1)\sum_{i=1}^{2 n}(-1)^i\left(\frac{H_i+H_{i+1}}{H_i-H_{i+1}}\right) is equal to
View answer and explanation
Correct option: C
2n2 n
Option B (−23Sin⁡−1(cos⁡3/2x)+c-\frac{2}{3} \operatorname{Sin}^{-1}(\cos ^{3 / 2} x)+c) is correct because after substitution and integration, the result matches this option.
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