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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Medium

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61 Medium
61 of 564
Analyzing +4 | -1
The sum of the following series 1+6+9(12+22+32)7+12(12+22+32+42)9+15(12+22+...+52)11+...1 + 6 + \frac{{9({1^2} + {2^2} + {3^2})}}{7} + \frac{{12({1^2} + {2^2} + {3^2} + {4^2})}}{9} + \frac{{15({1^2} + {2^2} + ... + {5^2})}}{{11}} + ... up to 15 terms is
View answer and explanation
Correct option: B
7820
Option B (7820) is correct because the sum of the given series up to 15 terms is calculated to be 7820 by finding the general term and using summation formulas.
62 Medium
62 of 564
Analyzing +4 | -1
The expression log⁡25−∑k=14log⁡2(sin⁡(kπ5)){\log _2}5 - \sum\limits_{k = 1}^4 {{{\log }_2}\left( {\sin \left( {\frac{{k\pi }}{5}} \right)} \right)} reduces to pq,\frac{p}{q}, where pp and qq are co-prime, the value of p2+q2p^2+q^2 is
View answer and explanation
Correct option: B
17
Option B (17) is correct because the expression simplifies to 4, and thus p=4p=4 and q=1q=1, giving p2+q2=17p^2+q^2 = 17.
63 Medium
63 of 564
Understanding +4 | -1
Assertion: If a,b,ca, b, c are in A.P., then b+c,c+a,a+bb+c, c+a, a+b are in A.P.

Reason: If a,b,ca, b, c are in A.P., then 10a,10b,10c10^{{a}}, 10^{{b}}, 10^{{c}} are in GP.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Option B is correct because both the assertion and reason are correct mathematical statements, but the reason does not directly explain the assertion.
64 Medium
64 of 564
Understanding +4 | -1
The value of the series x  log⁡ea+x33! (log⁡ea)3+x55!  (log⁡ea)5+...x\,\,{\log _e}a + \frac{{{x^3}}}{{3!}}\,{\left( {{{\log }_e}a} \right)^3} + \frac{{{x^5}}}{{5!}}\,\,{\left( {{{\log }_e}a} \right)^5} + ... Is
View answer and explanation
Correct option: C
sinh⁡  (xlog⁡e  a)\sinh \,\,\left( {x{{\log }_e}\,\,a} \right)
Option C (sinh⁡  (xlog⁡e a)\sinh \,\,\left( {x{{\log }_e}\,a} \right)) is correct because the given series is the Taylor series expansion of sinh(x log a).
65 Medium
65 of 564
Analyzing +4 | -1
α,β\alpha, \beta be the roots of x2−3x+a=0x^2-3 x+a=0 and γ,δ\gamma, \delta be the roots of x2−12x+b=0x^2-12 x+b=0 and numbers α,β,γ,δ\alpha, \beta, \gamma, \delta (in order) form an increasing G.P. then
View answer and explanation
Correct option: C
a=2, b=32\mathrm{a}=2, \mathrm{~b}=32
Option C (a=2, b=32) is correct because the roots form an increasing G.P. with a common ratio of 2, and the values of a and b satisfy the given conditions.
66 Medium
66 of 564
Analyzing +4 | -1
If λ=4+37b−81b4−3b\lambda=\sqrt{4+\sqrt{\frac{37 b-81 b^4-3}{b}}} where b>0b > 0, then sum of all possible integral values of λ\lambda is
View answer and explanation
Correct option: D
5
Option D (5) is correct because the sum of all possible integral values of λ\lambda is 5, derived from the AM-GM inequality and the given condition.
67 Medium
67 of 564
Application +4 | -1
(444…n(444 \ldots n times 4)+(6666…n times 6)2=)+(6666 \ldots n \text { times } 6)^2=
View answer and explanation
Correct option: D
49⋅10n(10n−1)\frac{4}{9} \cdot 10^n\left(10^n-1\right)
Option D is correct. The expression (444…n(444 \ldots n times 4)+(6666…n times 6)2)+(6666 \ldots n \text { times } 6)^2 simplifies to 49⋅10n(10n−1)\frac{4}{9} \cdot 10^n(10^n-1) after converting the repeating digits to fractions and simplifying.
68 Medium
68 of 564
Application +4 | -1
If the first, second and last terms of an AP are a,b,ca, b, c respectively, then the sum is
View answer and explanation
Correct option: C
(a+c)(b+c−2a)2(b−a)\frac{(a+c)(b+c-2 a)}{2(b-a)}
Option C ((a+c)(b+c−2a)2(b−a)\frac{(a+c)(b+c-2a)}{2(b-a)}) is correct because it is the sum of the AP, derived using the given first, second, and last terms.
69 Medium
69 of 564
Application +4 | -1
If roots of the equation x3−12  x2+39  x−28=0{x^3} - 12\;{x^2} + 39\;x - 28 = 0 are in A.P., then their common difference will be
View answer and explanation
Correct option: C
±3\pm 3
Option C (±3\pm 3) is correct because substituting the roots in A.P. into the cubic equation allows us to solve for the common difference.
70 Medium
70 of 564
Application +4 | -1
21/4⋅41/8⋅81/16⋅161/32……2^{1 / 4} \cdot 4^{1 / 8} \cdot 8^{1 / 16} \cdot 16^{1 / 32} \ldots \ldots is equal to
View answer and explanation
Correct option: B
2
Option B is correct because the product simplifies to 2 by expressing each term as a power of 2 and summing the exponents.
71 Medium
71 of 564
Application +4 | -1
If a,b,ca,b,c be in G.P. and a+x,b+x,c+xa + x,b + x,c + x in H.P., then the value of xx is (a,b,ca,b,c are distinct numbers)
View answer and explanation
Correct option: B
bb
Option B (bb) is correct because substituting b2=acb^2 = ac and simplifying the equation derived from the harmonic progression leads to x=bx = b.
72 Medium
72 of 564
Application +4 | -1
If log⁡105=x,{\log _{10}}5 = x, then log⁡51250{\log _5}1250 equals to
View answer and explanation
Correct option: C
3+1x3 + \frac{1}{x}
Option C is correct because by using the properties of logarithms and the given value of log₁₀5, we can express log₅1250 as 3 + 1/x.
73 Medium
73 of 564
Analyzing +4 | -1
Let a,ba, b and cc and be non-negative real numbers satisfying a+b+c=9.a+b+c=9. If the maximum value of the expression a2b3c4a^2 b^3 c^4 can be expressed as 2x3y,2^x3^y, where xx and yy are natural numbers, then the value of log⁡10(xy),{\log _{10}}\left( {{x^y}} \right), is:
View answer and explanation
Correct option: B
3
Option B, 3, is correct because the maximum value of the expression is 210332^{10}3^3, leading to x = 10 and y = 3, and thus log10(103) = 3.
74 Medium
74 of 564
Application +4 | -1
The unit digit of 171983+111983−7198317^{1983}+11^{1983}-7^{1983} is
View answer and explanation
Correct option: A
1
Option A (1) is correct because by analyzing the unit digits of each term separately and then summing them, we find the unit digit of the entire expression to be 1.
75 Medium
75 of 564
Application +4 | -1
The sum to nn terms of the series 22+42+62+...{2^2} + {4^2} + {6^2} + ... is
View answer and explanation
Correct option: B
2n(n+1)(2n+1)3\frac{{2n\left( {n + 1} \right)\left( {2n + 1} \right)}}{3}
Option B (2n(n+1)(2n+1)3\frac{{2n(n + 1)(2n + 1)}}{3}) is correct because the sum of the series 22+42+62+...{2^2} + {4^2} + {6^2} + ... to nn terms is equal to 44 times the sum of the squares of the first nn natural numbers.

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76 Medium
76 of 564
Application +4 | -1
If SrS_{\mathrm{r}} denotes the sum of the first rr terms of an A.P., then S3r−Sr−1 S2r−S2r−1\frac{\mathrm{S}_{3 \mathrm{r}}-\mathrm{S}_{\mathrm{r}-1}}{\mathrm{~S}_{2 \mathrm{r}}-\mathrm{S}_{2 \mathrm{r}-1}} is equal to
View answer and explanation
Correct option: B
2r+12 r+1
Option B (2r+12r+1) is correct because, after simplification and substitution, the expression equals 2r+12r+1.
77 Medium
77 of 564
Analyzing +4 | -1
The sum of the series 1+1+a2  !+1+a+a23  !+1+a+a2+a34  !+…,1 + \frac{{1 + a}}{{2\;!}} + \frac{{1 + a + {a^2}}}{{3\;!}} + \frac{{1 + a + {a^2} + {a^3}}}{{4\;!}} + \ldots , is
View answer and explanation
Correct option: A
ea−ea−1\frac{{{e^a} - e}}{{a - 1}}
Option A (ea−ea−1\frac{{{e^a} - e}}{{a - 1}}) is correct because the sum of the series can be simplified to this expression using the formula for the sum of a geometric series and the exponential series.
78 Medium
78 of 564
Analyzing +4 | -1
Let TrT_r be the rth r^{\text {th }} term of a sequence, for r=1,2,3……r=1,2,3 \ldots \ldots If 3Tr+1=Tr3 T_{r+1}=T_r and T7=1243T_7=\frac{1}{243}, then the value of ∑r=1∞(Tr⋅Tr+1)\sum\limits_{{r=1}}^{\infty}\left(T_r \cdot T_{r+1}\right) is:
View answer and explanation
Correct option: B
278\frac{27}{8}
The correct answer is B (27/8) because the series is a geometric progression, and the sum to infinity is calculated using the formula for an infinite GP.
79 Medium
79 of 564
Application +4 | -1
If the sum of any number of terms in a sequence is always three times the squared number of these terms, then the sequence is
View answer and explanation
Correct option: A
an AP
Option A (an AP) is correct because if the sum of any number of terms in a sequence is three times the squared number of these terms, the sequence is an arithmetic progression.
80 Medium
80 of 564
Application +4 | -1
Let a,ar,ar2,……\mathrm{a}, \mathrm{ar}, \mathrm{ar}^2, \ldots \ldots be an infinite G.P. If ∑n=0∞arn=57\sum\limits_{n=0}^{\infty} a r^n=57 and ∑n=0∞a3r3n=9747\sum\limits_{n=0}^{\infty} a^3 r^{3 n}=9747, then a+18ra+18 r is equal to :
View answer and explanation
Correct option: D
31
Option D (31) is correct because it is the value of a + 18r after solving the given infinite geometric progression equations.
81 Medium
81 of 564
Application +4 | -1
Let a1,a2,…,a10a_1, a_2, \ldots, a_{10} be in AP and h1,h2,…,h10h_1, h_2, \ldots, h_{10} be in HP. If a1=h1=2a_1=h_1=2 and a10=h10=3a_{10}=h_{10}=3, then a4h7a_4 h_7 is
View answer and explanation
Correct option: D
6
Option D (6) is correct because by using the properties of AP and HP, we can find the values of a_4 and h_7 and then calculate their product.
82 Medium
82 of 564
Application +4 | -1
If a1,a2,a3…an\mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3 \ldots \mathrm{a}_n are in AP\mathrm{AP} where ai>0\mathrm{a}_{\mathrm{i}}>0 for all i\mathrm{i} then the value of 1a1+a2+1a2+a3+…+1an−1+an=\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}=
View answer and explanation
Correct option: D
n−1a1+an\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}
Option D (n−1a1+an\frac{{n-1}}{{\sqrt{{a_1}}+\sqrt{{a_n}}}}) is correct because it is the simplified form of the given expression after rationalizing the denominators and using the properties of an arithmetic progression.
83 Medium
83 of 564
Analyzing +4 | -1
Sum of the series 11!50!+13!48!+15!46!+⋯+15!0!\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\cdots+\frac{1}{5 ! 0 !} is:
View answer and explanation
Correct option: D
25051!\frac{2^{50}}{51 !}
Option D (25051!\frac{2^{50}}{51 !}) is correct because the sum of the series can be expressed as a binomial coefficient sum, which simplifies to 25051!\frac{2^{50}}{51 !}.
84 Medium
84 of 564
Application +4 | -1
1⋅1!+2⋅2!+3⋅3!+…+n⋅n!=1 \cdot 1 !+2 \cdot 2 !+3 \cdot 3 !+\ldots+n \cdot n !=
View answer and explanation
Correct option: B
(n+1)!−1(n+1) !-1
Option B ((n+1)!−1(n+1)! - 1) is correct because the series can be simplified by expressing each term as a difference of factorials.
85 Medium
85 of 564
Application +4 | -1
The sum to 10 terms of the series 11+12+14+21+22+24+31+32+34+…\frac{1}{1+1^{2}+1^{4}}+\frac{2}{1+2^{2}+2^{4}}+\frac{3}{1+3^{2}+3^{4}}+\ldots is
View answer and explanation
Correct option: C
55111\frac{55}{111}
The correct answer is C (55111\frac{55}{111}) because the sum to 10 terms of the given series is correctly calculated using telescoping series method.
86 Medium
86 of 564
Application +4 | -1
The first term of a G.P of real numbers given by b1,b2,b3,………b_1, b_2, b_3, \ldots \ldots \ldots. is Unity. Then the minimum value of 4b2+5b34 b_2+5 b_3 is …\ldots.
View answer and explanation
Correct option: B
−4/5-4 / 5
Option B (-4/5) is correct because it represents the minimum value of the quadratic expression 5r2+4r5r^2 + 4r obtained by completing the square or using the vertex formula.
87 Medium
87 of 564
Application +4 | -1
If ss is the sum of an infinite GP and the first term is aa, then the common ratio rr given by
View answer and explanation
Correct option: B
s−as\frac{s-a}{s}
Option B (s−as\frac{s-a}{s}) is correct because it is derived directly from the formula for the sum of an infinite geometric progression.
88 Medium
88 of 564
Analyzing +4 | -1
If Sn=4+11+21+34+50+…S_n=4+11+21+34+50+\ldots to nn terms, then 160(S29−S9)\frac{1}{60}\left(S_{29}-S_9\right) is equal to
View answer and explanation
Correct option: A
223
Option A (223) is correct because the sum of the series Sn is calculated, and then the expression (S29 - S9)/60 is evaluated to obtain the result.
89 Medium
89 of 564
Application +4 | -1
If a,b,ca, b, c are in A.P., then 10ax+10,10bx+10,10cx+1010^{a x+10}, 10^{b x+10}, 10^{c x+10} will be in
View answer and explanation
Correct option: C
G.P. for all values of xx
Option C (G.P. for all values of x) is correct because if a, b, c are in A.P., then 10ax+10,10bx+10,10cx+1010^{ax+10}, 10^{bx+10}, 10^{cx+10} will always be in G.P. regardless of the value of x.
90 Medium
90 of 564
Understanding +4 | -1
In an AP the sum of any two terms, such that the distance of one of them from the beginning is same as that of the other from the end, is
View answer and explanation
Correct option: B
Sum of first and last terms
Option B is correct because in an arithmetic progression, the sum of terms equidistant from the beginning and end is constant and equal to the sum of the first and last terms.
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