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JEE Advanced · Mathematics · Sequences and Series

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1
Let R\mathbb{R} denote the set of all real numbers. Let f:R→Rf:\mathbb{R}\to\mathbb{R} be a function such that f(x)>0f(x)>0 for all x∈Rx\in\mathbb{R}, and f(x+y)=f(x)f(y)f(x+y)=f(x)f(y) for all x,y∈Rx,y\in\mathbb{R}. Let the real numbers a1,a2,…,a50a_1,a_2,\dots,a_{50} be in an arithmetic progression. If f(a31)=64f(a25)f(a_{31})=64f(a_{25}), and ∑i=150f(ai)=3(225+1)\displaystyle\sum_{i=1}^{50} f(a_i) = 3(2^{25}+1), then the value of ∑i=630f(ai)\displaystyle\sum_{i=6}^{30} f(a_i) is _________________.
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Your answer: —
Correct answer: 96
2
Let 75⋯5⏞r77\overbrace{5\cdots5}^{r}7 denote the (r+2)(r+2) digit number where the first and the last digits are 77 and the remaining rr digits are 55. Consider the sum S=77+757+7557+⋯+75⋯5⏞987S=77+757+7557+\cdots+7\overbrace{5\cdots5}^{98}7. If S=75⋯5⏞997+mnS=\dfrac{7\overbrace{5\cdots5}^{99}7+m}{n}, where mm and nn are natural numbers less than 30003000, then the value of m+nm+n is
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Your answer: —
Correct answer: 1219
3
Let α=∑k=1∞sin⁡2k(π6)\alpha=\displaystyle\sum_{k=1}^{\infty}\sin^{2k}\left(\dfrac{\pi}{6}\right). Let g:[0,1]→Rg:[0,1]\to\mathbb{R} be the function defined by g(x)=2αx+2α(1−x)g(x)=2^{\alpha x}+2^{\alpha(1-x)}. Then, which of the following statements is/are TRUE?
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A
The minimum value of g(x)g(x) is 27/62^{7/6}
B
The maximum value of g(x)g(x) is 1+21/31+2^{1/3}
C
The function g(x)g(x) attains its maximum at more than one point
D
The function g(x)g(x) attains its minimum at more than one point
4
Let a1,a2,a3,…a_1,a_2,a_3,\ldots be an arithmetic progression with a1=7a_1=7 and common difference 88. Let T1,T2,T3,…T_1,T_2,T_3,\ldots be such that T1=3T_1=3 and Tn+1−Tn=anT_{n+1}-T_n=a_n for n≥1n\ge1. Then, which of the following is/are TRUE?
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A
T20=1604T_{20}=1604
B
∑k=120Tk=10510\displaystyle\sum_{k=1}^{20}T_k=10510
C
T30=3454T_{30}=3454
D
∑k=130Tk=35610\displaystyle\sum_{k=1}^{30}T_k=35610
5
Let l1,l2,…,l100l_1,l_2,\ldots,l_{100} be consecutive terms of an arithmetic progression with common difference d1d_1, and let w1,w2,…,w100w_1,w_2,\ldots,w_{100} be consecutive terms of another arithmetic progression with common difference d2d_2, where d1d2=10d_1 d_2=10. For each i=1,2,…,100i=1,2,\ldots,100, let RiR_i be a rectangle with length lil_i, width wiw_i and area AiA_i. If A51−A50=1000A_{51}-A_{50}=1000, then the value of A100−A90A_{100}-A_{90} is __________.
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Your answer: —
Correct answer: 18900
6
Let mm be the minimum possible value of log⁡3(3y1+3y2+3y3)\log_3(3^{y_1}+3^{y_2}+3^{y_3}), where y1,y2,y3y_1, y_2, y_3 are real numbers for which y1+y2+y3=9y_1+y_2+y_3=9. Let MM be the maximum possible value of (log⁡3x1+log⁡3x2+log⁡3x3)(\log_3 x_1 + \log_3 x_2 + \log_3 x_3), where x1,x2,x3x_1, x_2, x_3 are positive real numbers for which x1+x2+x3=9x_1+x_2+x_3=9. Then the value of log⁡2(m3)+log⁡3(M2)\log_2(m^3) + \log_3(M^2) is
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Your answer: —
Correct answer: 8
7
Let a1,a2,a3,…a_1, a_2, a_3, \ldots be a sequence of positive integers in arithmetic progression with common difference 22. Also, let b1,b2,b3,…b_1, b_2, b_3, \ldots be a sequence of positive integers in geometric progression with common ratio 22. If a1=b1=ca_1 = b_1 = c, then the number of all possible values of cc, for which the equality 2(a1+a2+⋯+an)=b1+b2+⋯+bn2(a_1+a_2+\cdots+a_n) = b_1+b_2+\cdots+b_n holds for some positive integer nn, is
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Your answer: —
Correct answer: 1
8
Let α\alpha and β\beta be the roots of x2−x−1=0x^2-x-1=0, with α>β\alpha>\beta. For all positive integers nn, define an=αn−βnα−βa_n=\dfrac{\alpha^n-\beta^n}{\alpha-\beta}, n≥1n\ge1, b1=1b_1=1 and bn=an−1+an+1b_n=a_{n-1}+a_{n+1}, n≥2n\ge2. Then which of the following options is/are correct?
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A
a1+a2+a3+⋯+an=an+2−1a_1+a_2+a_3+\cdots+a_n=a_{n+2}-1 for all n≥1n\ge1
B
∑n=1∞an10n=1089\displaystyle\sum_{n=1}^{\infty}\dfrac{a_n}{10^n}=\dfrac{10}{89}
C
∑n=1∞bn10n=889\displaystyle\sum_{n=1}^{\infty}\dfrac{b_n}{10^n}=\dfrac{8}{89}
D
bn=αn+βnb_n=\alpha^n+\beta^n for all n≥1n\ge1
9
Let AP(a;d)AP(a;d) denote the set of all the terms of an infinite arithmetic progression with first term aa and common difference d>0d>0. If AP(1;3)∩AP(2;5)∩AP(3;7)=AP(a;d)AP(1;3)\cap AP(2;5)\cap AP(3;7)=AP(a;d) then a+da+d equals _____
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Your answer: —
Correct answer: 157
10
Let XX be the set consisting of the first 2018 terms of the arithmetic progression 1,6,11,…1,6,11,\ldots, and YY be the set consisting of the first 2018 terms of the arithmetic progression 9,16,23,…9,16,23,\ldots. Then, the number of elements in the set X∪YX\cup Y is ______.
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Your answer: —
Correct answer: 3748
11
Paragraph\textbf{Paragraph}: Let pp, qq be integers and let α\alpha, β\beta be the roots of the equation x2−x−1=0x^2 - x - 1 = 0, where α≠β\alpha \ne \beta. For n=0,1,2,…n = 0, 1, 2, \ldots, let an=pαn+qβna_n = p\alpha^n + q\beta^n. FACT\textbf{FACT}: If aa and bb are rational numbers and a+b5=0a + b\sqrt{5} = 0, then a=0=ba = 0 = b. a12=a_{12} =
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A
a11−a10a_{11} - a_{10}
B
a11+a10a_{11} + a_{10}
C
2a11+a102a_{11} + a_{10}
D
a11+2a10a_{11} + 2a_{10}
12
Paragraph\textbf{Paragraph}: Let pp, qq be integers and let α\alpha, β\beta be the roots of the equation x2−x−1=0x^2 - x - 1 = 0, where α≠β\alpha \ne \beta. For n=0,1,2,…n = 0, 1, 2, \ldots, let an=pαn+qβna_n = p\alpha^n + q\beta^n. FACT\textbf{FACT}: If aa and bb are rational numbers and a+b5=0a + b\sqrt{5} = 0, then a=0=ba = 0 = b. If a4=28a_4 = 28, then p+2q=p + 2q =
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A
2121
B
1414
C
77
D
1212
13
The sides of a right angled triangle are in arithmetic progression. If the triangle has area 2424, then what is the length of its smallest side?
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Your answer: —
Correct answer: 6
14
Let bi>1b_i>1 for i=1,2,…,101i=1,2,\ldots,101. Suppose log⁡eb1,log⁡eb2,…,log⁡eb101\log_e b_1,\log_e b_2,\ldots,\log_e b_{101} are in Arithmetic Progression (A.P.) with the common difference log⁡e2\log_e 2. Suppose a1,a2,…,a101a_1,a_2,\ldots,a_{101} are in A.P. such that a1=b1a_1=b_1 and a51=b51a_{51}=b_{51}. If t=b1+b2+⋯+b51t=b_1+b_2+\cdots+b_{51} and s=a1+a2+⋯+a51s=a_1+a_2+\cdots+a_{51}, then
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A
s>ts>t and a101>b101a_{101}>b_{101}
B
s>ts>t and a101<b101a_{101}<b_{101}
C
s<ts<t and a101>b101a_{101}>b_{101}
D
s<ts<t and a101<b101a_{101}<b_{101}
15
Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6:116:11 and the seventh term lies between 130130 and 140140, then the common difference of this A.P. is
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Your answer: —
Correct answer: 9
16
This question is part of a Column-I to Column-II matching set (JEE Advanced 2015 Section 3, Q.59). Column II lists the values 1,2,3,4,51,2,3,4,5; select every value that correctly matches this Column-I entry. Entry (D)\textbf{Entry (D)}: Let the harmonic mean of two positive real numbers aa and bb be 44. If qq is a positive real number such that a,5,q,ba,5,q,b is an arithmetic progression, then the value(s) of ∣q−a∣|q-a| is (are)
Not attempted
A
11
B
22
C
33
D
44
E
55
17
Let a,b,ca,b,c be positive integers such that ba\frac{b}{a} is an integer. If a,b,ca,b,c are in geometric progression and the arithmetic mean of a,b,ca,b,c is b+2b+2, then the value of a2+a−14a+1\dfrac{a^2+a-14}{a+1} is
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Your answer: —
Correct answer: 4
18
A pack contains nn cards numbered from 11 to nn. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 12241224. If the smaller of the numbers on the removed cards is kk, then k−20=k-20=
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Your answer: —
Correct answer: 5
19
Let Sn=∑k=14n(−1)k(k+1)2k2S_n=\displaystyle\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2. Then SnS_n can take value(s)
Not attempted
A
1056
B
1088
C
1120
D
1332
Work out the sign pattern of (−1)k(k+1)/2(-1)^{k(k+1)/2} for k=1,2,3,4k=1,2,3,4 (it repeats with period 4: −,−,+,+-,-,+,+), group the terms of SnS_n into blocks of 4 consecutive kk's, and simplify each block using difference of squares.
Deconstruction\textbf{Deconstruction}: A sum with an alternating sign that depends on k(k+1)/2 mod 2k(k+1)/2 \bmod 2; find a closed form for SnS_n and match it against the given numeric options. Strategy\textbf{Strategy}: Determine the sign pattern's period, group the sum into blocks of 4 consecutive terms, simplify each block via difference-of-squares, sum the resulting arithmetic-like block values, and solve for which nn reproduce the given options. Execution\textbf{Execution}: Checking k=1,…,8k=1,\ldots,8: k(k+1)/2=1,3,6,10,15,21,28,36k(k+1)/2 = 1,3,6,10,15,21,28,36, with parities odd, odd, even, even, odd, odd, even, even — a period-4 pattern: sign is −1-1 for k≡1,2(mod4)k\equiv1,2\pmod4 and +1+1 for k≡3,0(mod4)k\equiv3,0\pmod4. - Grouping k=4m+1,…,4m+4k=4m+1,\ldots,4m+4: block value =−(4m+1)2−(4m+2)2+(4m+3)2+(4m+4)2= -(4m+1)^2-(4m+2)^2+(4m+3)^2+(4m+4)^2 - =[(4m+3)2−(4m+1)2]+[(4m+4)2−(4m+2)2]=(2)(8m+4)+(2)(8m+6)=32m+20=\big[(4m+3)^2-(4m+1)^2\big]+\big[(4m+4)^2-(4m+2)^2\big] = (2)(8m+4)+(2)(8m+6) = 32m+20. - Sn=∑m=0n−1(32m+20)=32⋅n(n−1)2+20n=16n2−16n+20n=16n2+4n=4n(4n+1)S_n=\displaystyle\sum_{m=0}^{n-1}(32m+20)=32\cdot\dfrac{n(n-1)}{2}+20n=16n^2-16n+20n=16n^2+4n=4n(4n+1). - Test n=8n=8: S8=16(64)+32=1024+32=1056S_8=16(64)+32=1024+32=1056 — matches option (A). - Test n=9n=9: S9=16(81)+36=1296+36=1332S_9=16(81)+36=1296+36=1332 — matches option (D). - Checking (B) 10881088: 16n2+4n=1088⇒4n2+n−272=016n^2+4n=1088\Rightarrow 4n^2+n-272=0, discriminant =1+4352=4353=1+4352=4353, not a perfect square — no integer solution. - Checking (C) 11201120: 4n2+n−280=04n^2+n-280=0, discriminant =4481=4481, not a perfect square — no integer solution. Verification\textbf{Verification}: Direct brute-force summation of the original series for n=8n=8 (i.e. k=1k=1 to 3232) and n=9n=9 (k=1 to 36) using the derived closed form 4n(4n+1)4n(4n+1) reproduces 10561056 and 13321332 exactly, and no integer nn makes the formula equal 10881088 or 11201120, confirming those are not attainable. Final Answer\textbf{Final Answer}: (A) and (D) — 1056 (at n=8n=8) and 1332 (at n=9n=9)