JEE Mathematics Sequence and Series MCQs Set 08

Answer: (b) \( 10^5 \)

Explanation:
1. Let \( L=\log_{10}x \). The series gives \( n=L\left(1+\frac12+\frac14+\cdots\right)=2L \).
2. \( 1+3+\cdots=n^2 \) and \( 4+7+10+\cdots=\frac{n(3n+5)}2 \), so the ratio is \( \frac{2n}{3n+5} \).
3. \( \frac{4L}{6L+5}=\frac{20}{7L}\Rightarrow 7L^2-30L-25=0\Rightarrow L=5 \) or \( -\frac57 \).
4. \( n \) is a number of terms, so it is a positive integer: \( L=5 \) (then \( n=10 \)). Hence \( x=10^5 \).

Teacher's Note:
1. Sum the infinite geometric series first to link n and log x.
2. Reject roots that make the number of terms negative or non-integer.

Answer: (c) \( \frac5{36} \)

Explanation:
1. For \( |r|\lt 1 \): \( \sum_{n\ge1}(-1)^{n+1}nr^n=\frac r{(1+r)^2} \) (differentiate the series of \( \frac1{1+r} \) term by term).
2. With \( r=\frac15 \): \( \frac{1/5}{(6/5)^2}=\frac15\cdot\frac{25}{36}=\frac5{36} \).

Teacher's Note:
1. Differentiate the geometric series 1/(1 + r) term by term to get the formula.
2. Substitute r = 1/5.

Answer: (b) 20

Explanation:
1. The common terms form an A.P. with first term 21 and common difference lcm(4, 5) = 20.
2. The first sequence ends at 417 (the second goes to 466), so the largest common term is at most 417: 21, 41, ..., 401.
3. \( n=\frac{401-21}{20}+1=20 \).

Teacher's Note:
1. Common terms of two A.P.s form an A.P. with difference equal to the LCM of the differences.
2. The smaller upper limit decides where the common terms stop.

Answer: (a) \( a\in\left[\frac34,\infty\right) \)

Explanation:
1. Let the common ratio be \( r \): \( y=xr,\ z=xr^2 \). Then \( x(1+r+r^2)=ax \) so \( a=1+r+r^2 \) (\( x\ne0 \)).
2. \( 1+r+r^2=\left(r+\frac12\right)^2+\frac34\ge\frac34 \), and every value \( \ge\frac34 \) is attained for real \( r \).

Teacher's Note:
1. Express y and z using the common ratio and cancel x.
2. Find the range of a quadratic in r over all real r.

Answer: (b) \( \frac1{60} \)

Explanation:
1. \( s_k=\frac{2k-1}{1-\frac{2}{2k+1}}=\frac{(2k-1)(2k+1)}{2k-1}=2k+1 \). So \( s_1,s_2,\ldots=3,5,7,\ldots \).
2. The series is \( \sum_{k\ge1}\frac1{(2k+1)(2k+3)(2k+5)} \).
3. Using \( \frac1{(2k+1)(2k+3)(2k+5)}=\frac14\left[\frac1{(2k+1)(2k+3)}-\frac1{(2k+3)(2k+5)}\right] \), the sum telescopes to \( \frac14\cdot\frac1{3\cdot5}=\frac1{60} \).

Teacher's Note:
1. Find s_k as a closed form before doing anything else; it simplifies to 2k + 1.
2. Telescoping works because the three factors are in A.P.

Answer: (c) H.P.

Explanation:
1. Adding, \( p+q+r=0 \), so \( x=1 \) is a root of \( px^2+qx+r=0 \).
2. Equal roots means both roots are 1, so the product of roots \( \frac rp=1 \), i.e. \( p=r \).
3. \( p=a(b-c)(a+b+c) \) and \( r=c(a-b)(a+b+c) \), so \( a(b-c)=c(a-b) \Rightarrow ab+bc=2ac \).
4. Dividing by \( abc \): \( \frac1c+\frac1a=\frac2b \), so \( a,b,c \) are in H.P.

Teacher's Note:
1. Notice p + q + r = 0, so 1 is automatically a root.
2. Equal roots then force the product of roots to be 1.

Answer: (b) 1

Explanation:
1. The \( k \)-th term is \( \frac{k(k+1)/2}{[k(k+1)/2]^2}=\frac{2}{k(k+1)} \).
2. \( S_n=2\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)=2\left(1-\frac1{n+1}\right)=\frac{2n}{n+1} \).
3. For \( n\ge1 \): \( 1\le S_n\lt 2 \), so the integral part is 1.

Teacher's Note:
1. The sum of cubes equals the square of the sum of the first k natural numbers.
2. Telescope the result and bound it.

Answer: (b) 5050

Explanation:
1. The highest power in each factor is 1, 2, 3, ..., 100.
2. The highest exponent in the product is their sum: 1 + 2 + ... + 100 = 100·101/2 = 5050.

Teacher's Note:
1. Degrees add when polynomials are multiplied.
2. Use the formula for the sum of first n natural numbers.

Answer: (d) (11)! − 1

Explanation:
1. \( k\cdot k!=(k+1)!-k! \).
2. The sum telescopes: \( (2!-1!)+(3!-2!)+\cdots+(11!-10!)=11!-1! \).
3. So the sum is 11! − 1.

Teacher's Note:
1. Write k as (k + 1) − 1 inside the product.
2. Check with n = 1: 1·1! = 1 = 2! − 1.

Answer: (c) 31

Explanation:
1. Let there be \( 2m+1 \) stones, with \( m \) stones on each side of the middle one.
2. The stones at distance \( 10j \) m (two of them, for \( j=1,\ldots,m \)) each need a round trip of \( 20j \) m. Total distance \( =2\sum_{j=1}^m20j=20m(m+1) \).
3. \( 20m(m+1)=4800\Rightarrow m(m+1)=240\Rightarrow m=15 \).
4. Number of stones = 2·15 + 1 = 31.

Teacher's Note:
1. Each stone is picked up and brought to the middle, so every stone costs a return trip.
2. Convert the 4.8 km into metres before equating.

Answer: (b) x is a negative real

Explanation:
1. Reciprocals are in A.P.: \( \log_2(5\cdot2^x+1),\ \log_4(2^{1-x}+1)=\frac12\log_2(2^{1-x}+1),\ 1 \).
2. So \( \log_2(2^{1-x}+1)=\log_2(5\cdot2^x+1)+1=\log_2\big(2(5\cdot2^x+1)\big) \).
3. Let \( u=2^x \): \( \frac2u+1=10u+2\Rightarrow10u^2+u-2=0\Rightarrow u=\frac25 \).
4. \( x=\log_2\frac25=1-\log_25\lt 0 \), and it is irrational. So x is a negative real.

Teacher's Note:
1. Take reciprocals to turn the H.P. into an A.P. of logarithms.
2. Equal logarithms of positive numbers mean equal arguments.

Answer: (a) 6479

Explanation:
1. Row r contains r odd numbers, so the first 80 rows contain 1 + 2 + ... + 80 = 3240 odd numbers.
2. The last number in row 80 is the 3240th odd number: \( 2\cdot3240-1=6479 \).

Teacher's Note:
1. Count how many odd numbers have been used up to the required row.
2. The n-th odd number is 2n − 1.

Answer: (d) \( \sqrt[4]e \)

Explanation:
1. Let \( L=\ln a\ne0 \). The left side is a G.P. with first term and ratio \( 2L \): \( \frac{2L}{1-2L} \).
2. The right side is \( 3\cdot\frac{L}{1-L} \).
3. \( 2(1-L)=3(1-2L)\Rightarrow 4L=1\Rightarrow L=\frac14 \) (this satisfies \( |2L|\lt 1 \)).
4. So \( a=e^{1/4}=\sqrt[4]e \).

Teacher's Note:
1. Treat each side as an infinite G.P. in L.
2. Cancel L only after noting that L = 0 (a = 1) is excluded.

Answer: (b) G.P.

Explanation:
1. \( a_1=2^p-1 \) and for \( n\ge2 \): \( a_n=S_n-S_{n-1}=2^{np}-2^{(n-1)p}=2^{(n-1)p}(2^p-1) \).
2. This also holds for \( n=1 \), and \( \frac{a_{n+1}}{a_n}=2^p \) is constant. The sequence is a G.P.

Teacher's Note:
1. Find the n-th term from the difference of consecutive sums.
2. A constant ratio between consecutive terms identifies a G.P.

Answer: (d) \( [12,\infty) \)

Explanation:
1. A.P. gives \( a=5^{1+x}+5^{1-x}+25^x+25^{-x}=5\left(t+\frac1t\right)+\left(t^2+\frac1{t^2}\right) \), where \( t=5^x\gt 0 \).
2. Each bracket is at least 2 and 2 respectively, so \( a\ge10+2=12 \), with equality at \( t=1 \).
3. As \( t\to\infty \), \( a\to\infty \). So \( a\in[12,\infty) \).

Teacher's Note:
1. The middle term of an A.P. is the mean, so a/2 is the average of the other two.
2. Use t + 1/t ≥ 2 for t > 0.

Answer: (b) \( 5050\pi \)

Explanation:
1. The innermost disc is red, the annulus between radii 1 and 2 is green, 2 to 3 red, and so on. Green annuli lie between radii (1, 2), (3, 4), ..., (99, 100).
2. Green area \( =\pi\sum_{j=1}^{50}\big[(2j)^2-(2j-1)^2\big]=\pi\sum_{j=1}^{50}(4j-1) \).
3. \( =\pi\,(4\cdot1275-50)=5050\pi \).

Teacher's Note:
1. Identify which annuli are green and use (2j)² − (2j − 1)² = 4j − 1.
2. Sum of the first 50 natural numbers is 1275.

Answer: (a) equal to 0

Explanation:
1. \( S_{11}=S_{19}\Rightarrow \frac{11}2(2a+10d)=\frac{19}2(2a+18d) \Rightarrow 22a+110d=38a+342d \Rightarrow 16a+232d=0 \Rightarrow 2a+29d=0 \).
2. \( S_{30}=15(2a+29d)=0 \).

Teacher's Note:
1. Equate the two sums and simplify to a relation between a and d.
2. The relation 2a + 29d = 0 is exactly the factor appearing in S₃₀.

Answer: (d) 11 : 12

Explanation:
1. Let \( C \) have radius \( R \). Then \( P_0P_1=\frac R2 \), \( P_1P_2=\frac R4 \), \( P_2P_3=\frac R8,\ldots \)
2. Radii of the shaded circles: \( \frac R4,\frac R8,\frac R{16},\ldots \), so their areas form a G.P. with first term \( \frac{\pi R^2}{16} \) and ratio \( \frac14 \).
3. Shaded area \( =\frac{\pi R^2/16}{1-1/4}=\frac{\pi R^2}{12} \).
4. Unshaded area : area of C = (1 − 1/12) : 1 = 11 : 12.

Teacher's Note:
1. Each successive diameter is half the previous one.
2. Areas of circles with radii in G.P. form a G.P. with the square of the ratio.

Answer: (c) g₁ = 1

Explanation:
1. With \( g_1=a \) and ratio \( r \): \( a(1+r+r^2)=13 \) and \( a^2(1+r^2+r^4)=91 \).
2. Since \( 1+r^2+r^4=(1+r+r^2)(1-r+r^2) \), dividing gives \( a(1-r+r^2)=7 \).
3. Subtracting: \( 2ar=6\Rightarrow ar=3 \), so \( g_2=3 \). Then \( \frac3r+3+3r=13\Rightarrow 3r^2-10r+3=0\Rightarrow r=\frac13 \) (decreasing) or 3.
4. So the G.P. is 9, 3, 1, \( \frac13,\ldots \): the greatest term is 9, \( g_4=\frac13 \), \( 3g_4=g_3 \), \( g_2=3 \), but \( g_1=9\ne1 \).

Teacher's Note:
1. Use the factorisation 1 + r² + r⁴ = (1 + r + r²)(1 − r + r²).
2. A decreasing G.P. with positive terms needs 0 < r < 1.

Answer: (c) \( \frac{25}{54} \)

Explanation:
1. The series is \( \sum_{n\ge1}n^2x^{n-1} \) with \( x=-\frac15 \).
2. \( \sum_{n\ge1}n^2x^{n-1}=\frac{1+x}{(1-x)^3} \) for \( |x|\lt 1 \).
3. \( \frac{4/5}{(6/5)^3}=\frac45\cdot\frac{125}{216}=\frac{25}{54} \).

Teacher's Note:
1. Recognise an arithmetico-geometric type series with n² as the coefficient.
2. Use the standard result for the sum of n²x^(n−1).

Answer: (a) \( \left[2^{\frac{1-n}2}\right]r \)

Explanation:
1. The first square has side \( 2r \) and inradius \( r \).
2. Joining mid-points gives a square with side \( \frac1{\sqrt2} \) times the old side, so its inradius is also \( \frac1{\sqrt2} \) of the old inradius.
3. Inradii form a G.P. with ratio \( \frac1{\sqrt2} \): the n-th is \( r\left(\frac1{\sqrt2}\right)^{n-1}=2^{\frac{1-n}2}r \).

Teacher's Note:
1. The inradius of a square is half its side.
2. The mid-point square has half the area, so its side shrinks by √2.

Answer: (b) 16

Explanation:
1. The roots are primes \( p,q \): \( a=p+q \) is odd, so one prime is 2 and the other is an odd prime \( q \).
2. \( a+b=2+q+2q=35\Rightarrow q=11 \). So \( a=13,\ b=22 \) and \( f(x)=x^2-13x+22 \).
3. \( \sum_{k=1}^{10}f(k)=385-13\cdot55+220=-110 \) and \( f(10)=100-130+22=-8 \).
4. N = (−110)(−8) = 880; digit sum = 8 + 8 + 0 = 16.

Teacher's Note:
1. An odd sum of two primes forces one of them to be 2.
2. Use the formulas for the sum of k and of k² up to 10.