JEE Mathematics Sequence and Series MCQs Set 34

Answer: (c) 1 : 4

Explanation:
1. \( G=\sqrt{ab} \) and \( M=\frac{a+b}{2ab} \), so \( \frac1M=\frac{2ab}{a+b} \).
2. \( \frac{1/M}{G}=\frac{2\sqrt{ab}}{a+b}=\frac45\Rightarrow \frac{a+b}{\sqrt{ab}}=\frac52 \).
3. Let \( t=\sqrt{a/b} \): \( t+\frac1t=\frac52\Rightarrow t=2\ \text{or}\ \frac12 \), so \( a:b=t^2:1=4:1\ \text{or}\ 1:4 \).

Teacher's Note:
1. 1/M is the harmonic mean of a and b, so the ratio is HM/GM.
2. Put t = sqrt(a/b) to turn the condition into t + 1/t = 5/2.

Answer: (b) 8

Explanation:
1. Let there be \( 2m \) terms. The even-placed terms exceed the odd-placed ones by \( d \) each, so \( md=30-24=6 \).
2. Last minus first \( =(2m-1)d=10.5\Rightarrow 12-d=10.5\Rightarrow d=1.5 \).
3. \( m=\frac{6}{1.5}=4 \), so the number of terms is \( 2m=8 \).
4. Check: the terms are 1.5, 3, 4.5, ..., 12; odd places 1.5+4.5+7.5+10.5=24, even places 3+6+9+12=30.

Teacher's Note:
1. Pair each odd-placed term with the even-placed term after it; every pair differs by d.
2. Use md = 6 to turn the last-minus-first condition into one equation in d.

Answer: (c) \( \frac{20}{11} \)

Explanation:
1. The \( k \)th term is \( \frac{1}{1+2+\cdots+k}=\frac{2}{k(k+1)} \).
2. \( S_{10}=2\sum_{k=1}^{10}\left(\frac1k-\frac1{k+1}\right)=2\left(1-\frac1{11}\right)=\frac{20}{11} \)

Teacher's Note:
1. Each denominator is a triangular number k(k+1)/2.
2. The reciprocal series telescopes.

Answer: (b) \( 6n^2-n \)

Explanation:
1. \( a_1=S_1=5 \) and \( a_2=S_2-S_1=16-5=11 \), so \( d=6 \).
2. New AP: first term \( 5 \), common difference \( 12 \).
3. \( S_n=\frac n2\left[10+12(n-1)\right]=n(5+6n-6)=6n^2-n \)

Teacher's Note:
1. Get a1 and d from S1 and S2.
2. Substitute into S = n/2 (2a + (n - 1)D) with D = 12.

Answer: (c) \( \frac{11}{2} \)

Explanation:
1. The \( n \)th term is \( \frac{2n+1}{n(n+1)(2n+1)/6}=\frac{6}{n(n+1)} \).
2. \( S_{11}=6\left(1-\frac1{12}\right)=\frac{11}{2} \)

Teacher's Note:
1. Same pattern as before: the 2n + 1 cancels with the factor in the sum of squares.
2. 6/(n(n+1)) telescopes to 6(1 - 1/12).

Answer: (c) 12100

Explanation:
1. The \( k \)th term is \( k(2k)^2=4k^3 \).
2. \( S_{10}=4\left[\frac{10\cdot11}{2}\right]^2=4\times55^2=4\times3025=12100 \)

Teacher's Note:
1. Spot the pattern k times (2k) squared.
2. Use the sum of cubes formula (n(n+1)/2)^2.

Answer: (a) \( \frac{55}{336} \)

Explanation:
1. \( \frac{1}{n(n+1)(n+2)(n+3)}=\frac13\left[\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right] \).
2. Telescoping: \( \text{sum}=\frac13\left[\frac{1}{1\cdot2\cdot3}-\frac{1}{6\cdot7\cdot8}\right]=\frac13\cdot\frac{56-1}{336}=\frac13\cdot\frac{55}{336} \).
3. \( k=\frac{55}{336} \)

Teacher's Note:
1. Use the identity 1/(n(n+1)(n+2)(n+3)) = (1/3)[1/(n(n+1)(n+2)) - 1/((n+1)(n+2)(n+3))].
2. Only the first and the last pieces survive.

Answer: (c) \( 13m \)

Explanation:
1. \( a_4-a_7+a_{10}=(a+3d)-(a+6d)+(a+9d)=a+6d=a_7 \), so \( a_7=m \).
2. \( S_{13}=\frac{13}{2}(a_1+a_{13})=13\,a_7=13m \)

Teacher's Note:
1. The combination a4 - a7 + a10 collapses to a7.
2. The sum of 13 terms is 13 times the 7th (middle) term.

Answer: (a) 101

Explanation:
1. Convert the terms: \( 1\tfrac35=\tfrac85,\ 2\tfrac25=\tfrac{12}{5},\ 3\tfrac15=\tfrac{16}{5},\ 4=\tfrac{20}{5},\ 4\tfrac45=\tfrac{24}{5} \), i.e. the \( k \)th term is \( \left(\frac{4(k+1)}{5}\right)^2 \).
2. \( S_{10}=\frac{16}{25}\sum_{k=1}^{10}(k+1)^2=\frac{16}{25}\left(\sum_{j=1}^{11}j^2-1\right)=\frac{16}{25}(506-1)=\frac{16}{25}\times505 \).
3. \( \frac{16}{25}\times505=\frac{16}{5}\times101\Rightarrow m=101 \)

Teacher's Note:
1. Write every mixed number as an improper fraction with denominator 5; the numerators go 8, 12, 16, 20, 24.
2. Sum of squares up to 11 is 506.

Answer: (b) 6

Explanation:
1. The product of the second and second-last terms equals the product of the first and last terms: \( a\cdot l=128 \), with \( a+l=66 \).
2. \( a,l \) are roots of \( t^2-66t+128=0 \), so \( a=2,\ l=64 \) (increasing GP).
3. Sum \( \frac{lr-a}{r-1}=126\Rightarrow 64r-2=126r-126\Rightarrow r=2 \).
4. \( 64=2\cdot2^{n-1}\Rightarrow n=6 \)

Teacher's Note:
1. In a GP, terms equidistant from the ends have the same product.
2. Use S = (lr - a)/(r - 1) with the first and last terms.

Answer: (c) 3

Explanation:
1. \( \frac1a,\frac1b,\frac1c,\frac1d \) are in AP with common difference \( D \). Let \( \frac1a=A \).
2. \( \frac{A^2-(A+3D)^2}{(A+D)^2-(A+2D)^2}=\frac{-3D(2A+3D)}{-D(2A+3D)}=3 \)

Teacher's Note:
1. HP terms have reciprocals in AP.
2. Factor each numerator and denominator as a difference of squares.

Answer: (b) 10

Explanation:
1. The \( k \)th term is \( \frac{(k+1)^2-k}{k(k+1)}=\frac{k^2+k+1}{k(k+1)}=1+\frac{1}{k(k+1)} \).
2. \( S=10+\sum_{k=1}^{10}\left(\frac1k-\frac1{k+1}\right)=10+\frac{10}{11} \).
3. \( S=10\tfrac{10}{11} \), so \( [S]=10 \)

Teacher's Note:
1. Write the general term as 1 + 1/(k(k+1)).
2. The greatest integer of 10.909 is 10.

Answer: (b) 2

Explanation:
1. Let \( b=ar,\ c=ar^2 \). Then \( x=\frac{a(1+r)}{2} \) and \( y=\frac{ar(1+r)}{2} \).
2. \( \frac ax+\frac cy=\frac{2}{1+r}+\frac{2ar^2}{ar(1+r)}=\frac{2}{1+r}+\frac{2r}{1+r}=2 \)

Teacher's Note:
1. Write the three GP terms as a, ar, ar^2.
2. Both fractions share the denominator 1 + r.

Answer: (c) 6

Explanation:
1. The expression equals \( \left(\frac ab+\frac ba\right)+\left(\frac bc+\frac cb\right)+\left(\frac ca+\frac ac\right) \).
2. Each bracket is at least \( 2 \) by AM-GM, with equality when \( a=b=c \).
3. Minimum value \( =6 \)

Teacher's Note:
1. Regroup the six fractions into three reciprocal pairs.
2. x + 1/x is at least 2 for positive x.

Answer: (c) 3

Explanation:
1. Put \( x=b+c-a,\ y=c+a-b,\ z=a+b-c \) (all positive). Then \( a=\frac{y+z}{2},\ b=\frac{z+x}{2},\ c=\frac{x+y}{2} \).
2. The expression is \( \frac12\left[\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\right]\ge\frac12\times6=3 \) by the previous result.
3. Equality at \( x=y=z \), i.e. an equilateral triangle, so the minimum is \( 3 \)

Teacher's Note:
1. Triangle inequality makes b + c - a, c + a - b and a + b - c positive.
2. The substitution turns it into the same six-fraction sum as the previous question.

Answer: (c) 10

Explanation:
1. \( \sum_{k=1}^{j}1=j \) and \( \sum_{j=1}^{i}j=\frac{i(i+1)}{2} \).
2. \( \sum_{i=1}^{n}\frac{i(i+1)}{2}=\frac{n(n+1)(n+2)}{6}=220 \), so \( n(n+1)(n+2)=1320=10\times11\times12 \).
3. \( n=10 \)

Teacher's Note:
1. Evaluate the innermost sum first and work outwards.
2. Factor 1320 as three consecutive integers.

Answer: (b) 1

Explanation:
1. The first terms are \( 1,3,7,13,21,\ldots \) with differences \( 2,4,6,8,\ldots \), so \( t_n=n^2-n+1 \).
2. \( \sqrt{n^2-n+1}-n=\frac{-n+1}{\sqrt{n^2-n+1}+n}\to\frac{-n}{2n}=-\frac12 \).
3. \( S=-\frac12\Rightarrow 2|S|=1 \)

Teacher's Note:
1. The differences of the first terms form the AP 2, 4, 6, ..., so t_n is quadratic.
2. Rationalise the limit by multiplying by the conjugate.

Answer: (a) 1

Explanation:
1. \( U_n=\frac{1}{(n+2)(n+3)}=\frac1{n+2}-\frac1{n+3} \).
2. \( S_n=\frac13-\frac1{n+3}\to\frac13 \), so \( \lim3S_n=1 \)

Teacher's Note:
1. Cancel (n+1)! from the numerator and denominator first.
2. Telescoping gives S_n = 1/3 - 1/(n+3).

Answer: (a) 0

Explanation:
1. Let \( b=ar,\ c=ar^2 \). HP condition: \( \frac{2}{c-a}=\frac{1}{a-b}+\frac{1}{b-c} \), i.e. \( 2(a-b)(b-c)=(c-a)\left[(b-c)+(a-b)\right]=(c-a)(a-c) \).
2. Substituting \( b=ar,\ c=ar^2 \): \( 2a^2r(1-r)^2=-a^2(1-r)^2(1+r)^2 \), which gives \( a^2(1-r)^2(r^2+4r+1)=0 \).
3. The terms are distinct, so \( r\ne1 \) and \( r^2+4r+1=0 \). Hence \( a+4b+c=a(1+4r+r^2)=0 \)

Teacher's Note:
1. Write everything in terms of a and r.
2. a + 4b + c factors as a(1 + 4r + r^2), the same quadratic as in the condition.

Answer: (b) 2

Explanation:
1. The \( k \)th term is \( \frac{k(k+1)/2}{\left[k(k+1)/2\right]^2}=\frac{2}{k(k+1)} \).
2. \( S_n=2\sum_{k=1}^{n}\left(\frac1k-\frac1{k+1}\right)=2\left(1-\frac{1}{n+1}\right)<2 \).
3. \( S_n \) increases towards \( 2 \) but never exceeds it, while \( S_n>1 \) for \( n\ge2 \) (for example \( S_2=\frac43 \)). So the smallest such number is \( 2 \)

Teacher's Note:
1. The sum of cubes 1^3 + ... + n^3 equals the square of 1 + 2 + ... + n.
2. The sum approaches 2 from below as n grows.

Answer: (b) 5

Explanation:
1. Apply weighted AM-GM to \( \frac x2\ (\times2),\ \frac{2y}{4}\ (\times4),\ \frac{3z}{3}\ (\times3),\ 4w\ (\times1) \) (ten parts in all), whose sum is \( 50 \).
2. \( \frac{50}{10}\ge\left[\left(\frac x2\right)^2\left(\frac y2\right)^4z^3(4w)\right]^{1/10}=\left[\frac{x^2y^4z^3w}{16}\right]^{1/10} \).
3. So the expression is at most \( 5 \), attained at \( x=10,\ y=10,\ z=5,\ w=\frac54 \)

Teacher's Note:
1. Choose the parts so that each equals the same quantity: x/2, 2y/4, 3z/3 and 4w with the right multiplicities.
2. The 1/16 in the problem is exactly what the parts' product produces.

Answer: (c) 3

Explanation:
1. \( a^2+b^2-4a-6b+13=(a-2)^2+(b-3)^2=K \). Then \( f(0)=5K \) and \( f(1)=3K \), so \( f(0)=f(1)\Rightarrow K=0 \), i.e. \( a=2,\ b=3 \).
2. For the AP and HP between the same two numbers, \( A_rH_{11-r}=ab \) for every \( r \).
3. \( \frac{1}{10}\sum_{r=4}^{8}ab=\frac{5\times6}{10}=3 \)

Teacher's Note:
1. Complete the squares to find a and b.
2. The product A_r H_(n+1-r) equals ab, a standard property of an AP and HP between the same extremes.