JEE Mathematics Sequence and Series MCQs Set 23

Answer: (a) \( \{t_n\} \) is an A.P., \( \{s_n\} \) is a G.P.

Explanation:
1. \( t_n=(n+1)\log5-(n-1)\log3 \) is linear in n, so it is an A.P. with common difference \( \log\frac53 \).
2. \( s_n \) is a power of a fixed number, so it is a G.P.

Teacher's Note:
1. Expand the logarithm.

Answer: (b) G.P.

Explanation:
1. Expanding along the third row: \( 2\{b(b\alpha-c)-c(a\alpha-b)\}-\{a(b\alpha-c)-b(a\alpha-b)\} \).
2. \( =2\alpha(b^2-ac)-(b^2-ac)=(2\alpha-1)(b^2-ac) \).
3. Since \( \alpha\ne\frac12 \), \( b^2=ac \): G.P.

Teacher's Note:
1. Expand along the row containing the zero.

Answer: (c) H.P.

Explanation:
1. \( \frac{a_r-a_{r+1}}{a_ra_{r+1}}=\frac1{a_{r+1}}-\frac1{a_r}=k \).
2. The reciprocals have a constant difference, so the terms are in H.P.

Teacher's Note:
1. Divide through by the product.

Answer: (c) H.P.

Explanation:
1. Test the H.P. \( 1,\frac12,\frac13,\frac14 \): the three ratios are \( \frac{1/6}{1/4}=\frac23,\ \frac{5/6}{5/4}=\frac23,\ 3\cdot\frac{1/6}{3/4}=\frac23 \).
2. They agree, so the numbers are in H.P. (the relations are exactly the H.P. conditions).

Teacher's Note:
1. Check with a simple H.P.

Answer: (c) imaginary

Explanation:
1. The middle terms are \( a_n=A=\frac{a+b}2,\ b_n=G=\sqrt{ab},\ c_n=H=\frac{2ab}{a+b} \).
2. The discriminant is \( G^2-4AH=ab-4ab=-3ab\lt 0 \), so the roots are imaginary.

Teacher's Note:
1. The middle term of each progression is the corresponding mean.

Answer: (d) \( (\sqrt2-1)(b-a) \)

Explanation:
1. Let \( b=ar,\ c=ar^2 \), \( p=a+d,\ q=a+2d \). The G.P. condition \( (b+p)^2=2a(c+q) \) gives \( d^2+2ad(r-1)-a^2(r-1)^2=0 \).
2. So \( d=a(r-1)(-1\pm\sqrt2) \), and \( a(r-1)=b-a \). One value is \( d=(\sqrt2-1)(b-a) \) (the other is \( -(\sqrt2+1)(b-a) \)).

Teacher's Note:
1. Form a quadratic in d.

Answer: (a) \( 3 \)

Explanation:
1. The product of the three terms has logarithm \( \log x(\log y-\log z)+\log y(\log z-\log x)+\log z(\log x-\log y)=0 \), so the product is 1.
2. By A.M.-G.M., the sum is at least \( 3\sqrt[3]1=3 \).

Teacher's Note:
1. Take the logarithm of the product.

Answer: (c) \( 6 \)

Explanation:
1. The expression has 6 terms with product \( (abc)^6 \).
2. By A.M.-G.M., the sum is at least \( 6abc \), with equality at \( a=b=c \). So \( \lambda=6 \).

Teacher's Note:
1. Apply A.M.-G.M. to the six terms.

Answer: (b) \( [3,\infty) \)

Explanation:
1. All three ratios are positive and their product is 1.
2. By A.M.-G.M., the sum is at least 3, attained at \( x=y=z \).

Teacher's Note:
1. The ratios are positive.

Answer: (b) \( -55 \)

Explanation:
1. The sum of all ten numbers is 0 and the sum of their squares is \( 2(1+4+9+16+25)=110 \).
2. The sum of pair products is \( \frac{0^2-110}2=-55 \).

Teacher's Note:
1. Use (sum) squared = sum of squares + 2 (sum of pair products).

Answer: (c) Rs. 20500

Explanation:
1. The balance Rs. 10000 is cleared in 10 instalments of Rs. 1000. The interest paid with the instalments is \( 1000+900+\dots+100=5500 \).
2. Total \( =5000+10000+5500=20500 \).

Teacher's Note:
1. The interest is on the unpaid balance each year.

Answer: (d) \( 2632 \)

Explanation:
1. Total \( =5050 \). Multiples of 3: 1683. Multiples of 5: 1050. Multiples of 15: 315.
2. Divisible by 3 or 5: \( 1683+1050-315=2418 \).
3. Required sum \( =5050-2418=2632 \).

Teacher's Note:
1. Use inclusion and exclusion.

Answer: (d) never less than n

Explanation:
1. By A.M.-G.M., the sum is at least \( n\sqrt[n]1=n \).

Teacher's Note:
1. A.M. is at least G.M.

Answer: (a) \( 3048 \)

Explanation:
1. \( 960=2^6\cdot3\cdot5 \).
2. The sum of divisors is \( (2^7-1)(1+3)(1+5)=127\cdot4\cdot6=3048 \).

Teacher's Note:
1. Use the product of G.P. sums for each prime.

Answer: (d) \( 2^{\left(1-\frac1{\sqrt2}\right)} \)

Explanation:
1. By A.M.-G.M., the sum is at least \( 2\sqrt{2^{\sin\theta+\cos\theta}} \).
2. The minimum of \( \sin\theta+\cos\theta \) is \( -\sqrt2 \), so the sum is at least \( 2\cdot2^{-1/\sqrt2}=2^{1-1/\sqrt2} \), attained when \( \sin\theta=\cos\theta=-\frac1{\sqrt2} \).

Teacher's Note:
1. Minimise the exponent first.

Answer: (b) H.P.

Explanation:
1. The area is fixed, so each side is inversely proportional to the altitude on it: \( a=\frac{2\Delta}{h_a} \).
2. The reciprocals of the sides are proportional to the altitudes, which are in A.P., so the sides are in H.P.

Teacher's Note:
1. Side times altitude is twice the area.

Answer: (c) \( 11 \)

Explanation:
1. After each jump and slip the net gain is 1 m, so after 10 jumps the monkey is at 10 m.
2. The 11th jump takes it from 10 m to 12 m, the top.

Teacher's Note:
1. The last jump has no slip.

Answer: (c) \( 1540 \)

Explanation:
1. For n rows, \( \frac{n(n+1)}2+669=(n-8)^2 \), giving \( n^2-33n-1210=0 \), so \( n=55 \).
2. The number of balls is \( \frac{55\cdot56}2=1540 \).

Teacher's Note:
1. Form an equation in the number of rows.

Answer: (d) \( 6 \)

Explanation:
1. The expression equals \( \left(\frac ba+\frac ab\right)+\left(\frac ca+\frac ac\right)+\left(\frac cb+\frac bc\right) \), and each bracket is at least 2.
2. The minimum is 6.

Teacher's Note:
1. Pair reciprocal fractions.

Answer: (a) \( 9 \)

Explanation:
1. By A.M.-G.M. each factor is at least \( 3\sqrt[3]{abc} \) and \( \frac3{\sqrt[3]{abc}} \) respectively.
2. The product is at least 9, attained at \( a=b=c \).

Teacher's Note:
1. Apply A.M.-G.M. to each bracket.

Answer: (d) \( 24 \)

Explanation:
1. The power of 5 in 100! is \( \lfloor100/5\rfloor+\lfloor100/25\rfloor=20+4=24 \).
2. There are plenty of factors of 2, so there are 24 trailing zeros.

Teacher's Note:
1. Count factors of 5.

Answer: (b) \( 3 \)

Explanation:
1. The bracket is a G.P. with ratio 3x, so it equals \( \frac{1-(3x)^6}{1-3x} \).
2. Then \( (1-p)\frac{1-(3x)^6}{1-3x}=1-p^6 \) is satisfied by \( p=3x \). So \( \frac px=3 \).

Teacher's Note:
1. Sum the G.P. in brackets.

Answer: (c) \( (\sqrt2)^{n+1} \)

Explanation:
1. The ratio is 2, so \( x_n=2^{n-1} \).
2. \( y_n^2=4\cdot2^{n-1}=2^{n+1} \), and in the first quadrant \( y_n=2^{\frac{n+1}2}=(\sqrt2)^{n+1} \).

Teacher's Note:
1. Substitute x_n into the parabola.

Answer: (a) 12 cm

Explanation:
1. Let the edges be \( \frac6r,6,6r \) (volume \( =216 \) gives the middle edge 6).
2. Surface area: \( 2\cdot36\left(\frac1r+1+r\right)=252 \) gives \( r+\frac1r=\frac52 \), so \( r=2 \) or \( \frac12 \).
3. The edges are 3, 6, 12, so the longest is 12 cm.

Teacher's Note:
1. The middle edge is the cube root of the volume.

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

Explanation:
1. By similar triangles, if \( L_k \) is at distance \( \frac{k}{n+1}AB \) from B, then \( L_kM_k=\frac{ka}{n+1} \) (numbering from B).
2. The sum is \( \frac a{n+1}\cdot\frac{n(n+1)}2=\frac{an}2 \).

Teacher's Note:
1. The lengths form an A.P.