JEE Mathematics Sequence and Series MCQs Set 19

Answer: (b) \( 1 \)

Explanation:
1. The reciprocals are in A.P.: \( A+(p-1)d=\frac1q \) and \( A+(q-1)d=\frac1p \).
2. Subtracting, \( d=\frac1{pq} \), and then \( A=\frac1{pq} \).
3. The reciprocal of the \( pq^{th} \) term is \( A+(pq-1)d=1 \), so the term is 1.

Teacher's Note:
1. Solve for the first term and difference of the A.P.

Answer: (b) \( -1 \)

Explanation:
1. The H.M. is \( \frac{2ab}{a+b} \). For \( n=-1 \): \( \frac{2}{a^{-1}+b^{-1}}=\frac{2ab}{a+b} \).
2. So \( n=-1 \).

Teacher's Note:
1. Test the candidate values of n.

Answer: (b) \( 2 \)

Explanation:
1. With \( H=\frac{2ab}{a+b} \): \( \frac{H+a}{H-a}=\frac{3b+a}{b-a} \) and \( \frac{H+b}{H-b}=\frac{3a+b}{a-b} \).
2. The sum is \( \frac{(3b+a)-(3a+b)}{b-a}=2 \).

Teacher's Note:
1. Substitute H and simplify.

Answer: (a) \( 2 \)

Explanation:
1. \( H=\frac{2pq}{p+q} \), so \( \frac Hp+\frac Hq=\frac{H(p+q)}{pq}=2 \).

Teacher's Note:
1. Use H = 2pq / (p + q).

Answer: (c) \( a \)

Explanation:
1. \( H=\frac{2xy}{x+y} \) with \( xy=\frac{a^2}{1-a^2b^2} \) and \( x+y=\frac{2a}{1-a^2b^2} \).
2. \( H=\frac{2a^2}{2a}=a \).

Teacher's Note:
1. Compute the sum and product first.

Answer: (a) \( \frac{63}{120} \)

Explanation:
1. There are 6 means between them in the usual sense, so with reciprocals \( \frac13 \) and \( \frac{13}6 \) the A.P. has 8 terms and \( d=\frac{13/6-1/3}7=\frac{11}{42} \).
2. The sixth mean is the 7th term: \( \frac13+6\cdot\frac{11}{42}=\frac{40}{21} \) in reciprocal form, so the mean is \( \frac{21}{40}=\frac{63}{120} \).

Teacher's Note:
1. Convert to an A.P. of reciprocals.

Answer: (c) \( 14 \)

Explanation:
1. The reciprocals form an A.P. from 1 to 31 with \( d=\frac{30}{n+1} \).
2. The \( 7^{th} \) mean has reciprocal \( 1+7d \) and the \( (n-1)^{th} \) has \( 1+(n-1)d \). The ratio of the means is 9 : 5, so \( \frac{1+(n-1)d}{1+7d}=\frac95 \).
3. Substituting d: \( \frac{31n-29}{n+211}=\frac95 \), giving \( 146n=2044 \) and \( n=14 \).

Teacher's Note:
1. The ratio of reciprocals is the inverse of the ratio of the means.

Answer: (c) H.P.

Explanation:
1. Take \( a=1,b=\frac12,c=\frac13 \): the terms are \( \frac65,\frac38,\frac29 \).
2. Their reciprocals \( \frac56,\frac83,\frac92 \) have equal differences \( \frac{11}6 \), so the terms are in H.P. (In general, \( \frac{b+c}a=\frac{a+b+c}a-1 \) and \( \frac{a+b+c}{a} \) are in A.P. when 1/a, 1/b, 1/c are in A.P. after scaling by abc.)

Teacher's Note:
1. Check the reciprocals.

Answer: (c) H.P.

Explanation:
1. Let the coefficients be \( {}^nC_r,{}^nC_{r+1},{}^nC_{r+2},{}^nC_{r+3} \). Then \( \frac{a+b}a=\frac{n+1}{r+1} \), \( \frac{b+c}b=\frac{n+1}{r+2} \), \( \frac{c+d}c=\frac{n+1}{r+3} \).
2. The reciprocals \( \frac{r+1}{n+1},\frac{r+2}{n+1},\frac{r+3}{n+1} \) are in A.P., so the terms are in H.P.

Teacher's Note:
1. Use C(n, r+1) / C(n, r) = (n - r) / (r + 1).

Answer: (b) \( a^n+c^n>2b^n \)

Explanation:
1. For H.P., \( b=\frac{2ac}{a+c}\le\sqrt{ac} \) (H.M. is at most G.M.), with strict inequality for unequal terms.
2. By A.M.-G.M., \( a^n+c^n>2(ac)^{n/2}>2b^n \).

Teacher's Note:
1. Use H.M. less than G.M. less than A.M.

Answer: (c) H.P.

Explanation:
1. \( a^2-b^2=c^2-a^2 \).
2. \( \frac1{b+c}-\frac1{a+c}=\frac{a-b}{(b+c)(a+c)} \) and \( \frac1{a+b}-\frac1{b+c}=\frac{c-a}{(a+b)(b+c)} \).
3. Equality of these needs \( (a-b)(a+b)=(c-a)(c+a) \), which is the given condition. So the reciprocals are in A.P. and the terms are in H.P.

Teacher's Note:
1. Compare the differences of the reciprocals.

Answer: (c) \( (n-1)a_1a_n \)

Explanation:
1. If \( \frac1{a_{k+1}}-\frac1{a_k}=d \), then \( a_ka_{k+1}=\frac{a_k-a_{k+1}}d \).
2. The sum telescopes to \( \frac{a_1-a_n}d \). Also \( \frac1{a_n}-\frac1{a_1}=(n-1)d \) gives \( \frac{a_1-a_n}d=(n-1)a_1a_n \).

Teacher's Note:
1. Write each product as a difference.

Answer: (b) \( 2\log(x-z) \)

Explanation:
1. The expression is \( \log\{(x+z)^2-2y(x+z)\} \). With \( y=\frac{2xz}{x+z} \): \( (x+z)^2-4xz=(x-z)^2 \).
2. So the value is \( 2\log(x-z) \).

Teacher's Note:
1. Substitute y = 2xz / (x + z).

Answer: (c) both (a) and (b)

Explanation:
1. Let the reciprocals be \( p,q,r,s \) in A.P., so \( q+r=p+s \) and \( qr>ps \) (the product of the middle terms exceeds that of the ends).
2. So \( ad=\frac1{ps}>\frac1{qr}=bc \). Also by convexity of \( \frac1t \), \( \frac1p+\frac1s>\frac1q+\frac1r \), that is \( a+d>b+c \).
3. Both are true.

Teacher's Note:
1. Look at the A.P. of reciprocals.

Answer: (d) \( \frac1{(1-x)^2} \)

Explanation:
1. This is an A.G.P. \( S=\frac1{1-x}+\frac{x}{(1-x)^2}=\frac1{(1-x)^2} \).

Teacher's Note:
1. Subtract x S from S.

Answer: (b) \( \frac{2000}{9801} \)

Explanation:
1. The terms are \( 2(0.1),\ 4(0.1)^3,\ 6(0.1)^5,\dots \), that is \( 20k(0.01)^k \).
2. \( S=20\cdot\frac{0.01}{(1-0.01)^2}=\frac{0.2}{0.9801}=\frac{2000}{9801} \).

Teacher's Note:
1. Write each term as k times a power of 0.01.

Answer: (d) \( \frac{35}{16} \)

Explanation:
1. The terms are \( (3k-2)x^{k-1} \) with \( x=\frac15 \).
2. \( S=\frac{3}{(1-x)^2}-\frac2{1-x}=\frac{75}{16}-\frac52=\frac{35}{16} \).

Teacher's Note:
1. Split into k x^(k-1) and x^(k-1) parts.

Answer: (d) \( \frac1{(1-x)^3} \)

Explanation:
1. The coefficients are the triangular numbers \( \frac{(k+1)(k+2)}2 \), which are the coefficients in the expansion of \( (1-x)^{-3} \).

Teacher's Note:
1. Compare with the binomial series.

Answer: (b) \( 2 \)

Explanation:
1. The exponent is \( \frac14+\frac28+\frac3{16}+\frac4{32}+\dots=\frac12\sum_{k\ge1}\frac k{2^k}=\frac12\cdot2=1 \).
2. The value is 2.

Teacher's Note:
1. Add the exponents of 2.

Answer: (a) \( 50(1-i) \)

Explanation:
1. The \( k^{th} \) term is \( ki^k \).
2. Each group of four terms gives \( i-2-3i+4=2-2i \).
3. There are 25 groups, so \( S=50-50i=50(1-i) \).

Teacher's Note:
1. Group four terms at a time.

Answer: (c) \( \frac{n^2+n+2}2 \)

Explanation:
1. The differences 2, 3, 4, ... are in A.P.
2. \( T_n=2+\sum_{k=2}^{n}k=2+\frac{n(n+1)}2-1=\frac{n^2+n+2}2 \).

Teacher's Note:
1. Use the method of differences.

Answer: (d) \( 49^2+2 \)

Explanation:
1. The differences are 1, 3, 5, 7, ... so \( t_n=2+(n-1)^2 \).
2. \( t_{50}=49^2+2 \).

Teacher's Note:
1. The differences are odd numbers.

Answer: (b) \( 2(10^{n+1}-9n-10)/27 \)

Explanation:
1. The \( k^{th} \) term is \( \frac23(10^k-1) \).
2. \( S=\frac23\left[\frac{10(10^n-1)}9-n\right]=\frac{2(10^{n+1}-9n-10)}{27} \).

Teacher's Note:
1. Write each term as 6 times (10 to the k minus 1) / 9.

Answer: (d) \( 4(2^n-1)+8n \)

Explanation:
1. The differences 4, 8, 16, ... are a G.P., so \( t_n=8+2^{n+1} \).
2. \( S_n=8n+4(2^n-1) \).

Teacher's Note:
1. The terms are 8 plus a power of 2.

Answer: (c) \( \frac1{(1-b)(1-ab)} \)

Explanation:
1. The \( k^{th} \) term is \( b^k\frac{1-a^{k+1}}{1-a} \).
2. \( S=\frac1{1-a}\left[\frac1{1-b}-\frac a{1-ab}\right]=\frac1{(1-b)(1-ab)} \).

Teacher's Note:
1. Sum each G.P. separately.