JEE Mathematics Sequence and Series MCQs Set 37

Answer: (a) H.P.

Explanation:
1. \( x=\dfrac1{1-a},\ y=\dfrac1{1-b},\ z=\dfrac1{1-c} \), so \( \dfrac1x=1-a,\ \dfrac1y=1-b,\ \dfrac1z=1-c \).
2. Since a, b, c are in A.P., so are \( 1-a,1-b,1-c \). The reciprocals of x, y, z are in A.P., hence x, y, z are in H.P.

Teacher's Note:
1. Write each sum as 1/(1 - ratio).
2. Reciprocals in A.P. means the numbers are in H.P.

Answer: (a) \( 2n-H_n \)

Explanation:
1. The \( k^{th} \) term is \( \dfrac{2k-1}{k}=2-\dfrac1k \).
2. The sum is \( 2n-\left(1+\dfrac12+\dots+\dfrac1n\right)=2n-H_n \).

Teacher's Note:
1. Split each fraction into an integer part and a reciprocal.
2. The constants add up to 2n.

Answer: (a) 1

Explanation:
1. The sum of n A.M.s between a and b is \( \dfrac{n(a+b)}{2} \).
2. The reciprocals of the H.M.s are n A.M.s between \( \dfrac1a \) and \( \dfrac1b \), whose sum is \( \dfrac n2\left(\dfrac1a+\dfrac1b\right)=\dfrac{n(a+b)}{2ab} \).
3. Equating the two sums gives \( ab=1 \).

Teacher's Note:
1. Reciprocals of harmonic means are arithmetic means of the reciprocals.
2. The sum of n A.M.s depends only on the end terms.

Answer: (a) \( \dfrac{n}{\sqrt a+\sqrt{a+nx}} \)

Explanation:
1. Rationalise: \( \dfrac{1}{\sqrt{a+rx}+\sqrt{a+(r-1)x}}=\dfrac{\sqrt{a+rx}-\sqrt{a+(r-1)x}}{x} \).
2. The sum telescopes to \( \dfrac{\sqrt{a+nx}-\sqrt a}{x} \). Multiplying and dividing by \( \sqrt{a+nx}+\sqrt a \): \( \dfrac{(a+nx)-a}{x(\sqrt a+\sqrt{a+nx})}=\dfrac{n}{\sqrt a+\sqrt{a+nx}} \).

Teacher's Note:
1. Rationalising each term makes the series telescope.
2. Rationalise once more to match the given form.

Answer: (a) 2056

Explanation:
1. If the 2011th term is N, then \( N-\lfloor\sqrt N\rfloor=2011 \), since \( \lfloor\sqrt N\rfloor \) perfect squares are at most N.
2. Try \( N=2056 \): \( \lfloor\sqrt{2056}\rfloor=45 \) (as \( 45^2=2025\le2056<2116 \)) and \( 2056-45=2011 \). Also 2056 is not a perfect square.

Teacher's Note:
1. Count how many perfect squares are below the required term.
2. Verify the final answer is not itself a perfect square.

Answer: (a) \( \dfrac{3^{10}\cdot2^4}{7^7} \)

Explanation:
1. Write \( \dfrac a2,\dfrac a2,\dfrac b3,\dfrac b3,\dfrac b3,\dfrac c2,\dfrac c2 \), seven positive numbers with sum 3. By A.M.-G.M. their product is at most \( \left(\dfrac37\right)^7 \).
2. So \( \dfrac{a^2b^3c^2}{2^2\cdot3^3\cdot2^2}\le\dfrac{3^7}{7^7} \), i.e. \( a^2b^3c^2\le\dfrac{3^{10}\cdot2^4}{7^7} \), with equality at \( a=c=\dfrac67,\ b=\dfrac97 \).

Teacher's Note:
1. Split the variables into equal parts so that the weights match the exponents.
2. Equality holds when all the parts are equal.

Answer: (c) 5665

Explanation:
1. The \( k^{th} \) term is \( (2k-1)k^2=2k^3-k^2 \).
2. \( S_{10}=2\left(\dfrac{10\cdot11}{2}\right)^2-\dfrac{10\cdot11\cdot21}{6}=2\times3025-385=5665 \).

Teacher's Note:
1. Find the general term from the pattern of the factors.
2. Use the formulas for the sum of cubes and the sum of squares.

Answer: (a) Statement 1 is true, Statement 2 is true and Statement 2 is the correct explanation of Statement 1

Explanation:
1. Statement 2: for \( a,ar,ar^2 \) adding the middle term gives \( a(1+r),\ 2ar,\ ar(1+r) \). Their reciprocals \( \dfrac1{a(1+r)},\dfrac1{2ar},\dfrac1{ar(1+r)} \) satisfy \( \dfrac1{a(1+r)}+\dfrac1{ar(1+r)}=\dfrac{r+1}{ar(1+r)}=\dfrac1{ar}=2\cdot\dfrac1{2ar} \), so they are in A.P. and the new terms are in H.P. True.
2. Statement 1 is a particular case of this: adding 6 to 3, 6, 12 gives 9, 12, 18, which are in H.P. (reciprocals \( \tfrac19,\tfrac1{12},\tfrac1{18} \) have the common difference \( \tfrac1{36} \)). So Statement 1 is true and follows from Statement 2.

Teacher's Note:
1. Prove the general statement first and then apply it to numbers.
2. Statement 1 is just a special case of Statement 2.

Answer: (c) 23

Explanation:
1. The \( k^{th} \) term is \( \dfrac{(2k-1)(2k+1)}{2^k}=\dfrac{4k^2-1}{2^k} \).
2. Using \( \sum\dfrac{k^2}{2^k}=6 \) and \( \sum\dfrac1{2^k}=1 \): the sum is \( 4\times6-1=23 \).

Teacher's Note:
1. Split the general term into a k squared part and a constant part.
2. The sum of k squared over 2 to the k is 6.

Answer: (d) Statement 1 is false, Statement 2 is true

Explanation:
1. The differences 1, 2, 3, 4, 5, 6 form an A.P., so \( t_k=1+\dfrac{k(k-1)}{2} \), a quadratic in k. Statement 2 is true.
2. \( S_{30}=30+\dfrac12\left(\sum k^2-\sum k\right)=30+\dfrac12(9455-465)=30+4495=4525 \), not 4520. Statement 1 is false.

Teacher's Note:
1. Compute the sum directly before accepting a given value.
2. Second differences constant means a quadratic general term.

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

Explanation:
1. Let \( t=5^x+5^{-x}\ge2 \). The first term is \( 5t \) and the third term is \( 25^x+25^{-x}=t^2-2 \).
2. A.P. means \( a=5t+t^2-2 \), an increasing function of \( t\ge2 \), so \( a\ge10+4-2=12 \).

Teacher's Note:
1. Express everything in terms of 5 to the x plus 5 to the minus x.
2. Use the minimum value 2 of t from A.M.-G.M.

Answer: (d) \( \left(\dfrac{\sqrt a-\sqrt b}{\sqrt2}\right)^2 \)

Explanation:
1. \( P-Q=\dfrac{a+b}{2}-\sqrt{ab}=\dfrac{a+b-2\sqrt{ab}}{2} \).
2. This equals \( \dfrac{(\sqrt a-\sqrt b)^2}{2}=\left(\dfrac{\sqrt a-\sqrt b}{\sqrt2}\right)^2 \).

Teacher's Note:
1. Recognise the numerator as a perfect square.
2. This shows directly that P is never smaller than Q.

Answer: (b) \( x^3-3Ax^2+3\dfrac{G^3}{H}x-G^3=0 \)

Explanation:
1. \( a+b+c=3A \) and \( abc=G^3 \).
2. \( H=\dfrac{3}{\frac1a+\frac1b+\frac1c}=\dfrac{3abc}{ab+bc+ca}\Rightarrow ab+bc+ca=\dfrac{3G^3}{H} \).
3. The cubic is \( x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0 \), i.e. option (b).

Teacher's Note:
1. Find the sum, the sum of products in pairs and the product of the roots.
2. Use the formula for the harmonic mean to get the middle coefficient.

Answer: (d) in H.P.

Explanation:
1. The four products are \( \dfrac{abcd}{d},\ \dfrac{abcd}{c},\ \dfrac{abcd}{b},\ \dfrac{abcd}{a} \).
2. Since d, c, b, a are in A.P., the constant abcd times their reciprocals is an H.P. For example, with 1, 2, 3, 4 the products are 6, 8, 12, 24, whose reciprocals \( \tfrac16,\tfrac18,\tfrac1{12},\tfrac1{24} \) have the constant difference \( -\tfrac1{24} \).

Teacher's Note:
1. Test with a simple A.P. such as 1, 2, 3, 4.
2. The reciprocals of the products form an A.P.

Answer: (a) 100

Explanation:
1. The reciprocals are in A.P.: \( t_3=3,\ t_6=5 \), so \( 3d=2\Rightarrow d=\dfrac23 \).
2. The last term is \( \dfrac{203}{3} \): \( 3+(n-3)\dfrac23=\dfrac{203}3\Rightarrow n-3=97\Rightarrow n=100 \).

Teacher's Note:
1. Reciprocals of an H.P. form an A.P.
2. Use the general term of an A.P. for the last term.

Answer: (a) 144 cm

Explanation:
1. Each new triangle has half the side, hence half the perimeter. The perimeters are \( 72,36,18,\dots \).
2. Sum \( =\dfrac{72}{1-\frac12}=144 \) cm.

Teacher's Note:
1. The midpoint triangle has sides equal to half of the original sides.
2. Use the infinite G.P. sum formula.

Answer: (b) \( 3\left(1-\dfrac1{3^{20}}\right) \)

Explanation:
1. The sequence is a G.P. with first term 2 and ratio \( \dfrac13 \).
2. \( S_{20}=\dfrac{2\left(1-3^{-20}\right)}{1-\frac13}=3\left(1-\dfrac1{3^{20}}\right) \).

Teacher's Note:
1. The constant ratio shows that it is a geometric progression.
2. The first option is an A.P. formula and does not apply.

Answer: (c) 3

Explanation:
1. Here \( a=b+1 \) and the common difference is \( d=\dfrac{b-a}{7}=-\dfrac17 \).
2. \( A_1^2-A_2^2+A_3^2-\dots=(A_1-A_2)(A_1+A_2)+(A_3-A_4)(A_3+A_4)+(A_5-A_6)(A_5+A_6)=\dfrac17(A_1+A_2+\dots+A_6) \).
3. The sum of the six means is \( 3(a+b)=3(2b+1) \), so the expression equals \( \dfrac{3(2b+1)}{7} \). For \( b=3 \) it is 3, a prime; \( b=1,2,4 \) do not give an integer.

Teacher's Note:
1. Group the alternating squares in pairs and use difference of squares.
2. Test each option; only one gives a prime.

Answer: (a) \( 0<M\le1 \)

Explanation:
1. All numbers are positive, so \( M>0 \).
2. By A.M.-G.M. on \( (a+b) \) and \( (c+d) \): \( M\le\left(\dfrac{a+b+c+d}{2}\right)^2=1 \).
3. Hence \( 0<M\le1 \).

Teacher's Note:
1. Apply A.M.-G.M. to the two factors whose sum is fixed.
2. The lower bound is positive but can be arbitrarily close to 0.

Answer: (c) a prime number

Explanation:
1. The reciprocals of the roots are in A.P., so the middle root satisfies \( \dfrac2{\beta}=\dfrac1\alpha+\dfrac1\gamma \).
2. \( \dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma=\dfrac{\sum\alpha\beta}{\alpha\beta\gamma}=\dfrac{36}{36}=1 \), so \( \dfrac3\beta=1\Rightarrow\beta=3 \), a prime number.

Teacher's Note:
1. The middle term of an A.P. is the average of the three terms.
2. The sum of reciprocals of the roots comes from the coefficients.

Answer: (a) 9

Explanation:
1. The series is \( 3\sum_{k\ge0}\dfrac1{4^k}+d\sum_{k\ge1}\dfrac{k}{4^k} \).
2. \( \sum\dfrac1{4^k}=\dfrac43 \) and \( \sum_{k\ge1}\dfrac{k}{4^k}=\dfrac{1/4}{(1-1/4)^2}=\dfrac49 \).
3. So \( 4+\dfrac{4d}{9}=8\Rightarrow d=9 \).

Teacher's Note:
1. Separate the series into a geometric series and an arithmetico-geometric series.
2. Use the standard sum of k x^k.

Answer: (d) 16

Explanation:
1. The means are \( ar,ar^2,ar^3 \) with \( ar^4=b \).
2. \( \dfrac{G_3-G_2}{G_2-G_1}=\dfrac{ar^2(r-1)}{ar(r-1)}=r=2 \).
3. \( \dfrac ba=r^4=16 \).

Teacher's Note:
1. The ratio of successive differences of a G.P. equals its common ratio.
2. Four ratios separate a and b.

Answer: (d) 2

Explanation:
1. \( b=\dfrac{2ac}{a+c} \). Then \( b-a=\dfrac{a(c-a)}{a+c} \), \( b+a=\dfrac{a(3c+a)}{a+c} \), \( b-c=\dfrac{c(a-c)}{a+c} \), \( b+c=\dfrac{c(3a+c)}{a+c} \).
2. The expression is \( \dfrac{3c+a}{c-a}+\dfrac{3a+c}{a-c}=\dfrac{(3c+a)-(3a+c)}{c-a}=\dfrac{2(c-a)}{c-a}=2 \).

Teacher's Note:
1. Substitute the formula for the harmonic mean.
2. The factors c - a cancel at the end.

Answer: (c) \( a=2,\ b=32 \)

Explanation:
1. Let the G.P. be \( \alpha,\alpha r,\alpha r^2,\alpha r^3 \). Then \( \alpha(1+r)=3 \) and \( \alpha r^2(1+r)=12 \), so \( r^2=4,\ r=2 \).
2. \( \alpha=1 \), so the numbers are 1, 2, 4, 8. Then \( a=\alpha\beta=2 \) and \( b=\gamma\delta=32 \).

Teacher's Note:
1. Divide the two sum equations to find the common ratio.
2. The G.P. must be increasing, so r is taken as 2 and not as -2.