JEE Mathematics Sequence and Series MCQs Set 42

Answer: (c) \(\dfrac{n(4n^2-1)c^2}3\)

Explanation:
1. \(a_n=S_n-S_{n-1}=c(2n-1)\).
2. \(\sum a_k^2=c^2\sum(2k-1)^2=c^2\dfrac{n(4n^2-1)}3\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (d) \(\tan a_n-\tan a_1\)

Explanation:
1. \(\tan a_{k+1}-\tan a_k=\dfrac{\sin(a_{k+1}-a_k)}{\cos a_k\cos a_{k+1}}=\sin d\sec a_k\sec a_{k+1}\).
2. Summing from k = 1 to n - 1 telescopes to tan a_n - tan a_1.

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (a) -23

Explanation:
1. The middle terms are \(t_n=1\) and \(t_{n+1}=7\), so \(d=6\). Then \(t_1=1-6(n-1)=7-6n\) and \(t_{2n}=7+6(n-1)=6n+1\).
2. \((7-6n)(6n+1)+713\ge0\) gives \(-36n^2+36n+720\ge0\), i.e. \(n^2-n-20\le0\), so \(n\le5\).
3. For \(n=5\): \(t_1=7-30=-23\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (b) \(0<x<10\)

Explanation:
1. \(\dfrac x{1-r}=5\) gives \(r=1-\dfrac x5\).
2. Convergence needs \(|r|<1\): \(-1<1-\tfrac x5<1\), so \(0<x<10\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (c) \(c\Delta=0\)

Explanation:
1. Let \(s=\alpha+\beta\) and \(p=\alpha\beta\). G.P.: \((\alpha^2+\beta^2)^2=(\alpha+\beta)(\alpha^3+\beta^3)\), i.e. \((s^2-2p)^2=s(s^3-3sp)\).
2. This simplifies to \(s^2p=4p^2\), i.e. \(p(s^2-4p)=0\).
3. With \(p=c/a\) and \(s^2-4p=\Delta/a^2\): \(c\Delta=0\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (a) \(3-\dfrac32+\dfrac34-\dfrac38+\cdots\)

Explanation:
1. \(\dfrac a{1-r}=2\) and \(\dfrac{a^3}{1-r^3}=24\). With \(a=2(1-r)\): \(\dfrac{8(1-r)^3}{1-r^3}=24\) gives \((1-r)^2=3(1+r+r^2)\).
2. \(2r^2+5r+2=0\) so \(r=-\tfrac12\) (the other root has |r| > 1) and \(a=3\).
3. The series is \(3-\tfrac32+\tfrac34-\tfrac38+\cdots\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (b) obtuse angled

Explanation:
1. The A.P. condition gives \(\left(\dfrac{2b}{3c}\right)^2=\dfrac a{2b}\cdot\dfrac{3c}a=\dfrac{3c}{2b}\), so \(8b^3=27c^3\) and \(b=\dfrac32c\).
2. G.P.: \(b^2=ac\) gives \(a=\dfrac94c\). Sides in the ratio 9 : 6 : 4.
3. The largest angle: \(\cos A=\dfrac{36+16-81}{2\cdot6\cdot4}=-\dfrac{29}{48}<0\), so the triangle is obtuse.

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (c) -1

Explanation:
1. Sides \(a,ar,ar^2\) form a triangle iff \(a+ar>ar^2\), i.e. \(r^2-r-1<0\), so \(r<\dfrac{1+\sqrt5}2\approx1.618\).
2. So \(1<r<1.618\) is not an integer: \([r]=1\), \([-r]=-2\) and the sum is -1.

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (a) \(p^2-q^2=P^2-Q^2\)

Explanation:
1. \(p^2-P^2=\left(\dfrac{a+b}2\right)^2-ab=\dfrac{(a-b)^2}4\) and \(q^2-Q^2=\dfrac{(b-c)^2}4\).
2. In an A.P. \(a-b=b-c\), so \(p^2-P^2=q^2-Q^2\), i.e. \(p^2-q^2=P^2-Q^2\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (d) \(99\cdot2^{100}+1\)

Explanation:
1. Let \(S=\sum_{k=1}^{100}k2^{k-1}\). Then \(2S=\sum k2^k\).
2. \(2S-S=100\cdot2^{100}-(1+2+\cdots+2^{99})=100\cdot2^{100}-(2^{100}-1)=99\cdot2^{100}+1\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (a) \(1-\dfrac1{(n+1)2^n}\)

Explanation:
1. The k-th term is \(\dfrac{k+2}{k(k+1)}\cdot\dfrac1{2^k}=\left(\dfrac2k-\dfrac1{k+1}\right)\dfrac1{2^k}=\dfrac2{k2^k}-\dfrac2{(k+1)2^{k+1}}\).
2. The sum telescopes to \(\dfrac2{1\cdot2}-\dfrac2{(n+1)2^{n+1}}=1-\dfrac1{(n+1)2^n}\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (b) 0.6

Explanation:
1. \(\dfrac a{1-r}=20\) and \(\dfrac{a^2}{1-r^2}=100\).
2. With \(a=20(1-r)\): \(\dfrac{400(1-r)^2}{1-r^2}=100\) gives \(\dfrac{4(1-r)}{1+r}=1\), so \(r=\dfrac35=0.6\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (b) 189

Explanation:
1. \(abc=6^6\) and \(b^2=ac\) give \(b^3=6^6\), so \(b=36\) and \(ac=1296\).
2. \(b-a=36-a\) is a perfect cube less than 36: try 1, 8, 27. \(a=35,28\) do not divide 1296 (\(a=9\) works): \(a=9,\ b=36,\ c=144\) (ratio 4, \(b-a=27=3^3\)).
3. \(a+b+c=189\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (c) 3

Explanation:
1. After the first term, the terms are \(\dfrac{4k-2}{3^k}\) for \(k=1,2,\ldots\).
2. \(\sum\dfrac k{3^k}=\dfrac{1/3}{(1-1/3)^2}=\dfrac34\) and \(\sum\dfrac1{3^k}=\dfrac12\), so the sum is \(1+4\cdot\dfrac34-2\cdot\dfrac12=3\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (a) 8

Explanation:
1. Let the common difference be \(e\ne0\): \(b=a+e,\ c=a+2e,\ d=a+3e\). The equation becomes \(-2e+ke^2+8e^3=-6e+4e^2-e^3\).
2. Dividing by e: \(9e^2+(k-4)e+4=0\). A real nonzero e exists iff \((k-4)^2\ge144\), i.e. \(k\ge16\) or \(k\le-8\).
3. The smallest positive k is 16, and half of it is 8.

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (c) 1

Explanation:
1. \(27^S=3^{3S}=3^6\) gives \(S=2\). The series is a G.P.: \(S=\dfrac1{1-|\cos x|}\).
2. \(1-|\cos x|=\tfrac12\) so \(|\cos x|=\tfrac12\) and \(x=\tfrac\pi3,\tfrac{2\pi}3\); the sum is \(\pi\), so the value is 1.

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (b) 6

Explanation:
1. A.P.: \(2(2a+b)=(5a-b)+(a+2b)\) gives \(b=2a\).
2. G.P.: \((ab+1)^2=(b+1)^2(a-1)^2\) so \(ab+1=\pm(b+1)(a-1)\). With \(b=2a\): the + sign gives \(a=-2\) (rejected); the - sign gives \(4a^2-a=0\), so \(a=\tfrac14,\ b=\tfrac12\).
3. \(a^{-1}+b^{-1}=4+2=6\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (d) 2

Explanation:
1. Let the quantities be \(x,y\): \(p=\dfrac{2x+y}3\), \(q=\dfrac{x+2y}3\), \(G^2=xy\).
2. \(2p-q=x\) and \(2q-p=y\), so \((2p-q)(2q-p)=xy=G^2\): k = 2.

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (b) G.P.

Explanation:
1. By the Cauchy-Schwarz inequality \((a^2+b^2+c^2)(b^2+c^2+d^2)\ge(ab+bc+cd)^2\) always.
2. So the given inequality forces equality, which holds iff a/b = b/c = c/d: a, b, c, d are in G.P.

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (c) 6

Explanation:
1. The expression has six terms \(ab^2,ac^2,bc^2,ba^2,ca^2,cb^2\) whose product is \((abc)^6\).
2. By AM-GM their sum is at least \(6\sqrt[6]{(abc)^6}=6abc\), with equality at \(a=b=c\). So \(\lambda=6\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (d) \(2n\)

Explanation:
1. \(\tfrac1a,\tfrac1{H_1},\ldots,\tfrac1{H_n},\tfrac1b\) are in A.P. with common difference \(d\), and \(\tfrac1b-\tfrac1a=(n+1)d\).
2. \(\dfrac{H_1+a}{H_1-a}=\dfrac{\frac1a+\frac1{H_1}}{\frac1a-\frac1{H_1}}=\dfrac{2/a+d}{-d}\) and \(\dfrac{H_n+b}{H_n-b}=\dfrac{2/b-d}d\) (using \(\tfrac1{H_1}=\tfrac1a+d\) and \(\tfrac1{H_n}=\tfrac1b-d\)).
3. Sum \(=\dfrac{2(1/b-1/a)}d-2=2(n+1)-2=2n\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (d) \(a,b,-\dfrac c2\text{ are in G.P.}\)

Explanation:
1. A.P.: \(b=\dfrac{a+c}2\). H.P.: \(\dfrac2{b^2}=\dfrac1{a^2}+\dfrac1{c^2}\), so \(2a^2c^2=b^2(a^2+c^2)\).
2. With \(s=a+c,\ p=ac\): \(8p^2=s^2(s^2-2p)\), i.e. \((s^2-4p)(s^2+2p)=0\). \(s^2=4p\) means \(a=c\) (all equal, excluded), so \(s^2=-2p\).
3. Then \(b^2=\dfrac{s^2}4=-\dfrac{ac}2\): a, b, -c/2 are in G.P.

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (c) \(\dfrac{10}{(1+x)(1+11x)}\)

Explanation:
1. \(\dfrac1{(1+kx)(1+(k+1)x)}=\dfrac1x\left(\dfrac1{1+kx}-\dfrac1{1+(k+1)x}\right)\).
2. The sum telescopes: \(S_n=\dfrac1x\left(\dfrac1{1+x}-\dfrac1{1+(n+1)x}\right)=\dfrac n{(1+x)(1+(n+1)x)}\), so \(S_{10}=\dfrac{10}{(1+x)(1+11x)}\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (a) \(xyz=x+y+z\)

Explanation:
1. \(x=\dfrac1{\sin^2\phi}\), \(y=\dfrac1{\cos^2\phi}\), so \(x+y=\dfrac1{\sin^2\phi\cos^2\phi}=xy\).
2. \(z=\dfrac1{1-\sin^2\phi\cos^2\phi}=\dfrac{xy}{xy-1}\). Then \(xyz=\dfrac{(xy)^2}{xy-1}\) and \(x+y+z=xy+\dfrac{xy}{xy-1}=\dfrac{(xy)^2}{xy-1}\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (b) 6

Explanation:
1. A.P.: \(d=\dfrac19\), so \(a_4=2+\dfrac39=\dfrac73\).
2. H.P.: reciprocals go from \(\tfrac12\) to \(\tfrac13\) with common difference \(-\tfrac1{54}\), so \(\tfrac1{h_7}=\tfrac12-\tfrac6{54}=\tfrac7{18}\) and \(h_7=\tfrac{18}7\).
3. \(a_4h_7=\dfrac73\cdot\dfrac{18}7=6\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (c) 5

Explanation:
1. The product of \((1+x^{2^k})\) for \(k=0,\ldots,7\) equals \(\dfrac{1-x^{256}}{1-x}=1+x+\cdots+x^{255}\).
2. So \(n=255\), with unit digit 5.

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (d) 4

Explanation:
1. The k-th term (k = 0, 1, ...) is \(\dfrac{k!}{(k+5)!}=\dfrac14\left[\dfrac{k!}{(k+4)!}-\dfrac{(k+1)!}{(k+5)!}\right]\).
2. The sum of n terms telescopes to \(\dfrac14\left[\dfrac1{4!}-\dfrac{n!}{(n+4)!}\right]\), so a = 4, b = 4, c = 0, d = 4 and \(a+b-c-d=4\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (a) G.P.

Explanation:
1. \(\tfrac1a,\tfrac1b,\tfrac1c\) are in A.P., so \(-\tfrac1a,-\tfrac1b,-\tfrac1c\) are in A.P.
2. The exponential of an A.P. is a G.P.

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (b) \(\dfrac1{12}n(n+1)(3n^2+n+2)\)

Explanation:
1. \(V_r=\dfrac r2[2r+(r-1)(2r-1)]=\dfrac{r(2r^2-r+1)}2\).
2. \(\sum V_r=\dfrac12\left[2\sum r^3-\sum r^2+\sum r\right]=\dfrac{n(n+1)}{12}(3n^2+n+2)\) (check n = 1: V_1 = 1, and n = 2: 1 + 5 = 6, matching the formula).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (d) a composite number

Explanation:
1. \(V_{r+1}-V_r=\dfrac{(r+1)(2r^2+3r+2)-r(2r^2-r+1)}2=3r^2+2r+1\).
2. \(T_r=3r^2+2r-1=(3r-1)(r+1)\): a product of two integers each at least 2, so it is composite (it is not always odd or even, e.g. T_1 = 4, T_2 = 15).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (a) 7

Explanation:
1. \(\dfrac{n(n+1)}2=\dfrac15\cdot\dfrac{n(n+1)(2n+1)}6\).
2. \(\dfrac12=\dfrac{2n+1}{30}\) gives \(n=7\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (b) 4

Explanation:
1. The harmonic mean is \(\dfrac{2r_1r_2}{r_1+r_2}\).
2. \(r_1r_2=\dfrac{8+2\sqrt3}{5+\sqrt2}\) and \(r_1+r_2=\dfrac{4+\sqrt3}{5+\sqrt2}\), so the harmonic mean is \(\dfrac{2(8+2\sqrt3)}{4+\sqrt3}=\dfrac{4(4+\sqrt3)}{4+\sqrt3}=4\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (d) 1

Explanation:
1. \(ar^{n-1}=128\): \(a\cdot2^{n-1}=128\). Sum: \(a(2^n-1)=255\).
2. Dividing: \(\dfrac{2(2^n-1)}{2^n}=\dfrac{255}{128}\) gives \(2^n=256\), so n = 8 and \(a=128/128=1\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (a) 32

Explanation:
1. \(a+b=12\), \(G^2=ab\) and \(H=\dfrac{2ab}{a+b}=\dfrac{ab}6\).
2. \(ab+\dfrac{ab}2=48\) gives \(ab=32\), so \(G^2=32\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (d) 308

Explanation:
1. \(S_n=S_{n-1}\cdot(-1)+n^3\), so \(S_1=1,\ S_2=8-1=7,\ S_3=27-7=20,\ S_4=64-20=44,\ S_5=125-44=81\).
2. For odd \(n=2m+1\) group the terms: \(S_n=1+\sum_{j=1}^m[(2j+1)^3-(2j)^3]=1+\sum(12j^2+6j+1)=1+2m(m+1)(2m+1)+3m(m+1)+m\).
3. For \(n=39,\ m=19\): \(1+2\cdot19\cdot20\cdot39+3\cdot380+19=1+29640+1140+19=30800\), so \(S_{39}/100=308\).

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.

Answer: (c) 38

Explanation:
1. Multiples of 7 from 105 to 497: (497 - 105)/7 + 1 = 57.
2. Multiples of 21 from 105 to 483: (483 - 105)/21 + 1 = 19.
3. Divisible by 7 but not by 21: 57 - 19 = 38.

Teacher's Note:
1. Identify whether the terms form an A.P., G.P. or H.P. and use the corresponding standard formulas.
2. For series with a polynomial times a power of a number, use the difference trick or telescoping.