JEE Mathematics Sequence and Series MCQs Set 21

Answer: (a) \( \frac18n^3(n^2+1)(n^2+3) \)

Explanation:
1. The \( n^{th} \) group ends at the number \( T_n=\frac{n(n+1)}2 \) and starts after \( T_{n-1} \). Using \( \sum_{1}^{m}k^3=\left(\frac{m(m+1)}2\right)^2 \), the group sum is \( \left(\frac{T_n(T_n+1)}2\right)^2-\left(\frac{T_{n-1}(T_{n-1}+1)}2\right)^2 \).
2. Simplifying gives \( \frac18n^3(n^2+1)(n^2+3) \). Check: n = 2 gives 35 = 8 + 27.

Teacher's Note:
1. Test n = 1, 2.

Answer: (c) \( \pi^2/8 \)

Explanation:
1. The even terms give \( \frac1{4}\left(\frac1{1^2}+\frac1{2^2}+\dots\right)=\frac{\pi^2}{24} \).
2. Odd part \( =\frac{\pi^2}6-\frac{\pi^2}{24}=\frac{\pi^2}8 \).

Teacher's Note:
1. Subtract the sum over even numbers.

Answer: (b) \( \frac{\sqrt{a+nx}-\sqrt a}x \)

Explanation:
1. Rationalise: \( \frac1{\sqrt{a+rx}+\sqrt{a+(r-1)x}}=\frac{\sqrt{a+rx}-\sqrt{a+(r-1)x}}x \).
2. The sum telescopes to \( \frac{\sqrt{a+nx}-\sqrt a}x \).

Teacher's Note:
1. Multiply by the conjugate.

Answer: (a) \( f(2n)-16f(n) \) for all \( n\in N \)

Explanation:
1. The odd numbers \( 1,3,\dots,2n-1 \) are all the numbers up to 2n minus the even ones.
2. \( \sum_{r=1}^n(2r-1)^4=f(2n)-\sum_{r=1}^n(2r)^4=f(2n)-16f(n) \).
3. For n = 3, (b) gives 82 but the sum is 707, so (b) is false.

Teacher's Note:
1. Subtract the sum of fourth powers of even numbers.

Answer: (b) \( 2\sqrt{ab}\lt (a+b) \)

Explanation:
1. \( a+b-2\sqrt{ab}=(\sqrt a-\sqrt b)^2>0 \) for \( a\ne b \).

Teacher's Note:
1. A.M. is greater than G.M. for unequal numbers.

Answer: (d) \( a=b=c \)

Explanation:
1. A.P. gives \( 2b=a+c \) and G.P. gives \( b^2=ac \).
2. Then \( \left(\frac{a+c}2\right)^2=ac \), so \( (a-c)^2=0 \), \( a=c \) and hence \( a=b=c \).

Teacher's Note:
1. Eliminate b.

Answer: (a) A.P.

Explanation:
1. \( b^2=ac \) gives \( 2\log b=\log a+\log c \).

Teacher's Note:
1. Take logarithms of the G.P. condition.

Answer: (d) \( G^2=AH \)

Explanation:
1. \( A\cdot H=\frac{a+b}2\cdot\frac{2ab}{a+b}=ab=G^2 \).

Teacher's Note:
1. Multiply A and H.

Answer: (b) (4, 9)

Explanation:
1. \( ab=36 \) and \( a+b=13 \), so the numbers are the roots of \( t^2-13t+36=0 \): 4 and 9.

Teacher's Note:
1. Form the quadratic from the sum and product.

Answer: (d) \( 8 \)

Explanation:
1. Let the numbers be \( a,ar,ar^2,a \). The last three are in A.P. with difference 6: \( ar^2-ar=6 \) and \( a-ar^2=6 \).
2. Adding: \( a-ar=12 \), and the first gives \( -ar(1-r)=6 \), so \( r=-\frac12 \) and \( a=8 \).

Teacher's Note:
1. Use the two difference equations.

Answer: (b) \( 18 \)

Explanation:
1. \( G^2=AH=27\cdot12=324 \), so \( G=18 \).

Teacher's Note:
1. G squared is A times H.

Answer: (b) \( \frac{144}{15},12,15 \)

Explanation:
1. \( G^2=AH \) and \( \frac{144}{15}\cdot15=144=12^2 \), so \( G=12 \).
2. A.M. exceeds H.M., so \( A=15 \) and \( H=\frac{144}{15} \).

Teacher's Note:
1. Use G squared = A H and A > H.

Answer: (c) \( 12 \)

Explanation:
1. \( H=\frac{G^2}A=\frac{324}{27}=12 \).

Teacher's Note:
1. H = G squared over A.

Answer: (b) G.P.

Explanation:
1. Applying componendo and dividendo, \( \frac{a}{bx}=\frac b{cx}=\frac c{dx} \).
2. So \( \frac ab=\frac bc=\frac cd \): a G.P.

Teacher's Note:
1. Use componendo-dividendo.

Answer: (d) infinitely many A.P.'s and infinitely many G.P.'s

Explanation:
1. G.P.: any ratio \( r=4^{1/k} \) works, since \( 4=r^k \) and \( 16=r^{2k} \).
2. A.P.: common difference \( \frac3m \) works, since \( 4=1+m\cdot\frac3m \) and \( 16=1+4m\cdot\frac3m \).
3. So both are infinite in number.

Teacher's Note:
1. Allow ratios and differences that are fractions.

Answer: (c) \( -2 \)

Explanation:
1. \( 2a=ar+ar^2 \) gives \( r^2+r-2=0 \), so \( r=1 \) or \( -2 \).
2. Alternating signs need \( r\lt 0 \), so \( r=-2 \).

Teacher's Note:
1. The ratio must be negative.

Answer: (a) \( \frac32 \)

Explanation:
1. Let the numbers be a and b with \( a\lt b \), \( a+b=4 \) and \( \sqrt{a(b+1)}=2 \). Then \( (4-b)(b+1)=4 \) gives \( b(3-b)=0 \), so \( b=3,a=1 \).
2. \( H=\frac{2\cdot1\cdot3}{4}=\frac32 \).

Teacher's Note:
1. Form equations from the two conditions.

Answer: (a) G.P.

Explanation:
1. If the terms are in G.P. with ratio \( \rho \), then \( T_p-T_q=(p-q)D \) and so on, giving \( \frac{q-r}{p-q}=\frac{T_q-T_r}{T_p-T_q}=\rho \) and \( \frac{r-s}{q-r}=\rho \).
2. So the three differences are in G.P.

Teacher's Note:
1. Subtract consecutive terms of the G.P.

Answer: (b) \( >8abc \)

Explanation:
1. By A.M.-G.M., \( a+b\ge2\sqrt{ab} \) and similarly for the others, with strict inequality as the numbers are unequal.
2. The product exceeds \( 8abc \).

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

Answer: (b) G.P.

Explanation:
1. The left side is \( (ap-b)^2+(bp-c)^2+(cp-d)^2\le0 \).
2. Hence \( b=ap,\ c=bp,\ d=cp \): a G.P.

Teacher's Note:
1. Write the expression as a sum of squares.

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

Explanation:
1. The \( n^{th} \) term is the middle one of the \( 2n-1 \) terms, so a, b, c are the A.M., G.M. and H.M. of the end terms.
2. So \( a\ge b\ge c \) and \( b^2=ac \). Option (b) is false.

Teacher's Note:
1. The middle term is the mean of the end terms.

Answer: (a) \( -\frac2n \)

Explanation:
1. Terms: \( A+mD,A+nD,A+rD \) in G.P. gives \( AD(2n-m-r)=D^2(mr-n^2) \).
2. From H.P., \( mr=\frac{n(m+r)}2 \), so \( mr-n^2=\frac n2(m+r-2n) \).
3. Then \( -A=\frac{nD}2 \), so \( \frac DA=-\frac2n \).

Teacher's Note:
1. Use the H.P. relation to simplify mr minus n squared.

Answer: (b) obtuse angled

Explanation:
1. With \( b=ar,\ c=ar^2 \), the A.P. condition is \( \frac4{9r^2}=\frac{3r}2 \), so \( r^3=\frac8{27} \) and \( r=\frac23 \).
2. The sides are in the ratio 9 : 6 : 4. A triangle exists since \( 6+4>9 \), and \( 9^2=81>6^2+4^2=52 \), so the largest angle is obtuse.

Teacher's Note:
1. Compare the square of the longest side with the other two.

Answer: (c) G.P.

Explanation:
1. \( 2b=a+c \), \( c^2=bd \) and \( d=\frac{2ce}{c+e} \).
2. \( c^2=\frac{a+c}2\cdot\frac{2ce}{c+e} \) gives \( c(c+e)=e(a+c) \), so \( c^2=ae \).
3. So a, c, e are in G.P.

Teacher's Note:
1. Eliminate b and d.

Answer: (a) \( b^2=-\frac{ac}2 \)

Explanation:
1. \( 2b=a+c \) and \( b^2=\frac{2a^2c^2}{a^2+c^2} \). Putting \( s=a^2+c^2 \): \( (s+2ac)s=8a^2c^2 \), so \( (s-2ac)(s+4ac)=0 \).
2. \( s=2ac \) gives \( a=c \) (all equal, excluded). So \( s=-4ac \) and \( b^2=\frac{(a+c)^2}4=\frac{s+2ac}4=-\frac{ac}2 \).

Teacher's Note:
1. Factorise the quadratic in a squared plus c squared.