JEE Mathematics Sequence and Series MCQs Set 35

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

Explanation:
1. From \( S_n=cn^2 \): \( a_n=S_n-S_{n-1}=c(2n-1) \).
2. Sum of squares \( =c^2\sum(2k-1)^2=c^2\cdot\frac{n(4n^2-1)}3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \frac1{12}n(n+1)(3n^2+n+2) \)

Explanation:
1. \( V_r=\frac r2[2r+(r-1)(2r-1)]=\frac{r(2r^2-r+1)}{2} \).
2. Summing: \( \sum V_r=\frac12\left[2\cdot\frac{n^2(n+1)^2}{4}-\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2\right]=\frac{n(n+1)(3n^2+n+2)}{12} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( G_1=G_2=G_3=\dots \)

Explanation:
1. Since \( A_nH_n=G_n^2 \) for any two positive numbers, \( G_n=\sqrt{A_{n-1}H_{n-1}}=G_{n-1} \).
2. Hence \( G_1=G_2=G_3=\dots \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( Q_1,Q_2,Q_3,\dots \) are in AP with common difference 6

Explanation:
1. With \( T_r=3r^2+2r-1 \): \( Q_r=T_{r+1}-T_r=6r+5 \).
2. So \( Q_{r+1}-Q_r=6 \) and the sequence is an AP with common difference 6.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) 4

Explanation:
1. Note \( \frac1{(4r-1)^2}-\frac1{(4r+3)^2}=\frac{8(4r+1)}{(4r-1)^2(4r+3)^2} \).
2. The sum telescopes to \( \frac18\left(\frac19-\frac1{(4n+3)^2}\right) \).
3. Require \( \frac19-\frac1{(4n+3)^2}>\frac8{75} \), i.e. \( (4n+3)^2>225 \), so \( 4n+3>15 \) and \( n>3 \). The least n is 4.

Teacher's Note:
1. Spot the difference of squares in the denominators to telescope the sum.
2. Solve the resulting inequality in n.

Answer: (d) \( \left(3,\frac14\right) \)

Explanation:
1. \( \frac{a}{1-r}=4 \) and \( ar=\frac34 \).
2. Then \( a=4(1-r) \) and \( 4r(1-r)=\frac34 \), i.e. \( 16r^2-16r+3=0 \), so \( r=\frac14 \) or \( \frac34 \).
3. The pair \( (3,\frac14) \) appears in the options; the other root gives \( (1,\frac34) \).

Teacher's Note:
1. Sum of an infinite GP is a/(1 - r).
2. Check the options against both conditions.

Answer: (b) a composite number

Explanation:
1. \( T_r=V_{r+1}-V_r-2=3r^2+2r-1=(r+1)(3r-1) \).
2. For \( r\ge1 \) both factors are at least 2, so \( T_r \) is composite (e.g. \( T_1=4,\ T_2=15 \), so it is not always odd or even).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) HP

Explanation:
1. Write the products as \( \frac{P}{d},\frac Pc,\frac Pb,\frac Pa \) with \( P=abcd \).
2. Their reciprocals are proportional to \( d,c,b,a \), which are in AP, so the products are in HP.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( H_1<H_2<H_3<\dots \)

Explanation:
1. \( H_n=\frac{G_{n-1}^2}{A_{n-1}} \) with \( G_{n-1}=G_1 \) constant and \( A_{n-1} \) decreasing, so \( H_n \) increases.
2. Thus \( H_1<H_2<H_3<\dots \)

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. With ratio \( r\ne1 \): \( b_1=a,\ b_2=a(1+r),\ b_3=a(1+r+r^2),\ b_4=a(1+r+r^2+r^3) \). Differences \( ar,\ ar^2,\ ar^3 \) are not equal (not AP) and ratios \( 1+r,\ \frac{1+r+r^2}{1+r} \) are not equal (not GP): Statement 1 true.
2. For HP \( \frac1{b_i} \) would be in AP; with \( r=2 \): \( b=a(1,3,7,15) \) and \( 1,\frac13,\frac17,\frac1{15} \) are not in AP: Statement 2 false.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( A_1>A_2>A_3>\dots \)

Explanation:
1. \( A_n=\frac{A_{n-1}+H_{n-1}}2 \) and \( H_{n-1}<A_{n-1} \) (distinct numbers), so \( A_n<A_{n-1} \).
2. Thus the arithmetic means decrease: \( A_1>A_2>A_3>\dots \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 5

Explanation:
1. \( a_n=\frac37\left[1-\left(-\frac34\right)^n\right] \). \( b_n>a_n \) means \( a_n<\frac12 \), i.e. \( \left(-\frac34\right)^n>-\frac16 \).
2. For even \( n \) this always holds. For odd \( n \): \( \left(\frac34\right)^n<\frac16 \) needs \( n\ge7 \) (\( (\tfrac34)^5\approx0.237 \) fails, \( (\tfrac34)^7\approx0.133 \) holds).
3. So the condition holds for every \( n>5 \) but fails at \( n=5 \): \( n_0=5 \).

Teacher's Note:
1. Sum the alternating geometric series.
2. Check odd and even values of n separately.

Answer: (c) \( n(2c)^{1/n} \)

Explanation:
1. By AM-GM on the \( n \) numbers \( a_1,\dots,a_{n-1},2a_n \): \( \frac{a_1+\dots+a_{n-1}+2a_n}{n}\ge(2c)^{1/n} \).
2. So the minimum is \( n(2c)^{1/n} \), reached when \( a_1=\dots=a_{n-1}=2a_n \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. By AM-GM, \( (a+b)(c+d)\le\left(\frac{a+b+c+d}{2}\right)^2=1 \).
2. Both factors are positive, so \( M>0 \). Hence \( 0<M\le1 \).

Teacher's Note:
1. AM-GM on the two sums a + b and c + d.
2. Equality when a + b = c + d = 1.

Answer: (b) 11

Explanation:
1. First sum: \( \frac{2n}{2}[4+(2n-1)3]=n(6n+1) \). Second sum: \( \frac n2[114+2(n-1)]=n(n+56) \).
2. Equating: \( 6n+1=n+56 \), so \( n=11 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) 0

Explanation:
1. The recurrence makes the terms an AP with common difference \( d \) and mean \( 15+5d \); \( 27-2a_2>0 \) gives \( d<-\frac32 \).
2. Mean of squares \( =(15+5d)^2+10d^2=90 \): \( 7d^2+30d+27=0 \), so \( d=-3 \) or \( -\frac97 \); only \( d=-3 \) qualifies.
3. Mean \( =15-15=0 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) (ii), (iii) and (iv) only

Explanation:
1. \( \sum_{k=1}^mk=\frac{m(m+1)}2 \); summing over \( m \) gives \( \frac{p(p+1)(p+2)}{6} \); summing over \( p \) gives \( \frac{n(n+1)(n+2)(n+3)}{24} \).
2. Expanding: \( \frac{n^4+6n^3+11n^2+6n}{24} \), so \( a_0=\frac1{24},\ a_1=\frac14,\ a_2=\frac{11}{24},\ a_3=\frac14,\ a_4=0 \).
3. Statements (ii), (iii), (iv) are correct; (i) is false.

Teacher's Note:
1. Nested sums give successive figurate (tetrahedral and higher) numbers.
2. The coefficient a0 is 1/24, not 1/6.

Answer: (d) 1

Explanation:
1. \( \tan^{-1}\frac1{2n^2}=\tan^{-1}\frac{2}{4n^2}=\tan^{-1}\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)} \) because \( 1+(2n+1)(2n-1)=4n^2 \).
2. So the term is \( \tan^{-1}(2n+1)-\tan^{-1}(2n-1) \) and the series telescopes to \( \lim_{n\to\infty}\tan^{-1}(2n+1)-\tan^{-1}1=\frac\pi2-\frac\pi4=\frac\pi4 \).
3. Hence \( \tan t=1 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.