JEE Mathematics Sequence and Series MCQs Set 13

Answer: (d) 43, 45, ..., 89

Explanation:
1. The sum of the first k odd numbers is \( k^2 \). So \( 45^2-21^2 \) is the sum of the 22nd to the 45th odd numbers.
2. The 22nd odd number is 43 and the 45th is 89.

Teacher's Note:
1. Think of the sum of the first 45 odd numbers minus the first 21.

Answer: (b) \( \frac12m(m^2+1) \)

Explanation:
1. The \( m^{th} \) term of the k th A.P. is \( 1+(m-1)k \).
2. Sum over \( k=1 \) to m: \( m+(m-1)\frac{m(m+1)}2=\frac m2(2+m^2-1)=\frac m2(m^2+1) \).

Teacher's Note:
1. Sum the k-dependent part with the formula for the sum of natural numbers.

Answer: (a) \( 551\times129 \)

Explanation:
1. The numbers are 103, 110, ..., 999 (since \( 103=98+5 \) and \( 999=994+5 \)).
2. Number of terms \( =\frac{999-103}7+1=129 \).
3. \( S=\frac{129}2(103+999)=129\times551 \).

Teacher's Note:
1. Find the first and last such numbers.

Answer: (c) \( mn \)

Explanation:
1. For an odd number of terms the sum is the number of terms times the middle term.
2. \( S=n\cdot m \).

Teacher's Note:
1. Use the average equals middle term property.

Answer: (b) \( 37 \)

Explanation:
1. The sum of n terms is \( n^2 \).
2. \( 36^2=1296\lt 1357 \) and \( 37^2=1369>1357 \), so \( n=37 \).

Teacher's Note:
1. Take the square root of 1357.

Answer: (c) \( 9 \)

Explanation:
1. The angle sum is \( (n-2)180=\frac n2\{240+5(n-1)\} \), giving \( n^2-25n+144=0 \), so \( n=9 \) or \( 16 \).
2. For \( n=16 \) the largest angle would be \( 120+75=195^\circ>180^\circ \), impossible for a convex polygon.
3. So \( n=9 \) (largest angle \( 160^\circ \)).

Teacher's Note:
1. Reject the root that gives an angle above 180 degrees.

Answer: (b) \( 3050 \)

Explanation:
1. Multiples of 2: 2550. Multiples of 5: 1050. Multiples of 10: 550.
2. Sum \( =2550+1050-550=3050 \).

Teacher's Note:
1. Subtract the multiples of 10, which were counted twice.

Answer: (a) \( 0 \)

Explanation:
1. \( S_m-S_n=0 \) gives \( \frac{(m-n)}2\{2a+(m+n-1)d\}=0 \).
2. Since \( m\ne n \), \( 2a+(m+n-1)d=0 \), so \( S_{m+n}=\frac{m+n}2\{2a+(m+n-1)d\}=0 \).

Teacher's Note:
1. Subtract the two sums and factor.

Answer: (d) \( 4n^3 \)

Explanation:
1. Before row n there are \( 2+4+\dots+2(n-1)=n(n-1) \) odd numbers, and through row n there are \( n(n+1) \).
2. The sum of the first k odd numbers is \( k^2 \), so the row sum is \( n^2(n+1)^2-n^2(n-1)^2=n^2\cdot4n=4n^3 \).

Teacher's Note:
1. Check n = 1: 1 + 3 = 4.

Answer: (a) \( 4n+3 \)

Explanation:
1. \( T_n=S_n-S_{n-1}=2n^2+5n-2(n-1)^2-5(n-1)=4n+3 \).

Teacher's Note:
1. T_n = S_n - S_(n-1).

Answer: (b) \( 7 \)

Explanation:
1. Take \( a-3d,a-d,a+d,a+3d \): \( 2a=8 \), \( a=4 \).
2. \( 16-d^2=15 \) gives \( d=\pm1 \) and the terms 1, 3, 5, 7 in some order.
3. The greatest is 7.

Teacher's Note:
1. Compare with the earlier problem on the least number.

Answer: (c) \( \frac{2m-1}{2n-1} \)

Explanation:
1. \( \frac{S_m}{S_n}=\frac{2a+(m-1)d}{2a+(n-1)d}\cdot\frac mn=\frac{m^2}{n^2} \), so \( \frac{2a+(m-1)d}{2a+(n-1)d}=\frac mn \).
2. Hence \( 2a=d \) (cross-multiplying gives \( (m-n)(2a-d)=0 \)).
3. \( \frac{T_m}{T_n}=\frac{a+(m-1)d}{a+(n-1)d}=\frac{(2m-1)d/2}{(2n-1)d/2}=\frac{2m-1}{2n-1} \).

Teacher's Note:
1. Use the fact that the sum is a perfect square multiple, that is, 2a = d.

Answer: (d) \( a^{1/a} \)

Explanation:
1. \( \log_{a^{1/k}}x=k\log_ax \), so the left side is \( (1+2+\dots+a)\log_ax=\frac{a(a+1)}2\log_ax \).
2. Equating to \( \frac{a+1}2 \): \( \log_ax=\frac1a \), so \( x=a^{1/a} \).

Teacher's Note:
1. Convert each logarithm to base a.

Answer: (b) \( 6 \)

Explanation:
1. \( S_{2n}=3S_n \) gives \( 2(2a+(2n-1)d)=3(2a+(n-1)d) \), so \( a=\frac{(n+1)d}2 \).
2. \( S_{3n}=\frac{3n}2\cdot4nd=6n^2d \) and \( S_n=\frac n2\cdot2nd=n^2d \).
3. The ratio is 6.

Teacher's Note:
1. Express a in terms of n and d.

Answer: (b) \( 17 \)

Explanation:
1. \( T_2=S_2-S_1=(20+4)-(5+2)=17 \).

Teacher's Note:
1. T2 = S2 - S1.

Answer: (d) None of these

Explanation:
1. Each term of \( S_1 \) exceeds the corresponding term of \( S_2 \) by d, so \( S_1-S_2=100d \).
2. Thus \( d=\frac{S_1-S_2}{100} \), which is not given in the options.

Teacher's Note:
1. Pair the terms a2 - a1, a4 - a3, and so on.

Answer: (a) \( \frac15 \)

Explanation:
1. The condition is \( S_{4n}=2S_{3n} \), which gives \( a=\frac{(1-n)d}2 \).
2. Then \( S_{2n}=n\{2a+(2n-1)d\}=n\{(1-n)d+(2n-1)d\}=n^2d \) and \( S_{4n}=2n\{(1-n)d+(4n-1)d\}=6n^2d \).
3. The next 2n terms sum to \( 6n^2d-n^2d=5n^2d \), so the ratio is \( \frac15 \).

Teacher's Note:
1. Write the next n terms as S4n - S3n.

Answer: (d) \( \frac32,\frac23 \)

Explanation:
1. \( \frac12\left(x+\frac1x\right)=\frac{13}{12} \) gives \( 6x^2-13x+6=0 \), so \( x=\frac32 \) or \( \frac23 \).

Teacher's Note:
1. Form a quadratic in x.

Answer: (b) \( \frac{n+1}2 \)

Explanation:
1. The sum is \( \frac{n(n+1)}2 \), so the mean is \( \frac{n+1}2 \).

Teacher's Note:
1. Divide the sum by n.

Answer: (b) 7, 11, 15, 19

Explanation:
1. With six terms, \( d=\frac{23-3}5=4 \).
2. The means are 7, 11, 15, 19.

Teacher's Note:
1. d = (b - a) / (n + 1).

Answer: (a) \( 6 \)

Explanation:
1. With \( d=\frac{14}{n+1} \), the first mean is \( 3+d \) and the last is \( 17-d \).
2. \( 17-d=3(3+d) \) gives \( d=2 \), so \( n+1=7 \) and \( n=6 \).

Teacher's Note:
1. Express the first and last means using d.

Answer: (a) \( \frac{n(a+b)}2 \)

Explanation:
1. The n means are equally spaced between a and b, so their average is \( \frac{a+b}2 \).
2. The sum is \( n\cdot\frac{a+b}2 \).

Teacher's Note:
1. The means are symmetric about the midpoint.

Answer: (d) n only

Explanation:
1. \( A=\frac{a+b}2 \) and \( S=\frac{n(a+b)}2 \).
2. \( S/A=n \), which depends only on n.

Teacher's Note:
1. Compute both and divide.

Answer: (a) \( \frac{41}2,19,\frac{35}2 \)

Explanation:
1. The series has 13 terms, so \( d=\frac{10-28}{12}=-\frac32 \).
2. The 6th, 7th and 8th terms (the middle three) are \( 28-\frac{15}2=\frac{41}2 \), \( 28-9=19 \) and \( 28-\frac{21}2=\frac{35}2 \).

Teacher's Note:
1. The middle term of 13 terms is the 7th.

Answer: (c) \( 0 \)

Explanation:
1. Put \( u=x+y,v=x-y \): \( f(u,v)=\frac{u^2-v^2}4 \).
2. \( f(x,y)=\frac{x^2-y^2}4 \) and \( f(y,x)=\frac{y^2-x^2}4 \), whose mean is 0.

Teacher's Note:
1. Find f(u, v) first.