JEE Mathematics Sequence and Series MCQs Set 36

Answer: (c) 2, 6, 18

Explanation:
1. Take the numbers as \( \tfrac ar,\ a,\ ar \). The product \( a^3=216\Rightarrow a=6 \).
2. Sum of products in pairs: \( a^2\left(\tfrac1r+r+1\right)=156\Rightarrow 36\left(r+\tfrac1r+1\right)=156\Rightarrow r+\tfrac1r=\tfrac{10}{3} \), so \( r=3 \) or \( \tfrac13 \).
3. The numbers are 2, 6, 18 (or 18, 6, 2).

Teacher's Note:
1. Writing the three terms as a/r, a, ar makes the product simple.
2. The answer is the same sequence read in either direction.

Answer: (d) 16

Explanation:
1. Common difference \( d=\dfrac{54-3}{n+1}=\dfrac{51}{n+1} \).
2. 8th mean \( =3+8d \) and \( (n-2) \)th mean \( =3+(n-2)d \). Then \( \dfrac{3+8d}{3+(n-2)d}=\dfrac35 \Rightarrow 15+40d=9+3(n-2)d \Rightarrow 6=(3n-46)d \).
3. Substituting \( d=\tfrac{51}{n+1} \): \( 6(n+1)=51(3n-46)\Rightarrow 147n=2352\Rightarrow n=16 \).

Teacher's Note:
1. Express both means through the common difference.
2. Verify: for n = 16, d = 3 and the means are 27 and 45, in ratio 3 : 5.

Answer: (b) 211

Explanation:
1. \( \dfrac{243}{2}=\dfrac{32}{3}r^6\Rightarrow r^6=\dfrac{729}{64}\Rightarrow r=\dfrac32 \).
2. The means are 16, 24, 36, 54, 81, whose sum is 211.

Teacher's Note:
1. Five means between two numbers means the ratio applies six times.
2. Add the five means only, not the end terms.

Answer: (d) \( p^3+(p+1)^3 \)

Explanation:
1. Common difference \( d=1 \), number of terms \( 2p+1 \).
2. \( S=\dfrac{2p+1}{2}\left[2(p^2+1)+2p\right]=(2p+1)(p^2+p+1) \).
3. Expanding, \( (2p+1)(p^2+p+1)=2p^3+3p^2+3p+1=p^3+(p+1)^3 \).

Teacher's Note:
1. Use the sum formula with d = 1.
2. Expand and compare with the options.

Answer: (c) 11

Explanation:
1. First sum \( =\dfrac{2n}{2}\left[4+(2n-1)3\right]=n(6n+1) \).
2. Second sum \( =\dfrac n2\left[114+(n-1)2\right]=n(n+56) \).
3. Equating: \( 6n+1=n+56\Rightarrow n=11 \).

Teacher's Note:
1. Write both sums with the standard formula and cancel n.
2. Check: for n = 11 both sums equal 737.

Answer: (b) 128

Explanation:
1. Let the first term be a and the ratio r. Then \( ar^2=a^2\Rightarrow a=r^2 \).
2. Second term \( ar=r^3=8\Rightarrow r=2,\ a=4 \).
3. Sixth term \( =ar^5=4\times32=128 \).

Teacher's Note:
1. Use the given relation to connect a and r.
2. Then solve for r from the second term.

Answer: (b) 2 and 8

Explanation:
1. For two numbers \( G^2=AH \), so \( G^2=\dfrac{16A}{5} \). Then \( 2A+\dfrac{16A}{5}=26\Rightarrow A=5,\ G^2=16 \).
2. The numbers have sum 10 and product 16, i.e. 2 and 8.

Teacher's Note:
1. For two numbers the G.M. is the geometric mean of A.M. and H.M.
2. Form a quadratic from the sum and the product.

Answer: (a) \( n\cdot2^{n+2}-2^{n+1}+2 \)

Explanation:
1. Let \( S=3\cdot2+5\cdot2^2+\dots+(2n+1)2^n \). Then \( 2S=3\cdot2^2+\dots+(2n-1)2^n+(2n+1)2^{n+1} \).
2. Subtracting: \( -S=6+2(2^2+2^3+\dots+2^n)-(2n+1)2^{n+1}=6+2(2^{n+1}-4)-(2n+1)2^{n+1} \).
3. So \( S=(2n+1)2^{n+1}-2^{n+2}+2=n\cdot2^{n+2}-2^{n+1}+2 \). Check n = 1: 6.

Teacher's Note:
1. This is an arithmetico-geometric series, so multiply by the common ratio and subtract.
2. Check the result with n = 1.

Answer: (c) 3050

Explanation:
1. Divisible by 2: \( 2+4+\dots+100=2550 \). Divisible by 5: \( 5+10+\dots+100=1050 \).
2. Divisible by both (multiples of 10): \( 10+20+\dots+100=550 \).
3. Required sum \( =2550+1050-550=3050 \).

Teacher's Note:
1. Use inclusion-exclusion so that multiples of 10 are not counted twice.
2. Each sum is an A.P. sum.

Answer: (d) \( \dfrac1{11} \)

Explanation:
1. The reciprocals form an A.P.: \( t_7=20,\ t_{13}=38 \), so \( 6d=18\Rightarrow d=3 \).
2. \( t_4=20-3\times3=11 \), so the 4th term of the H.P. is \( \dfrac1{11} \).

Teacher's Note:
1. Convert an H.P. problem to an A.P. problem by taking reciprocals.
2. Do not forget to take the reciprocal again at the end.

Answer: (a) 128

Explanation:
1. The smallest multiple of 7 above 100 is 105 and the largest below 1000 is 994.
2. Number of terms \( =\dfrac{994-105}{7}+1=127+1=128 \).

Teacher's Note:
1. Find the first and the last multiple inside the range.
2. Add 1 after dividing the gap by the common difference.

Answer: (c) \( \dfrac{14}{11},\ \dfrac{14}{8},\ \dfrac{14}{5} \)

Explanation:
1. The reciprocals \( 1,\ x_1,\ x_2,\ x_3,\ \tfrac17 \) form an A.P. with \( d=\dfrac{\frac17-1}{4}=-\dfrac{3}{14} \).
2. The reciprocals of the means are \( \dfrac{11}{14},\dfrac{8}{14},\dfrac{5}{14} \), so the means are \( \dfrac{14}{11},\dfrac{14}{8},\dfrac{14}{5} \).

Teacher's Note:
1. Insert arithmetic means between the reciprocals of the end terms.
2. Reciprocate the results to get the harmonic means.

Answer: (d) 900

Explanation:
1. Terms equidistant from the ends have the same sum: \( a_1+a_{24}=a_5+a_{20}=a_{10}+a_{15} \).
2. So \( 3(a_1+a_{24})=225\Rightarrow a_1+a_{24}=75 \).
3. \( S_{24}=\dfrac{24}{2}(a_1+a_{24})=12\times75=900 \).

Teacher's Note:
1. In an A.P. the sum of terms symmetric about the centre is constant.
2. Then apply the sum formula.

Answer: (a) 3, 7, 11 or 12, 7, 2

Explanation:
1. Let the numbers be \( 7-d,\ 7,\ 7+d \) (sum 21).
2. The new numbers are \( 7-d,\ 6,\ 8+d \) and are in G.P.: \( 36=(7-d)(8+d)\Rightarrow d^2+d-20=0\Rightarrow d=4 \) or \( -5 \).
3. For \( d=4 \): 3, 7, 11. For \( d=-5 \): 12, 7, 2.

Teacher's Note:
1. The middle term of an A.P. with sum 21 is always 7.
2. A quadratic in d gives two valid progressions.

Answer: (b) \( 2^{n+2}-3n-4 \)

Explanation:
1. The differences 4, 8, 16, 32 are powers of 2, so the \( n^{th} \) term is \( t_n=2^{n+1}-3 \).
2. \( S_n=\sum(2^{k+1}-3)=2(2^{n+1}-2)-3n=2^{n+2}-3n-4 \).

Teacher's Note:
1. When successive differences form a G.P., the general term has a power of the ratio plus a constant.
2. Check with n = 1: sum = 1.

Answer: (d) \( \dfrac{1}{27}\left(10^{n+1}-9n-10\right) \)

Explanation:
1. \( t_k=3\cdot\dfrac{10^k-1}{9}=\dfrac{10^k-1}{3} \).
2. \( S_n=\dfrac13\left(\dfrac{10(10^n-1)}{9}-n\right)=\dfrac{10^{n+1}-9n-10}{27} \).

Teacher's Note:
1. Write each term as a multiple of a repunit, which is a geometric sum.
2. Check with n = 1: the value is 3.

Answer: (d) 160

Explanation:
1. The terms form an A.P. with first term \( \tfrac17+2 \) and last term \( \tfrac{35}{7}+2=7 \).
2. Sum \( =\sum_{p=1}^{35}\left(\tfrac p7+2\right)=\tfrac{1}{7}\cdot\tfrac{35\cdot36}{2}+70=90+70=160 \).

Teacher's Note:
1. Split the general term into two simple sums.
2. The sum of the first n natural numbers is n(n+1)/2.

Answer: (a) 4, 9, 14

Explanation:
1. Let the numbers be \( a-d,\ a,\ a+d \). Sum \( =3a=27\Rightarrow a=9 \).
2. Product \( 9(81-d^2)=504\Rightarrow d^2=25\Rightarrow d=\pm5 \), so the numbers are 4, 9, 14.

Teacher's Note:
1. Choosing the middle term as a symbol simplifies the sum.
2. Check: 4 × 9 × 14 = 504.

Answer: (d) \( \dfrac83 \)

Explanation:
1. The \( k^{th} \) term is \( \dfrac{2^k-1}{4^{k-1}}=4\left(\dfrac1{2^k}-\dfrac1{4^k}\right) \).
2. Sum \( =4\left(\dfrac{1/2}{1-1/2}-\dfrac{1/4}{1-1/4}\right)=4\left(1-\dfrac13\right)=\dfrac83 \).

Teacher's Note:
1. Spot the pattern in the numerators 1, 3, 7, 15, 31 as powers of 2 minus 1.
2. Break the general term into two geometric series.

Answer: (c) 19668

Explanation:
1. The numbers are \( 103,119,\dots,791 \) (each is \( 16k+7 \)).
2. Number of terms \( =\dfrac{791-103}{16}+1=44 \).
3. Sum \( =\dfrac{44}{2}(103+791)=22\times894=19668 \).

Teacher's Note:
1. Such numbers form an A.P. with common difference 16.
2. Use sum = n/2 (first + last).

Answer: (c) \( \dfrac1{201} \)

Explanation:
1. Divide numerator and denominator by \( x^{100} \): the expression is \( \dfrac{1}{x^{-100}+x^{-99}+\dots+x^{100}} \).
2. By A.M.-G.M. on 201 positive terms (they pair as \( x^k+x^{-k}\ge2 \)), the denominator is at least 201, with equality at \( x=1 \).
3. So the greatest value is \( \dfrac1{201} \).

Teacher's Note:
1. Pair each power with its reciprocal to apply A.M.-G.M.
2. Equality holds when x equals 1.

Answer: (b) \( 2,\ 5,\ 8,\dots \)

Explanation:
1. \( a+2d=4a\Rightarrow 2d=3a \) and \( a+5d=17 \).
2. Substituting \( d=\tfrac{3a}{2} \): \( a+\tfrac{15a}{2}=17\Rightarrow a=2,\ d=3 \).

Teacher's Note:
1. Write each given condition as an equation in a and d.
2. Check: the third term is 8 = 4 × 2 and the sixth term is 17.