JEE Mathematics Sequence and Series MCQs Set 33

Answer: (a) \( \left(\frac{q^{p}}{p^{q}}\right)^{\frac{1}{p-q}} \)

Explanation:
1. \( ar^{p-1}=q,\ ar^{q-1}=p \), and the required term is \( T=ar^{p+q-1} \).
2. Dividing \( T \) by each given term: \( T/q=r^{q} \) and \( T/p=r^{p} \).
3. Hence \( (T/q)^{p}=r^{pq}=(T/p)^{q} \), so \( T^{p-q}=\frac{q^{p}}{p^{q}} \).
4. \( T=\left(\frac{q^{p}}{p^{q}}\right)^{\frac{1}{p-q}} \)

Teacher's Note:
1. Write the two given terms as ar^(p-1) and ar^(q-1), then express the required term using each of them.
2. Eliminating r by raising to the powers p and q is the standard trick for such GP questions.

Answer: (a) \( \frac{1}{6}\left[1-(0.1)^{20}\right] \)

Explanation:
1. First term \( a=0.15 \), common ratio \( r=0.1 \), \( n=20 \).
2. \( S_{20}=\frac{0.15\left(1-0.1^{20}\right)}{1-0.1}=\frac{0.15}{0.9}\left(1-0.1^{20}\right) \).
3. \( \frac{0.15}{0.9}=\frac{1}{6} \), so \( S_{20}=\frac{1}{6}\left[1-(0.1)^{20}\right] \)

Teacher's Note:
1. Use S = a(1 - r^n)/(1 - r) when |r| < 1.
2. Simplify 0.15/0.9 first; it makes the final form obvious.

Answer: (b) \( x^2+2x-15=0 \)

Explanation:
1. Let \( \frac{1}{a_r} \) be in AP with common difference \( d \). Then \( \frac{1}{a_{r+1}}-\frac{1}{a_r}=d\Rightarrow a_ra_{r+1}=\frac{a_r-a_{r+1}}{d} \).
2. \( \sum_{r=1}^{3}a_ra_{r+1}=\frac{a_1-a_4}{d} \). Also \( \frac{1}{a_4}-\frac{1}{a_1}=3d\Rightarrow a_1-a_4=3d\,a_1a_4 \).
3. So the sum equals \( 3a_1a_4 \) and the given expression equals \( 3 \).
4. \( x=3 \) satisfies \( x^2+2x-15=0 \) since \( 9+6-15=0 \).

Teacher's Note:
1. Reciprocals of HP terms form an AP, so a_r - a_(r+1) = d a_r a_(r+1).
2. Check each option by substituting x = 3.

Answer: (c) HP

Explanation:
1. Here \( f(k)=S-a_k \) where \( S=\sum_{r=1}^{n}a_r \) is a constant. Then \( \frac{a_k}{f(k)}=\frac{a_k}{S-a_k} \).
2. Take reciprocals: \( \frac{S-a_k}{a_k}=\frac{S}{a_k}-1 \).
3. \( \frac{1}{a_k} \) is in AP, so \( \frac{S}{a_k}-1 \) is in AP.
4. Hence the reciprocals are in AP, so the original terms are in HP.

Teacher's Note:
1. f(k) is the total of all other terms, so S is common to every k.
2. If the reciprocals of a sequence form an AP, the sequence is an HP.

Answer: (a) \( \alpha/\beta \)

Explanation:
1. Each even-indexed term is \( r \) times the preceding odd-indexed term: \( a_{2n}=r\,a_{2n-1} \).
2. Summing over \( n=1 \) to \( 100 \): \( \alpha=r\beta \).
3. \( r=\alpha/\beta \)

Teacher's Note:
1. Pair every odd-indexed term with the even-indexed term right after it.
2. Since each pair has ratio r, the two sums have ratio r as well.

Answer: (d) \( \frac{1}{3}\cdot2^{2n}-2^{n}+\frac{2}{3} \)

Explanation:
1. Sum of products two at a time \( =\frac{S^2-\sum x_i^2}{2} \) where \( S=\sum x_i \).
2. \( S=1+2+\cdots+2^{n-1}=2^{n}-1 \) and \( \sum x_i^2=1+4+\cdots+4^{n-1}=\frac{4^{n}-1}{3} \).
3. \( \frac12\left[(2^n-1)^2-\frac{4^n-1}{3}\right]=\frac{1}{2}\cdot\frac{2\cdot4^{n}-6\cdot2^{n}+4}{3}=\frac13\cdot2^{2n}-2^{n}+\frac23 \)

Teacher's Note:
1. Use (sum of products in pairs) = (S squared - sum of squares)/2.
2. Check with n = 2: the numbers 1, 2 give product 2, and the formula gives 16/3 - 4 + 2/3 = 2.

Answer: (a) 1/24

Explanation:
1. The general term is \( j\cdot\frac{(n-j+1)(n-j+2)}{2} \) for \( j=1,\ldots,n \); for large \( n \) this behaves like \( \frac{j(n-j)^2}{2} \).
2. Hence \( L=\lim\frac{1}{n^4}\sum\frac{j(n-j)^2}{2}=\frac12\int_0^1x(1-x)^2\,dx \) with \( x=j/n \).
3. \( \int_0^1x(1-x)^2dx=\frac{1}{2}-\frac23+\frac14=\frac{1}{12} \), so \( L=\frac{1}{24} \)

Teacher's Note:
1. Write the sum as a single sum over j and convert it to a Riemann integral.
2. Only the leading power n^4 matters, so lower-order terms in the bracket can be dropped.

Answer: (d) \( 2\sqrt2 \)

Explanation:
1. The indices pair up to the same total: \( 1+25=5+21=10+16=26 \), so \( t_1t_{25}=t_5t_{21}=t_{10}t_{16}=t_{13}^2 \).
2. \( t_{13}^6=512=2^9 \), so \( t_{13}=2^{3/2}=2\sqrt2 \)

Teacher's Note:
1. In a GP, t_m t_n = t_k^2 whenever m + n = 2k.
2. Take the sixth root of 512 to get t13.

Answer: (a) 24

Explanation:
1. The sum of the first \( m \) odd numbers is \( m^2 \), so \( p^2+q^2=r^2 \).
2. With \( p>5 \) the smallest Pythagorean triple (with \( p\ge 6 \) as one leg) is \( (6,8,10) \): \( 6^2+8^2=10^2 \).
3. \( p+q+r=6+8+10=24 \). (The triples with legs 3, 4, 5 are excluded because \( p>5 \) cannot be satisfied by them.)

Teacher's Note:
1. Use the fact that the first m odd numbers add up to m squared.
2. List Pythagorean triples in increasing order and apply the condition p > 5.

Answer: (c) 212.16

Explanation:
1. This is a GP with first term \( 1.05 \), ratio \( 1.05 \) and 49 terms.
2. \( S=\frac{1.05\left(1.05^{49}-1\right)}{0.05}=\frac{1.05^{50}-1.05}{0.05} \).
3. \( S=\frac{11.658-1.05}{0.05}=\frac{10.608}{0.05}=212.16 \)

Teacher's Note:
1. Notice that 1.05 times 1.05^49 is 1.05^50, which is given.
2. Divide by 0.05 carefully: 10.608/0.05 = 212.16.

Answer: (a) 20

Explanation:
1. Common terms form an AP with common difference \( \mathrm{lcm}(4,5)=20 \). The first common term is \( 21 \).
2. Terms: \( 21+20k\le 417 \) (the smaller upper limit), so \( k\le 19.8 \), giving \( k=0,1,\ldots,19 \).
3. That is 20 common terms (the last is 401).

Teacher's Note:
1. Common terms of two APs form an AP whose difference is the LCM of the two differences.
2. The last common term must not exceed the smaller of the two last terms.

Answer: (b) \( \frac{2^{21}-1}{2^{10}}+19 \)

Explanation:
1. \( t_r^2=2^r+2^{-r}+2 \).
2. \( \sum 2^r=2^{11}-2 \) and \( \sum2^{-r}=1-2^{-10} \).
3. Total \( =2^{11}-2+1-2^{-10}+20=2^{11}-2^{-10}+19 \). Since \( 2^{11}-2^{-10}=\frac{2^{21}-1}{2^{10}} \), the sum is \( \frac{2^{21}-1}{2^{10}}+19 \)

Teacher's Note:
1. Square first: (2^(r/2) + 2^(-r/2))^2 contains the constant 2.
2. Two GPs plus ten copies of 2 give the sum.

Answer: (a) \( \frac14\left(\sqrt{4n+3}-\sqrt3\right) \)

Explanation:
1. Rationalise each term: \( \frac{1}{\sqrt{4k-1}+\sqrt{4k+3}}=\frac{\sqrt{4k+3}-\sqrt{4k-1}}{4} \).
2. The sum telescopes: \( \frac14\left(\sqrt{4n+3}-\sqrt3\right) \)

Teacher's Note:
1. Multiply and divide by the conjugate; the denominators differ by 4 every time.
2. Intermediate roots cancel in a telescoping series.

Answer: (a) \( (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)>9 \)

Explanation:
1. By AM-GM, \( a+b+c\ge3\sqrt[3]{abc} \) and \( \frac1a+\frac1b+\frac1c\ge\frac{3}{\sqrt[3]{abc}} \).
2. Multiplying, the product is \( \ge9 \), with equality only if \( a=b=c \).
3. The numbers are distinct, so equality never occurs and the product is strictly greater than 9.

Teacher's Note:
1. AM-GM applied to both factors gives the lower bound 9.
2. Both option (a) and (c) look similar; the word 'distinct' decides which is exactly correct (the strict inequality).

Answer: (d) \( n^2\sqrt2\ \text{cm} \)

Explanation:
1. The side of the square is \( n \) cm. A segment at distance \( h \) from the diagonal has length \( n\sqrt2-2h \).
2. With \( h=\frac{k}{\sqrt2} \): length \( n\sqrt2-\sqrt2\,k=\sqrt2(n-k) \), for \( k=1,2,\ldots,n-1 \) on each side.
3. Total \( =n\sqrt2+2\sqrt2\sum_{k=1}^{n-1}(n-k)=n\sqrt2+2\sqrt2\cdot\frac{n(n-1)}{2}=\sqrt2\,n^2 \)

Teacher's Note:
1. A segment parallel to the diagonal at distance h is shorter by 2h on its length.
2. Both sides of the diagonal contribute the same sum, which gives the factor 2.

Answer: (b) \( b_7=27 \)

Explanation:
1. \( \sum_{r=1}^{9}a_r=\frac92(a_1+a_9)=369\Rightarrow a_9=81 \). So \( b_9=81 \) and \( r^8=81\Rightarrow r^2=3 \) (taking a real common ratio).
2. \( b_7=r^6=(r^2)^3=27 \).
3. Others: \( b_6=r^5=9\sqrt3 \), \( b_8=r^7=27\sqrt3 \), \( b_9=81 \) (not 18). Only \( b_7=27 \) holds.

Teacher's Note:
1. Use the AP sum formula to find a9, which equals b9.
2. Even powers of the ratio are integers here, so only b7 comes out as a clean number.

Answer: (d) 216

Explanation:
1. By weighted AM-GM on \( \frac x3\ (\times3),\ \frac y4\ (\times4),\ \frac z5\ (\times5) \): \( \frac{x+y+z}{12}\ge\left[\left(\frac x3\right)^3\left(\frac y4\right)^4\left(\frac z5\right)^5\right]^{1/12} \).
2. So \( x^3y^4z^5\le 3^34^45^5\left(\frac{12}{12}\right)^{12}=27\cdot256\cdot3125=2.16\times10^{7} \).
3. The given product is \( 0.1\times600^3=2.16\times10^7 \), which equals the maximum, so equality holds: \( \frac x3=\frac y4=\frac z5=1 \) giving \( x=3,y=4,z=5 \).
4. \( x^3+y^3+z^3=27+64+125=216 \)

Teacher's Note:
1. The product given is exactly the maximum allowed by AM-GM, which forces equality.
2. Use weights 3, 4 and 5 for the three variables.

Answer: (b) \( 10\times(11!) \)

Explanation:
1. Try the telescoping form \( (r^2+1)\,r!=r\,(r+1)!-(r-1)\,r! \): the right side is \( r(r+1)r!-(r-1)r!=r!\,(r^2+r-r+1)=(r^2+1)\,r! \), which is correct.
2. Define \( g(r)=(r-1)\,r! \). Then \( g(r+1)=r\,(r+1)! \), so the term is \( g(r+1)-g(r) \).
3. \( \sum_{r=1}^{10}\left[g(r+1)-g(r)\right]=g(11)-g(1)=10\times11!-0=10\times 11! \)

Teacher's Note:
1. Look for a function g with g(r+1) - g(r) equal to the general term; g(r) = (r - 1) r! works.
2. Telescoping leaves only g(11) minus g(1), and g(1) = 0.

Answer: (b) 7780

Explanation:
1. \( (r+2)(r-3)=r^2-r-6 \).
2. Use \( \sum_{r=16}^{30}=\sum_{1}^{30}-\sum_{1}^{15} \): \( \sum r^2=9455-1240=8215 \) and \( \sum r=465-120=345 \) (15 terms).
3. \( 8215-345-6\times15=8215-345-90=7780 \)

Teacher's Note:
1. Expand the product and use the formulas for the sums of r and r squared.
2. The range 16 to 30 has 15 terms, so the constant 6 is subtracted 15 times.

Answer: (c) 20

Explanation:
1. \( S_9=9a_5 \), so \( 200<9a_5<220\Rightarrow a_5=23\ \text{or}\ 24 \) (integers).
2. \( a_2=12 \): \( a_5=a_2+3d=12+3d \). For \( a_5=23 \) we get \( d=\frac{11}{3} \) which is not an integer, so the terms would not all be integers.
3. \( a_5=24\Rightarrow d=4 \), and \( a_4=12+2\times4=20 \)

Teacher's Note:
1. S9 is nine times the middle term a5.
2. Use the integer condition to reject the non-integer common difference.

Answer: (a) 120

Explanation:
1. The \( n \)th term is \( \frac{2n+1}{\frac{n(n+1)(2n+1)}{6}}=\frac{6}{n(n+1)} \).
2. \( S_{20}=6\sum_{n=1}^{20}\left(\frac1n-\frac1{n+1}\right)=6\left(1-\frac1{21}\right)=\frac{120}{21} \).
3. \( k=120 \)

Teacher's Note:
1. The numerators are 2n + 1 and the denominators are sums of squares n(n+1)(2n+1)/6.
2. The result 6/(n(n+1)) telescopes.

Answer: (b) 4020

Explanation:
1. The first series has difference 4 and the second has difference 5, so the common terms form an AP with difference \( 20 \). The first common term is \( 11 \).
2. \( S_{20}=\frac{20}{2}\left[2(11)+19(20)\right]=10\times402=4020 \)

Teacher's Note:
1. Find the first common term by listing; then the common difference is the LCM of 4 and 5.
2. Apply the AP sum formula with a = 11 and d = 20.