JEE Mathematics Sequence and Series MCQs Set 40

Answer: (b) \( 2^{25} \)

Explanation:
1. The coefficients 2, 5, 8, ... form an AP with general term \( 3r+2 \), so the sum is \( \sum_{r=0}^{20}(3r+2)\,{}^{20}C_r \).
2. \( \sum r\,{}^{20}C_r=20\cdot2^{19} \) and \( \sum{}^{20}C_r=2^{20} \), so the sum is \( 60\cdot2^{19}+2\cdot2^{20}=15\cdot2^{21}+2^{21}=16\cdot2^{21}=2^{25} \). Answer: (b)

Teacher's Note:
1. Write the general coefficient as \( 3r+2 \).
2. Use \( r\,{}^nC_r=n\,{}^{n-1}C_{r-1} \).

Answer: (a) \( 2-\frac{11}{2^{19}} \)

Explanation:
1. Let \( S=\frac12+\frac2{2^2}+\cdots+\frac{20}{2^{20}} \). Then \( \frac S2=\frac1{2^2}+\cdots+\frac{19}{2^{20}}+\frac{20}{2^{21}} \).
2. Subtract: \( \frac S2=\frac12+\frac1{2^2}+\cdots+\frac1{2^{20}}-\frac{20}{2^{21}}=1-\frac1{2^{20}}-\frac{10}{2^{20}}=1-\frac{11}{2^{20}} \).
3. \( S=2-\frac{11}{2^{19}} \). Answer: (a)

Teacher's Note:
1. Multiply by the common ratio and subtract (AGP method).
2. The remaining terms form a GP.

Answer: (b) \( \frac da,\frac eb,\frac fc \) are in AP

Explanation:
1. \( b^2=ac \), so the discriminant of \( ax^2+2bx+c \) is \( 4(b^2-ac)=0 \) and its (repeated) root is \( x=-\frac ba \).
2. This is also a root of \( dx^2+2ex+f=0 \): \( d\frac{b^2}{a^2}-\frac{2eb}a+f=0 \). Using \( b^2=ac \): \( \frac{dc}a-\frac{2eb}a+f=0 \).
3. Multiply by \( a \): \( dc-2eb+af=0 \), so \( 2eb=dc+af \). Divide by \( b^2=ac \): \( \frac{2e}b=\frac da+\frac fc \), so \( \frac da,\frac eb,\frac fc \) are in AP. Answer: (b)

Teacher's Note:
1. A GP condition makes the first quadratic a perfect square.
2. Substitute the repeated root in the second equation.

Answer: (c) 3

Explanation:
1. \( f(x+y)=f(x)f(y) \) and \( f(1)=2 \) give \( f(x)=2^x \).
2. \( \sum_{k=1}^{10}2^{a+k}=2^a\left(2+2^2+\cdots+2^{10}\right)=2^a\cdot2(2^{10}-1)=2^{a+1}(2^{10}-1) \).
3. \( 2^{a+1}=16=2^4\Rightarrow a=3 \). Answer: (c)

Teacher's Note:
1. Find \( f(x) \) from the functional equation.
2. Sum the GP.

Answer: (a) \( (A,\ 50+46A) \)

Explanation:
1. \( S_n=\frac d2n^2+\left(a_1-\frac d2\right)n \). Compare with \( \frac A2n^2+\left(50-\frac{7A}2\right)n \): \( d=A \) and \( a_1-\frac A2=50-\frac{7A}2 \), so \( a_1=50-3A \).
2. \( a_{50}=a_1+49d=50-3A+49A=50+46A \). So \( (d,a_{50})=(A,\ 50+46A) \). Answer: (a)

Teacher's Note:
1. Write \( S_n \) as a quadratic in \( n \).
2. Compare coefficients of \( n^2 \) and \( n \).

Answer: (b) 190

Explanation:
1. Let each side of the triangle have \( n \) balls: total \( \frac{n(n+1)}2 \). The square has side \( n-2 \) so \( (n-2)^2 \) balls.
2. \( \frac{n(n+1)}2+99=(n-2)^2\Rightarrow n^2-9n-190=0\Rightarrow(n-19)(n+10)=0 \), so \( n=19 \).
3. Balls in the triangle \( =\frac{19\cdot20}2=190 \). Answer: (b)

Teacher's Note:
1. Triangular numbers have the form \( \frac{n(n+1)}2 \).
2. Reject the negative root.

Answer: (b) 76

Explanation:
1. \( a_1+a_4+\cdots+a_{16} \) has 6 terms: \( 6a_1+45d=114 \), so \( 2a_1+15d=38 \).
2. \( a_1+a_6+a_{11}+a_{16}=4a_1+30d=2(2a_1+15d)=76 \). Answer: (b)

Teacher's Note:
1. Write each term using \( a_1 \) and \( d \).
2. The required sum is a multiple of \( 2a_1+15d \).

Answer: (c) 660

Explanation:
1. The \( n \)th term is \( T_n=\frac{(2n+1)(1^3+\cdots+n^3)}{1^2+\cdots+n^2}=\frac{(2n+1)\left[\frac{n(n+1)}2\right]^2}{\frac{n(n+1)(2n+1)}6}=\frac{3n(n+1)}2 \).
2. \( S_{10}=\frac32\left(\sum n^2+\sum n\right)=\frac32(385+55)=660 \). Answer: (c)

Teacher's Note:
1. Find the general term using the formulas for sums of cubes and squares.
2. Then sum \( n^2+n \).

Answer: (a) 620

Explanation:
1. \( \frac{1^3+\cdots+n^3}{1+\cdots+n}=\frac{n(n+1)}2 \), so the series part is \( \sum_{n=1}^{15}\frac{n(n+1)}2=\frac12(1240+120)=680 \).
2. \( \frac12(1+2+\cdots+15)=\frac12\cdot120=60 \).
3. Value \( =680-60=620 \). Answer: (a)

Teacher's Note:
1. Simplify each fraction using \( \sum n^3=(\sum n)^2 \).
2. Subtract the final half-sum.

Answer: (a) \( \frac85 \)

Explanation:
1. \( a_1=2-5d \), \( a_4=2-2d \), \( a_5=2-d \), so the product is \( f(d)=(2-5d)(2-2d)(2-d)=-10d^3+34d^2-32d+8 \).
2. \( f'(d)=-30d^2+68d-32=0\Rightarrow15d^2-34d+16=0\Rightarrow d=\frac23 \) or \( \frac85 \).
3. \( f''(d)=-60d+68 \): at \( d=\frac23 \) it is \( 28\gt0 \) (minimum); at \( d=\frac85 \) it is \( -28\lt0 \) (maximum). So \( d=\frac85 \). Answer: (a)

Teacher's Note:
1. Express the three terms through \( d \) using \( a_6=2 \).
2. Use the second derivative test.

Answer: (b) \( -133 \)

Explanation:
1. The terms are \( \left[-\frac13-\frac k{100}\right] \) for \( k=0,1,\ldots,99 \) (100 terms). The floor equals \( -1 \) while \( \frac13+\frac k{100}\le1 \), i.e. for \( k=0,\ldots,66 \) (67 terms).
2. For \( k=67,\ldots,99 \) (33 terms) \( \frac13+\frac k{100}\gt1 \) and the floor is \( -2 \).
3. Sum \( =67(-1)+33(-2)=-133 \). Answer: (b)

Teacher's Note:
1. Find where the value of the floor changes from \( -1 \) to \( -2 \).
2. Count the terms in each range.

Answer: (c) \( -320 \)

Explanation:
1. \( S_4=2(2a+3d)=16\Rightarrow2a+3d=8 \) and \( S_6=3(2a+5d)=-48\Rightarrow2a+5d=-16 \).
2. Subtracting: \( 2d=-24\Rightarrow d=-12 \), and \( 2a=8+36=44 \).
3. \( S_{10}=5(2a+9d)=5(44-108)=-320 \). Answer: (c)

Teacher's Note:
1. Write \( S_4 \) and \( S_6 \) in terms of \( 2a \) and \( d \).
2. Subtract the two equations.

Answer: (a) 200

Explanation:
1. \( 3a_1+21d=40 \), so \( a_1+7d=\frac{40}3 \).
2. \( S_{15}=\frac{15}2(2a_1+14d)=15(a_1+7d)=15\cdot\frac{40}3=200 \). Answer: (a)

Teacher's Note:
1. Add the three terms in terms of \( a_1 \) and \( d \).
2. \( S_{15} \) is 15 times the middle (8th) term.

Answer: (b) 2

Explanation:
1. Let \( b=ar \), \( c=ar^2 \) with \( r\ne1 \) (distinct). Then \( a(1+r+r^2)=xar\Rightarrow x=1+r+\frac1r \).
2. For \( r\gt0 \), \( r+\frac1r\ge2 \), so \( x\ge3 \); for \( r\lt0 \), \( r+\frac1r\le-2 \), so \( x\le-1 \). Hence \( x\in(-\infty,-1]\cup[3,\infty) \).
3. The value \( x=2 \) lies in the excluded gap, so \( x \) cannot be 2. Answer: (b)

Teacher's Note:
1. Express \( x \) as \( 1+r+\frac1r \).
2. Use AM-GM for positive and negative \( r \).

Answer: (c) 4

Explanation:
1. Let the first term be \( A \) and the common difference \( d \): \( a=A+6d \), \( b=A+10d \), \( c=A+12d \).
2. GP condition \( b^2=ac \): \( (A+10d)^2=(A+6d)(A+12d)\Rightarrow2Ad+28d^2=0\Rightarrow d(A+14d)=0 \). Since \( d\ne0 \), \( A=-14d \).
3. Then \( a=-8d \), \( c=-2d \) and \( \frac ac=4 \). Answer: (c)

Teacher's Note:
1. Write the three terms using the first term and common difference.
2. Apply \( b^2=ac \).

Answer: (b) 7820

Explanation:
1. The \( r \)th term is \( T_r=\frac{3r\,(1^2+\cdots+r^2)}{2r+1}=\frac{3r\cdot\frac{r(r+1)(2r+1)}6}{2r+1}=\frac12(r^3+r^2) \).
2. \( S_{15}=\frac12\left[\left(\frac{15\cdot16}2\right)^2+\frac{15\cdot16\cdot31}6\right]=\frac12(14400+1240)=7820 \). Answer: (b)

Teacher's Note:
1. Find the general term; the factor \( 2r+1 \) cancels.
2. Use the formulas for \( \sum r^3 \) and \( \sum r^2 \).

Answer: (c) 1356

Explanation:
1. Remainder 2: \( 16,23,\ldots,93 \) (12 terms), sum \( =\frac{12}2(16+93)=654 \).
2. Remainder 5: \( 12,19,\ldots,96 \) (13 terms), sum \( =\frac{13}2(12+96)=702 \).
3. Total \( =654+702=1356 \). Answer: (c)

Teacher's Note:
1. Write the numbers as \( 7k+2 \) and \( 7k+5 \) and list the two-digit ones.
2. Each list is an AP.

Answer: (b) \( \frac23 \)

Explanation:
1. \( \frac a{1-r}=3 \), so \( a=3(1-r) \). The cubes form a GP with first term \( a^3 \) and ratio \( r^3 \): \( \frac{a^3}{1-r^3}=\frac{27}{19} \).
2. \( \frac{27(1-r)^3}{1-r^3}=\frac{27}{19}\Rightarrow19(1-r)^2=1+r+r^2\Rightarrow6r^2-13r+6=0 \), i.e. \( (3r-2)(2r-3)=0 \).
3. For a convergent series with positive terms \( 0\lt r\lt1 \), so \( r=\frac23 \). Answer: (b)

Teacher's Note:
1. Use \( S_\infty=\frac{a}{1-r} \) for both the series and its cubes.
2. Reject the root that is not less than 1.

Answer: (d) 3 : 1

Explanation:
1. \( a+18d=0\Rightarrow a=-18d \).
2. \( \frac{t_{49}}{t_{29}}=\frac{a+48d}{a+28d}=\frac{30d}{10d}=3 \). Answer: (d)

Teacher's Note:
1. Use the given term to express \( a \) in terms of \( d \).
2. Substitute in the ratio.

Answer: (a) \( 2^{100} \)

Explanation:
1. Use \( S_k=\frac{1-q^{k+1}}{1-q} \). The left side is \( \frac1{1-q}\sum_{k=1}^{101}{}^{101}C_k\left(1-q^{k}\right) \) (with \( S_{k-1} \) multiplying \( {}^{101}C_k \)).
2. \( =\frac1{1-q}\left[(2^{101}-1)-((1+q)^{101}-1)\right]=\frac{2^{101}-(1+q)^{101}}{1-q} \).
3. \( T_{100}=\frac{1-\left(\frac{q+1}2\right)^{101}}{1-\frac{q+1}2}=\frac{2^{101}-(q+1)^{101}}{2^{100}(1-q)} \), so the left side \( =2^{100}T_{100} \) and \( \alpha=2^{100} \). Answer: (a)

Teacher's Note:
1. Write \( S_k \) as a ratio and sum with the binomial theorem.
2. Compare with the closed form of \( T_{100} \).

Answer: (b) 28

Explanation:
1. Let the terms be \( \frac ar,\ a,\ ar \). Product \( =a^3=512\Rightarrow a=8 \).
2. After adding 4: \( \frac8r+4,\ 12,\ 8r \) are in AP, so \( 24=\frac8r+4+8r \), i.e. \( 2r^2-5r+2=0 \) and \( r=2 \) or \( \frac12 \).
3. The terms are 4, 8, 16 (or 16, 8, 4), with sum \( 28 \). Answer: (b)

Teacher's Note:
1. Take the middle term as \( a \) so that the product is \( a^3 \).
2. Use the AP condition \( 2\times\text{middle}=\text{sum of the others} \).

Answer: (b) 27

Explanation:
1. The terms are \( \left(\frac34\right)^3,\left(\frac64\right)^3,\left(\frac94\right)^3,\left(\frac{12}4\right)^3,\ldots \), i.e. \( \left(\frac{3j}4\right)^3 \) for \( j=1,\ldots,15 \).
2. Sum \( =\frac{27}{64}\left(1^3+2^3+\cdots+15^3\right)=\frac{27}{64}\left(\frac{15\cdot16}2\right)^2=\frac{27}{64}\cdot14400=27\cdot225 \).
3. So \( k=27 \). Answer: (b)

Teacher's Note:
1. Write all the terms with denominator 4.
2. Use \( \sum j^3=\left(\frac{n(n+1)}2\right)^2 \).

Answer: (b) 248

Explanation:
1. In the first 20 terms the even-numbered terms carry a factor 2: \( A=(1^2+2^2+\cdots+20^2)+(2^2+4^2+\cdots+20^2)=\frac{20\cdot21\cdot41}6+4\cdot\frac{10\cdot11\cdot21}6=2870+1540=4410 \).
2. Similarly \( B=(1^2+\cdots+40^2)+4(1^2+\cdots+20^2)=22140+11480=33620 \).
3. \( B-2A=33620-8820=24800=100\times248 \), so \( \lambda=248 \). Answer: (b)

Teacher's Note:
1. Split each even-numbered term as one copy plus one extra copy.
2. Even-numbered squares are \( 4(1^2+2^2+\cdots) \).

Answer: (b) \( b,c \) and \( a \) are in AP

Explanation:
1. Bring all terms to one side: \( 225a^2+9b^2+25c^2-75ac-45ab-15bc=0 \), which equals \( \frac12\left[(15a-3b)^2+(3b-5c)^2+(5c-15a)^2\right]=0 \).
2. So \( 15a=3b=5c \), i.e. \( a:b:c=1:5:3 \). Taking \( a=\lambda,\ b=5\lambda,\ c=3\lambda \).
3. The numbers \( b,c,a \) are \( 5\lambda,3\lambda,\lambda \), an AP with common difference \( -2\lambda \). (The order \( a,b,c \) is \( 1,5,3 \), not an AP.) Answer: (b)

Teacher's Note:
1. Write the equation as a sum of three squares.
2. Check each option with \( a:b:c=1:5:3 \).