JEE Mathematics Sequence and Series MCQs Set 38

Answer: (d) \( 7+4\sqrt3:1 \)

Explanation:
1. \( \frac{a+b}{2}=2\sqrt{ab} \). With \( t=\frac ab \): \( t+1=4\sqrt t \), so \( \sqrt t=2\pm\sqrt3 \).
2. Since \( a \gt b \), \( t \gt 1 \): \( \sqrt t=2+\sqrt3 \) and \( t=7+4\sqrt3 \).
3. So \( a:b=7+4\sqrt3:1 \), which is the same as \( 2+\sqrt3:2-\sqrt3 \).

Teacher's Note:
1. Setting \( \sqrt{a/b}=s \) gives \( s^2-4s+1=0 \). The root larger than 1 corresponds to \( a \gt b \).

Answer: (b) \( s \gt t \) and \( a_{101} \lt b_{101} \)

Explanation:
1. The \( b_i \) form a G.P. with ratio \( 2 \): \( b_i=b_12^{i-1} \). Then \( t=b_1(2^{51}-1) \).
2. \( a_{51}=b_{51}=b_12^{50} \), so \( s=\frac{51}{2}\left(b_1+b_12^{50}\right)=25.5\,b_1(1+2^{50}) \). Compare with \( t\approx2\cdot2^{50}b_1 \): \( s \gt t \).
3. The common difference of the \( a_i \) is \( \frac{b_1(2^{50}-1)}{50} \), so \( a_{101}=b_1+2b_1(2^{50}-1)=b_1(2^{51}-1) \), which is far less than \( b_{101}=b_12^{100} \).

Teacher's Note:
1. An A.P. through the first and 51st terms of a G.P. lies above the G.P. in between and below it afterwards. Compare totals using the explicit formulas.

Answer: (c) \( 101 \)

Explanation:
1. The terms are \( 1.6,2.4,3.2,4,4.8,\dots=\frac{8}{5},\frac{12}{5},\frac{16}{5},\frac{20}{5},\dots=\frac{4(k+1)}{5} \) for \( k=1,2,3,\dots \)
2. Sum \( =\frac{16}{25}\sum_{k=1}^{10}(k+1)^2=\frac{16}{25}\left(2^2+\dots+11^2\right)=\frac{16}{25}\cdot505 \).
3. \( =\frac{16}{5}\cdot101 \), so \( m=101 \).

Teacher's Note:
1. Convert the mixed numbers to decimals to spot the pattern \( \frac{4(k+1)}{5} \). The sum of squares from \( 2^2 \) to \( 11^2 \) is \( 505 \).

Answer: (d) \( a=3,\ r=\frac14 \)

Explanation:
1. \( \frac{a}{1-r}=4 \) and \( ar=\frac34 \), so \( 4(1-r)r=\frac34 \), i.e. \( 16r^2-16r+3=0 \).
2. \( r=\frac14 \) or \( r=\frac34 \), with \( a=4(1-r)=3 \) or \( 1 \) respectively.
3. Among the given pairs only \( a=3 \), \( r=\frac14 \) satisfies both conditions.

Teacher's Note:
1. Use \( S_\infty=\frac{a}{1-r} \) together with the second term \( ar \). Check each option against both conditions.

Answer: (d) (ii) and (iii) only

Explanation:
1. Let the sides be \( a,ar,ar^2 \) with \( r \gt 1 \). Pythagoras: \( r^4=1+r^2 \), so \( r^2=\frac{1+\sqrt5}{2} \).
2. The smallest angle is opposite side \( a \) and adjacent to \( ar \): \( \tan\theta=\frac1r=\sqrt{\frac{2}{1+\sqrt5}} \).
3. Since \( \frac{2}{\sqrt5+1}=\frac{\sqrt5-1}{2} \), the value equals \( \sqrt{\frac{\sqrt5-1}{2}} \) also. So (ii) and (iii).

Teacher's Note:
1. Rationalising \( \frac{2}{\sqrt5+1} \) shows that two of the printed forms are the same number. The other two are its reciprocal-type values.

Answer: (b) \( \frac43 \)

Explanation:
1. \( (a+4d)^2=(a+d)(a+8d) \) gives \( 8d^2=ad \), so \( a=8d \) (as \( d\ne0 \)).
2. The terms are \( a+d=9d \), \( a+4d=12d \), \( a+8d=16d \).
3. The common ratio is \( \frac{12}{9}=\frac43 \).

Teacher's Note:
1. Equate the square of the middle term to the product of the others. The A.P. is a multiple of \( 9,12,16,\dots \) regardless of \( d \).

Answer: (a) \( 3 \)

Explanation:
1. A.P.: \( f(x)=\frac{2\left(2^x+2^{-x}\right)+\left(3^x+3^{-x}\right)}{2} \).
2. By A.M.-G.M., \( 2^x+2^{-x}\ge2 \) and \( 3^x+3^{-x}\ge2 \), with equality at \( x=0 \).
3. \( f(x)\ge\frac{2\cdot2+2}{2}=3 \). The minimum value is \( 3 \).

Teacher's Note:
1. Both terms are of the form \( t+\frac1t \), so each is at least \( 2 \), with equality at \( x=0 \).

Answer: (a) \( 0 \)

Explanation:
1. The recurrence makes the terms an A.P. with common difference \( d=a_2-15 \). The condition \( 27-2a_2 \gt 0 \) gives \( a_2 \lt 13.5 \), so \( d \lt -1.5 \).
2. \( \sum a_k^2=11\cdot225+2\cdot15\cdot55d+385d^2 \), so \( 225+150d+35d^2=90 \), i.e. \( 7d^2+30d+27=0 \) and \( d=-3 \) or \( -\frac97 \). Only \( d=-3 \) satisfies \( d \lt -1.5 \).
3. The mean of 11 terms of an A.P. is the middle term \( a_6=15+5d=0 \).

Teacher's Note:
1. The inequality is there to reject one of the two roots of the quadratic. The mean of an odd number of terms in an A.P. is its middle term.

Answer: (c) (i), (ii) and (iii) only

Explanation:
1. \( x=\frac{a+b}2 \), \( y=\sqrt{ab} \), \( z=\frac{2ab}{a+b} \). From \( a=5z \): \( a+b=10b \), so \( a=9b \).
2. \( x=y+2 \): \( 5b=3b+2 \), so \( b=1 \), \( a=9 \), and \( x=5 \), \( y=3 \), \( z=\frac95 \).
3. Check: \( xz=9=y^2 \) and \( 5 \gt 3 \gt \frac95 \). So (i), (ii), (iii) are true and (iv) is false.

Teacher's Note:
1. Write \( x,y,z \) in terms of \( a \) and \( b \), then use the two given relations to find \( a \) and \( b \). In general \( y^2=xz \) for the three means.

Answer: (b) \( 46 \)

Explanation:
1. Convergence needs \( |r| \lt 1 \), and \( \frac{T_1}{1-r}=\frac73 \), so \( T_1=\frac73\left(1-\frac p7\right)=\frac{7-p}{3} \).
2. \( T_1 \) must be a natural number: \( 7-p\in\{3,6,9,12\} \) with \( |p| \lt 7 \), so \( p\in\{4,1,-2,-5\} \).
3. \( 16+1+4+25=46 \).

Teacher's Note:
1. Use the infinite sum formula to express the first term through \( p \). Integer and range conditions restrict \( p \) to four values.

Answer: (a) \( 100 \)

Explanation:
1. Divide by \( 10^9 \): \( 1+2x+3x^2+\dots+10x^9 \) with \( x=\frac{11}{10} \), so \( k=\sum_{r=1}^{10}rx^{r-1} \).
2. The sum is \( \frac{1-11x^{10}+10x^{11}}{(1-x)^2} \). With \( x=1.1 \) this equals \( 100\left[1-11x^{10}+10x^{11}\right]=100\left[1-11x^{10}+11x^{10}\right]=100 \) because \( 10x^{11}=11x^{10} \).
3. So \( k=100 \).

Teacher's Note:
1. This is an arithmetico-geometric series. Note that \( 10x=11 \), which makes the numerator collapse to \( 1 \).

Answer: (c) (i), (ii), (iii) and (iv)

Explanation:
1. The repunit \( R_n=\frac{10^n-1}{9} \) is divisible by \( R_m \) whenever \( m\mid n \).
2. \( 91=7\times13 \), \( 81=9\times9 \), \( 75=3\times25 \) and \( 105=3\times35 \) are all composite, so each number is divisible by \( R_7 \), \( R_9 \), \( R_3 \) and \( R_3 \) respectively (each larger than 1).
3. So all four numbers are composite.

Teacher's Note:
1. A repunit with a composite number of digits is always composite, because \( R_m \) divides \( R_{km} \). Prime digit counts need not give a prime.

Answer: (b) \( 34 \)

Explanation:
1. The terms \( a_1,a_5,\dots,a_{49} \) are \( 13 \) terms with middle term \( a_{25} \): \( 13a_{25}=416 \), so \( a_{25}=32 \). Also \( a_9+a_{43}=2a_{26}=66 \), so \( a_{26}=33 \).
2. Hence \( d=1 \) and \( a_k=7+k \).
3. \( \sum_{k=1}^{17}(k+7)^2=\sum_{j=8}^{24}j^2=4900-140=4760=140\cdot34 \), so \( m=34 \).

Teacher's Note:
1. The sum of a symmetric set of terms of an A.P. equals the number of terms times the middle term. Use it to get the common difference.

Answer: (c) \( a \)

Explanation:
1. A.P. condition: \( 2\cdot7b=3a+15c \), i.e. \( 14ar=3a+15ar^2 \), so \( 15r^2-14r+3=0 \).
2. \( r=\frac35 \) or \( r=\frac13 \); with \( r\le\frac12 \) we get \( r=\frac13 \). The terms are \( 3a,\frac{7a}{3},\frac{5a}{3} \) with \( d=-\frac{2a}{3} \).
3. The fourth term is \( \frac{5a}3-\frac{2a}3=a \).

Teacher's Note:
1. Use \( 2\times\text{middle}=\text{sum of extremes} \) for the A.P. The restriction on \( r \) chooses the root.

Answer: (b) (i) and (ii) only

Explanation:
1. Dividing by \( a_ka_{k-1}a_{k-2} \): \( \frac{1}{a_{k-2}}+\frac1{a_k}=\frac{2}{a_{k-1}} \), so \( \frac1{a_k} \) is in A.P. with first term \( 1 \) and common difference \( d \).
2. \( \frac{S_{2p}}{S_p}=\frac{2(2-d+2pd)}{2-d+pd} \) is independent of \( p \) only if \( d=0 \) (ratio \( 2 \)) or \( 2-d=0 \) (ratio \( 4 \)).
3. If \( d=0 \): \( \frac1{a_{2016}}=1 \). If \( d=2 \): \( \frac{1}{a_{2016}}=1+2015\cdot2=4031 \). So (i) and (ii).

Teacher's Note:
1. Taking reciprocals turns the given relation into an A.P. The ratio of sums is constant only when numerator and denominator are proportional as functions of \( p \).

Answer: (a) (ii) and (iv) only

Explanation:
1. G.P.: \( a^2=\frac{b}{16} \), so \( b=16a^2 \). H.P.: \( \frac2b=\frac1a+6 \).
2. Substituting: \( \frac{1}{8a^2}=\frac{1+6a}{a} \), so \( 48a^2+8a-1=0 \) and \( a=\frac1{12} \) or \( a=-\frac14 \).
3. Then \( b=\frac19 \) or \( b=1 \) respectively. So (ii) and (iv).

Teacher's Note:
1. Write one relation from the G.P. and one from the H.P. and eliminate \( b \). Two values of \( a \) arise from the quadratic.

Answer: (c) (i) and (iv) only

Explanation:
1. Sum of the roots is \( 8 \) and their product is \( 16 \). By A.M.-G.M., \( \frac{\text{sum}}{4}=2\ge16^{1/4}=2 \), so equality holds and all four roots equal \( 2 \).
2. The equation is \( (x-2)^4=x^4-8x^3+24x^2-32x+16 \).
3. So \( a=24 \) and \( b=32 \).

Teacher's Note:
1. The A.M.-G.M. inequality is an equality here, which forces equal roots. Expand \( (x-2)^4 \) to read \( a \) and \( b \).

Answer: (a) \( 330 \)

Explanation:
1. Putting \( x=y=0 \) gives \( f(0)=0 \), so \( c=0 \). Comparing the \( xy \) terms in \( f(x+y)=f(x)+f(y)+xy \): \( 2a=1 \), so \( a=\frac12 \).
2. \( a+b=3 \) gives \( b=\frac52 \), so \( f(n)=\frac{n^2+5n}{2} \).
3. \( \sum_{n=1}^{10}f(n)=\frac12(385+275)=330 \).

Teacher's Note:
1. Substituting special values determines \( c \) and \( a \). The functional equation then gives the full polynomial.

Answer: (a) \( \frac{7}{81}\left(179+10^{-20}\right) \)

Explanation:
1. The \( k \)-th term is \( \frac79\left(1-10^{-k}\right) \): \( 0.7=\frac79(0.9) \), \( 0.77=\frac79(0.99) \), and so on.
2. Sum of \( 20 \) terms: \( \frac79\left[20-\left(10^{-1}+\dots+10^{-20}\right)\right]=\frac79\left[20-\frac{1-10^{-20}}{9}\right] \).
3. \( =\frac{7}{81}\left[180-1+10^{-20}\right]=\frac7{81}\left(179+10^{-20}\right) \).

Teacher's Note:
1. Write each term as \( \frac79 \) times a number of the form \( 0.99\ldots9=1-10^{-k} \). The sum then splits into an A.P.-type and a G.P.-type part.

Answer: (d) \( 8 \)

Explanation:
1. The sum \( f(a)=a^{-5}+a^{-4}+3a^{-3}+1+a^8+a^{10} \) has \( f'(a)=-5a^{-6}-4a^{-5}-9a^{-4}+8a^7+10a^9 \), which equals \( -18+18=0 \) at \( a=1 \).
2. Apply A.M.-G.M. to the eight positive terms \( a^{-5},a^{-4},a^{-3},a^{-3},a^{-3},a^8,a^{10},1 \): the product is \( a^{-5-4-9+8+10}=1 \), so the sum is at least \( 8 \).
3. Equality at \( a=1 \), so the minimum value is \( 8 \).

Teacher's Note:
1. Count the sum as eight terms (\( 3a^{-3} \) counts as three equal terms), whose product is \( a^0=1 \). A.M.-G.M. then gives the minimum \( 8 \) attained at \( a=1 \).

Answer: (d) \( 25 \)

Explanation:
1. The reciprocals form an A.P. with first term \( \frac15 \) and \( \frac1{a_{20}}=\frac1{25} \), so \( d=\frac{\frac1{25}-\frac15}{19}=-\frac{4}{475} \).
2. \( \frac{1}{a_n} \lt 0 \) when \( \frac15+(n-1)\left(-\frac4{475}\right) \lt 0 \), i.e. \( n-1 \gt \frac{475}{20}=23.75 \).
3. The least integer is \( n-1=24 \), so \( n=25 \).

Teacher's Note:
1. A harmonic progression changes sign exactly when its reciprocal A.P. does. Solve the linear inequality for the term number.