JEE Mathematics Sequence and Series MCQs Set 41

Answer: (b) \( \frac43 \)

Explanation:
1. The terms are \( a+d,\ a+4d,\ a+8d \). GP condition: \( (a+4d)^2=(a+d)(a+8d) \).
2. \( a^2+8ad+16d^2=a^2+9ad+8d^2\Rightarrow8d^2=ad\Rightarrow a=8d \) (as \( d\ne0 \)).
3. Ratio \( =\frac{a+4d}{a+d}=\frac{12d}{9d}=\frac43 \). Answer: (b)

Teacher's Note:
1. Write the three terms using \( a \) and \( d \).
2. Use \( b^2=ac \) for the GP.

Answer: (b) \( 4lm^2n \)

Explanation:
1. \( l+n=2m \). The terms \( l,G_1,G_2,G_3,n \) are in GP with ratio \( r \): \( G_1=lr \), \( G_2=lr^2 \), \( G_3=lr^3 \), \( n=lr^4 \).
2. \( G_1^4+2G_2^4+G_3^4=l^4r^4(1+2r^4+r^8)=l^4r^4(1+r^4)^2 \).
3. With \( r^4=\frac nl \): \( =l^4\cdot\frac nl\cdot\frac{(l+n)^2}{l^2}=ln(l+n)^2=ln\cdot4m^2=4lm^2n \). Answer: (b)

Teacher's Note:
1. Treat \( l,G_1,G_2,G_3,n \) as five terms of a GP.
2. Use \( r^4=n/l \).

Answer: (c) 100

Explanation:
1. Divide by \( 10^9 \): \( K=1+2\left(\frac{11}{10}\right)+3\left(\frac{11}{10}\right)^2+\cdots+10\left(\frac{11}{10}\right)^9 \), an arithmetico-geometric series with ratio \( \frac{11}{10} \).
2. Multiply by \( \frac{11}{10} \) and subtract: \( K\left(1-\frac{11}{10}\right)=1+\frac{11}{10}+\cdots+\left(\frac{11}{10}\right)^9-10\left(\frac{11}{10}\right)^{10} \).
3. The GP sum is \( 10\left[\left(\frac{11}{10}\right)^{10}-1\right] \), so \( -\frac K{10}=10\left(\frac{11}{10}\right)^{10}-10-10\left(\frac{11}{10}\right)^{10}=-10 \) and \( K=100 \). Answer: (c)

Teacher's Note:
1. Divide by \( 10^9 \) to get an AGP.
2. Multiply by the common ratio and subtract.

Answer: (d) \( 2+\sqrt3 \)

Explanation:
1. Let the GP be \( a,\ ar,\ ar^2 \) with \( r\gt1 \). After doubling the middle term: \( a,\ 2ar,\ ar^2 \) are in AP, so \( 4ar=a+ar^2 \).
2. \( r^2-4r+1=0\Rightarrow r=2\pm\sqrt3 \).
3. The GP is increasing, so \( r\gt1 \) and \( r=2+\sqrt3 \). Answer: (d)

Teacher's Note:
1. Write the AP condition \( 2\times\text{middle}=\text{sum of the extremes} \).
2. Reject the root that is less than 1.

Answer: (c) \( \frac7{81}\left(179+10^{-20}\right) \)

Explanation:
1. \( 0.77\ldots7\ (k\text{ digits})=\frac79\left(1-10^{-k}\right) \).
2. \( S_{20}=\frac79\left[20-\left(10^{-1}+10^{-2}+\cdots+10^{-20}\right)\right]=\frac79\left[20-\frac19\left(1-10^{-20}\right)\right] \).
3. \( =\frac79\cdot\frac{179+10^{-20}}9=\frac7{81}\left(179+10^{-20}\right) \). Answer: (c)

Teacher's Note:
1. Write each term as \( \frac79(1-10^{-k}) \).
2. Sum the GP of the \( 10^{-k} \) terms.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The \( r \)th bracket is \( (r-1)^2+(r-1)r+r^2=\frac{r^3-(r-1)^3}{r-(r-1)}=r^3-(r-1)^3 \) (the first bracket \( 1 \) is \( r=1 \)). There are 20 brackets (last first-term is \( 19^2=361 \), last term \( 20^2=400 \)).
2. By telescoping the sum is \( 20^3-0=8000 \), so the Assertion is true. The Reason (telescoping sum) is true and is exactly the argument used. Answer: (a)

Teacher's Note:
1. Recognise each bracket as a difference of cubes.
2. The sum telescopes.

Answer: (c) 21 months

Explanation:
1. After the first 3 months he has saved Rs. 600. In the next \( n \) months the savings are \( 240,280,320,\ldots \), an AP with first term 240 and difference 40.
2. \( 600+\frac n2\left[2(240)+(n-1)40\right]=11040\Rightarrow n^2+11n-522=0\Rightarrow(n-18)(n+29)=0 \), so \( n=18 \).
3. Total time \( =3+18=21 \) months. Answer: (c)

Teacher's Note:
1. Treat the first three months separately.
2. Subsequent savings form an AP.

Answer: (c) \( \frac{\alpha-\beta}{100} \)

Explanation:
1. Subtract: \( \alpha-\beta=(a_2-a_1)+(a_4-a_3)+\cdots+(a_{200}-a_{199}) \).
2. Each bracket equals \( d \) and there are 100 of them, so \( 100d=\alpha-\beta \) and \( d=\frac{\alpha-\beta}{100} \). Answer: (c)

Teacher's Note:
1. Subtract the two sums term by term.
2. Each difference is \( d \).

Answer: (b) 34 min

Explanation:
1. In the first 10 minutes he counts \( 10\times150=1500 \) notes, leaving 3000.
2. From the 11th minute the counts are \( 148,146,\ldots \). For \( n \) more minutes: \( \frac n2[2(148)+(n-1)(-2)]=3000\Rightarrow n^2-149n+3000=0\Rightarrow(n-24)(n-125)=0 \).
3. \( n=125 \) would make later counts negative (\( 148-2\cdot124\lt0 \)), so \( n=24 \). Total time \( =10+24=34 \) min. Answer: (b)

Teacher's Note:
1. Subtract the notes counted in the first 10 minutes.
2. Reject the root that gives negative counts.

Answer: (c) \( -12 \)

Explanation:
1. \( a(1+r)=12 \) and \( ar^2(1+r)=48 \). Dividing: \( r^2=4 \).
2. Alternate signs need \( r\lt0 \), so \( r=-2 \).
3. \( a(1-2)=12\Rightarrow a=-12 \). Answer: (c)

Teacher's Note:
1. Divide the two equations to find \( r \).
2. Alternating signs mean a negative ratio.

Answer: (d) \( \sqrt2 \)

Explanation:
1. By AM-GM, \( \frac{p^2+q^2}2\ge pq \), so \( pq\le\frac12 \).
2. \( (p+q)^2=p^2+q^2+2pq=1+2pq\le1+1=2 \).
3. So \( p+q\le\sqrt2 \), attained at \( p=q=\frac1{\sqrt2} \). Answer: (d)

Teacher's Note:
1. Use AM-GM on \( p^2 \) and \( q^2 \).
2. Square \( p+q \).

Answer: (b) \( e^{-1} \)

Explanation:
1. \( e^{-1}=1-1+\frac1{2!}-\frac1{3!}+\frac1{4!}-\cdots \).
2. The first two terms \( 1-1 \) cancel, so \( \frac1{2!}-\frac1{3!}+\frac1{4!}-\cdots=e^{-1} \). Answer: (b)

Teacher's Note:
1. Recall the series for \( e^{-1} \).
2. The first two terms cancel.

Answer: (c) \( \frac{11}{41} \)

Explanation:
1. \( \frac{\frac p2[2a_1+(p-1)d]}{\frac q2[2a_1+(q-1)d]}=\frac{p^2}{q^2}\Rightarrow\frac{(2a_1-d)+pd}{(2a_1-d)+qd}=\frac pq \).
2. Cross-multiplying gives \( (2a_1-d)(q-p)=0 \), so \( 2a_1=d \), i.e. \( a_1=\frac d2 \).
3. \( \frac{a_6}{a_{21}}=\frac{a_1+5d}{a_1+20d}=\frac{\frac d2+5d}{\frac d2+20d}=\frac{11}{41} \). Answer: (c)

Teacher's Note:
1. Write the sums with the AP formula.
2. Cross-multiply and use \( p\ne q \).

Answer: (a) HP

Explanation:
1. \( x=\frac1{1-a} \), \( y=\frac1{1-b} \), \( z=\frac1{1-c} \).
2. \( a,b,c \) in AP implies \( 1-a,1-b,1-c \) are in AP, so their reciprocals \( x,y,z \) are in HP. Answer: (a)

Teacher's Note:
1. Sum each infinite GP.
2. Reciprocals of an AP are in HP.

Answer: (a) 0

Explanation:
1. \( a+(m-1)d=\frac1n \) and \( a+(n-1)d=\frac1m \). Subtracting: \( (m-n)d=\frac1n-\frac1m=\frac{m-n}{mn} \), so \( d=\frac1{mn} \).
2. Then \( a=\frac1n-\frac{m-1}{mn}=\frac1{mn} \).
3. \( a-d=0 \). Answer: (a)

Teacher's Note:
1. Subtract the two equations to find \( d \).
2. Substitute back for \( a \).

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

Explanation:
1. For odd \( n=2m+1 \), \( S_n=S_{2m}+n\text{th term} \), and the \( n \)th term (odd position) is \( n^2 \).
2. \( S_{n-1}=\frac{(n-1)n^2}2 \) (using the even formula for \( n-1 \)). So \( S_n=\frac{(n-1)n^2}2+n^2=n^2\cdot\frac{n+1}2 \). Answer: (b) (Check: \( n=3 \) gives \( 1+8+9=18=\frac{9\cdot4}2 \).)

Teacher's Note:
1. Add the last (odd-position) term to the sum of the first \( n-1 \) terms.
2. Apply the even-\( n \) formula to \( n-1 \).

Answer: (b) \( \frac{(e-1)^2}{2e} \)

Explanation:
1. \( e+e^{-1}=2\left(1+\frac1{2!}+\frac1{4!}+\cdots\right) \).
2. So \( \frac1{2!}+\frac1{4!}+\cdots=\frac{e+e^{-1}}2-1=\frac{e^2+1-2e}{2e}=\frac{(e-1)^2}{2e} \). Answer: (b)

Teacher's Note:
1. Add the series for \( e \) and \( e^{-1} \).
2. Subtract the leading 1.

Answer: (d) \( \log_e\left(\frac4e\right) \)

Explanation:
1. \( \frac1{n(n+1)}=\frac1n-\frac1{n+1} \), so the series is \( \left(1-\frac12\right)-\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)-\cdots \).
2. \( =1-2\cdot\frac12+2\cdot\frac13-2\cdot\frac14+\cdots=2\left(1-\frac12+\frac13-\frac14+\cdots\right)-1=2\ln2-1 \).
3. \( 2\ln2-1=\ln4-\ln e=\log_e\frac4e \). Answer: (d)

Teacher's Note:
1. Use partial fractions on each term.
2. Recognise \( 1-\frac12+\frac13-\cdots=\ln2 \).

Answer: (b) \( 1-\log_34 \)

Explanation:
1. AP condition: \( 2\log_3\sqrt{3^{1-x}+2}=1+\log_3(4\cdot3^x-1) \), so \( \log_3(3^{1-x}+2)=\log_3\left[3(4\cdot3^x-1)\right] \).
2. \( 3^{1-x}+2=12\cdot3^x-3 \). Let \( t=3^x \): \( \frac3t+2=12t-3\Rightarrow12t^2-5t-3=0\Rightarrow(3t+1)(4t-3)=0 \).
3. \( t=3^x\gt0 \) gives \( t=\frac34 \), so \( x=\log_3\frac34=1-\log_34 \). Answer: (b)

Teacher's Note:
1. Use \( 2b=a+c \) and combine logs.
2. Reject the negative value of \( 3^x \).

Answer: (b) 2

Explanation:
1. The product is \( 2^{\frac14+\frac28+\frac3{16}+\cdots} \). The exponent \( \frac14+\frac28+\frac3{16}+\cdots=\sum_{k\ge1}\frac k{2^{k+1}} \).
2. \( \sum_{k\ge1}\frac k{2^k}=2 \), so the exponent is \( \frac12\cdot2=1 \).
3. The value is \( 2^1=2 \). Answer: (b)

Teacher's Note:
1. Combine the powers of 2 into one exponent.
2. The exponent is an arithmetico-geometric series.

Answer: (b) 512

Explanation:
1. \( ar^4=2 \). The product of the first 9 terms is \( a\cdot ar\cdot ar^2\cdots ar^8=a^9r^{36}=(ar^4)^9 \).
2. \( =2^9=512 \). Answer: (b)

Teacher's Note:
1. The 5th term is the middle term of the first nine.
2. Pair terms equidistant from the middle.

Answer: (b) \( x \)

Explanation:
1. \( \sum\frac{t^n}{n!}=e^t \) with \( t=\log_ex \).
2. \( e^{\log_ex}=x \). Answer: (b)

Teacher's Note:
1. Recognise the exponential series.
2. \( e^{\ln x}=x \).

Answer: (c) \( x \)

Explanation:
1. The exponent is the series \( \ln(1+(x-1))=\ln x \) (valid for \( |x-1|\lt1 \), i.e. \( 0\lt x\lt2 \)).
2. \( e^{\ln x}=x \). Answer: (c)

Teacher's Note:
1. Spot the series for \( \ln(1+t) \) with \( t=x-1 \).
2. Then use \( e^{\ln x}=x \).