JEE Mathematics Sequence and Series MCQs Set 17

Answer: (d) \( \left(\frac SR\right)^n \)

Explanation:
1. \( P=a^nr^{n(n-1)/2} \), so \( P^2=a^{2n}r^{n(n-1)} \).
2. \( R=\frac{1}{a r^{n-1}}\cdot S \) (the reciprocals form a G.P. with ratio 1/r), so \( \frac SR=a^2r^{n-1} \).
3. Hence \( \left(\frac SR\right)^n=a^{2n}r^{n(n-1)}=P^2 \).

Teacher's Note:
1. Compare S/R with the product of the first and last terms.

Answer: (d) None of these

Explanation:
1. The sum is \( \frac{3^{n+1}-1}2>1000 \), so \( 3^{n+1}>2001 \).
2. \( 3^7=2187>2001 \) and \( 3^6=729 \), so \( n+1=7 \) and \( n=6 \).
3. 6 is not among a, b, c.

Teacher's Note:
1. Be careful: the series goes up to 3 to the n.

Answer: (d) \( \frac{\sqrt5+1}2 \)

Explanation:
1. \( ar^2=ar+a \) gives \( r^2-r-1=0 \).
2. The positive root is \( r=\frac{1+\sqrt5}2 \).

Teacher's Note:
1. Form the quadratic and keep the positive root.

Answer: (c) \( 212.16 \)

Explanation:
1. The sum is \( \frac{1.05\left(1.05^{49}-1\right)}{0.05}=\frac{1.05^{50}-1.05}{0.05} \).
2. \( =\frac{11.658-1.05}{0.05}=212.16 \).

Teacher's Note:
1. Multiply 1.05 into the bracket first.

Answer: (b) \( \left(\frac13,1\right)\cup(1,3) \)

Explanation:
1. For \( a,ar,ar^2 \): \( \alpha^2=\frac{(1+r+r^2)^2}{1+r^2+r^4}=\frac{r^2+r+1}{r^2-r+1} \).
2. For real r this takes values in \( \left[\frac13,3\right] \), with \( \frac13 \) at \( r=-1 \) and 3 at \( r=1 \) (repeated terms, not distinct).
3. The value 1 occurs only at \( r=0 \), which gives a zero term repeated. So \( \alpha^2\in\left(\frac13,1\right)\cup(1,3) \).

Teacher's Note:
1. Use 1 + r squared + r to the fourth = (1 + r + r squared)(1 - r + r squared).

Answer: (d) None of these

Explanation:
1. The ratio is \( \frac2x \) and the sum is finite when \( \left|\frac2x\right|\lt 1 \), that is \( |x|>2 \).
2. This means \( x>2 \) or \( x\lt -2 \), which is not any of a, b, c alone.

Teacher's Note:
1. The condition is on the modulus of the ratio.

Answer: (d) \( \frac y{1-y} \)

Explanation:
1. \( y=\frac x{1+x} \), so \( y+xy=x \).
2. \( x=\frac y{1-y} \).

Teacher's Note:
1. Sum the infinite G.P. with ratio -x.

Answer: (c) \( \frac{2355}{999} \)

Explanation:
1. \( 2.\overline{357}=2+\frac{357}{999}=\frac{1998+357}{999}=\frac{2355}{999} \).

Teacher's Note:
1. A block of three repeating digits is over 999.

Answer: (c) \( 4 \)

Explanation:
1. \( ar=2 \) and \( \frac a{1-r}=8 \). So \( a=\frac2r \) and \( \frac2{r(1-r)}=8 \).
2. \( 4r^2-4r+1=0 \), so \( r=\frac12 \) and \( a=4 \).

Teacher's Note:
1. Substitute a = 2/r.

Answer: (c) \( \frac{x^2-y}{x^2+y} \)

Explanation:
1. \( x=\frac a{1-r} \) and \( y=\frac{a^2}{1-r^2} \), so \( \frac{x^2}y=\frac{1+r}{1-r} \).
2. Solving, \( r=\frac{x^2-y}{x^2+y} \).

Teacher's Note:
1. Divide x squared by y to remove a.

Answer: (d) only for \( -1\lt r\lt 1\ (r\ne0) \)

Explanation:
1. The terms \( r^n \) tend to 0 only when \( |r|\lt 1 \). Then the sum is \( \frac a{1-r} \).
2. So the condition is \( -1\lt r\lt 1 \) (r non-zero for a genuine G.P.).

Teacher's Note:
1. It is the convergence condition.

Answer: (a) \( \frac7{16} \)

Explanation:
1. \( 1-r=\frac{3/4}{4/3}=\frac9{16} \).
2. \( r=\frac7{16} \).

Teacher's Note:
1. 1 - r = a / S.

Answer: (a) \( 2 \)

Explanation:
1. The exponent is \( \frac13+\frac19+\dots=\frac{1/3}{2/3}=\frac12 \).
2. The value is \( 4^{1/2}=2 \).

Teacher's Note:
1. Add the exponents.

Answer: (d) \( 6.64474 \)

Explanation:
1. The ratio is \( \frac{1.212}{5.05}=0.24 \) and \( \frac{0.29088}{1.212}=0.24 \).
2. \( S=\frac{5.05}{0.76}=6.64474 \).

Teacher's Note:
1. Check the ratio on two pairs.

Answer: (b) \( \frac35 \)

Explanation:
1. \( a=20(1-r) \) and \( \frac{a^2}{1-r^2}=100 \).
2. \( \frac{400(1-r)}{1+r}=100 \) gives \( 4-4r=1+r \), so \( r=\frac35 \).

Teacher's Note:
1. The squared series has ratio r squared.

Answer: (c) \( \frac13 \)

Explanation:
1. The remaining terms sum to \( \frac{ar}{1-r} \), so \( a=\frac{2ar}{1-r} \).
2. \( 1-r=2r \), so \( r=\frac13 \).

Teacher's Note:
1. Sum of the remaining terms is S minus a.

Answer: (d) \( \frac1{1+e^x} \)

Explanation:
1. The first term is \( e^{-x} \) and the ratio is \( -e^{-x} \).
2. \( S=\frac{e^{-x}}{1+e^{-x}}=\frac1{e^x+1} \).

Teacher's Note:
1. Multiply the numerator and denominator by e to the x.

Answer: (c) \( \frac{40}{99} \)

Explanation:
1. The ratio is 0.01, so \( S=\frac{0.4}{0.99}=\frac{40}{99} \).

Teacher's Note:
1. The ratio is 1/100.

Answer: (a) \( -\frac12 \)

Explanation:
1. The ratio is \( \frac C{1+C} \), and \( \left|\frac C{1+C}\right|\lt 1 \) means \( C^2\lt (1+C)^2 \), that is \( 2C+1>0 \).
2. So the series has a finite sum when \( C>-\frac12 \).

Teacher's Note:
1. Square both sides of the modulus condition.

Answer: (b) \( \left(\frac{A-1}A\right)^{1/z} \)

Explanation:
1. \( A=\frac1{1-r^z} \), so \( r^z=1-\frac1A=\frac{A-1}A \).
2. \( r=\left(\frac{A-1}A\right)^{1/z} \).

Teacher's Note:
1. The ratio is r to the z.

Answer: (c) \( \frac72 \)

Explanation:
1. Apart from the 2, the terms form two G.P.'s: \( \frac12+\frac1{2^2}+\dots=1 \) and \( \frac13+\frac1{3^2}+\dots=\frac12 \).
2. Total \( =2+1+\frac12=\frac72 \).

Teacher's Note:
1. Split into two interleaved G.P.'s.

Answer: (d) \( 4 \)

Explanation:
1. \( x=\frac{1/4}{1/2}=\frac12 \).
2. \( \log_{\sqrt5}\frac12=-2\log_52 \), so \( a^{\log_bx}=5^{-1\cdot(-2\log_52)}=5^{\log_54}=4 \).

Teacher's Note:
1. Write 0.2 as 5 to the minus 1.

Answer: (a) \( \frac32,\frac34,\frac38,\frac3{16},\dots \)

Explanation:
1. \( a=3(1-r) \) and \( \frac{a^2}{1-r^2}=3 \), so \( \frac{9(1-r)}{1+r}=3 \).
2. \( r=\frac12,\ a=\frac32 \), giving \( \frac32,\frac34,\frac38,\dots \).

Teacher's Note:
1. Use the sums of the series and its squares.

Answer: (d) \( 3\pi/4 \)

Explanation:
1. \( \frac1{1-\cos\alpha}=2-\sqrt2 \) gives \( 1-\cos\alpha=\frac{2+\sqrt2}2 \).
2. \( \cos\alpha=-\frac{\sqrt2}2 \), so \( \alpha=\frac{3\pi}4 \).

Teacher's Note:
1. Rationalise 1 / (2 - root 2).

Answer: (c) \( 64 \)

Explanation:
1. The sum is \( \frac{n^{128}-1}{n-1}=(n+1)(n^2+1)(n^4+1)\cdots(n^{64}+1) \).
2. The largest factor of the form \( n^m+1 \) is \( n^{64}+1 \), so \( m=64 \).

Teacher's Note:
1. Factorise n to the 128 minus 1 repeatedly.