JEE Mathematics Sequence and Series MCQs Set 16

Answer: (d) \( 125\sqrt2 \)

Explanation:
1. The common ratio is \( \frac{\sqrt{10}}{\sqrt2}=\sqrt5 \).
2. \( T_7=\sqrt2(\sqrt5)^6=125\sqrt2 \).

Teacher's Note:
1. Find the ratio first, then use ar^(n-1).

Answer: (b) \( \frac12 \)

Explanation:
1. \( r^4=\frac{16/243}{1/3}=\frac{16}{81} \), so \( r=\frac23 \).
2. \( T_4=\frac{T_5}{r}=\frac13\cdot\frac32=\frac12 \).

Teacher's Note:
1. Divide the fifth term by r to step back one term.

Answer: (c) \( 128 \)

Explanation:
1. \( ar^2=a^2 \) gives \( a=r^2 \), and \( ar=8 \) gives \( r^3=8 \).
2. So \( r=2,\ a=4 \) and \( T_6=4\cdot2^5=128 \).

Teacher's Note:
1. Use ar^2 = a^2 to link a and r.

Answer: (b) \( 2 \)

Explanation:
1. \( r^2=\frac{128}{32}=4 \), so the positive ratio is 2.

Teacher's Note:
1. Divide the two terms.

Answer: (d) \( 1+\frac{\log l-\log a}{\log r} \)

Explanation:
1. \( l=ar^{n-1} \), so \( (n-1)\log r=\log l-\log a \).
2. \( n=1+\frac{\log l-\log a}{\log r} \).

Teacher's Note:
1. Take logarithms of l = a r^(n - 1).

Answer: (c) \( 32 \)

Explanation:
1. Going backwards divides by 2 each step, so the 5th from the last is \( \frac{512}{2^4}=32 \).

Teacher's Note:
1. Reverse the G.P. to get ratio 1/2.

Answer: (c) \( 13^{th} \) term

Explanation:
1. \( 3\cdot2^{n-1}=12288 \) gives \( 2^{n-1}=4096=2^{12} \).
2. So \( n=13 \).

Teacher's Note:
1. Divide by 3 and write the result as a power of 2.

Answer: (a) \( 12 \)

Explanation:
1. \( r^2=4 \), so \( r=\pm2 \). Then \( t_1(r+r^4)=216 \).
2. For \( r=2 \): \( 18t_1=216 \), so \( t_1=12 \). For \( r=-2 \): \( 14t_1=216 \) gives a non-integer.
3. So \( t_1=12 \).

Teacher's Note:
1. The integer condition rejects r = -2.

Answer: (b) \( \sqrt{mn} \)

Explanation:
1. The \( p^{th} \) term lies midway between the \( (p-q)^{th} \) and \( (p+q)^{th} \) terms.
2. Hence \( T_p^2=T_{p-q}T_{p+q}=mn \) and \( T_p=\sqrt{mn} \).

Teacher's Note:
1. Equally spaced terms: the middle one is the geometric mean.

Answer: (c) \( 4^5 \)

Explanation:
1. The product is \( \frac{a}{r^2}\cdot\frac ar\cdot a\cdot ar\cdot ar^2\) with \( a=T_3=4 \), which is \( a^5=4^5 \).

Teacher's Note:
1. Pair the terms equidistant from T3.

Answer: (a) \( -\frac25 \)

Explanation:
1. \( 4a_2+5a_3=4r+5r^2 \).
2. This quadratic in r is least at \( r=-\frac{4}{10}=-\frac25 \).

Teacher's Note:
1. Use the vertex of the parabola.

Answer: (a) \( 11 \)

Explanation:
1. \( r=-\frac12 \), so \( 5\left(-\frac12\right)^{n-1}=\frac5{1024} \).
2. \( \left(-\frac12\right)^{n-1}=\frac1{2^{10}} \), so \( n-1=10 \) (even, so the sign is positive) and \( n=11 \).

Teacher's Note:
1. Check the sign when the ratio is negative.

Answer: (a) \( 1-\left(\frac1{10}\right)^{100} \)

Explanation:
1. \( a=0.9,\ r=0.1 \): \( S_{100}=0.9\cdot\frac{1-(0.1)^{100}}{0.9}=1-\left(\frac1{10}\right)^{100} \).

Teacher's Note:
1. Cancel the 0.9 with 1 - r.

Answer: (b) \( \frac32 \)

Explanation:
1. Take \( \frac ar,a,ar \): \( a^3=216 \), so \( a=6 \).
2. \( 6\left(r+1+\frac1r\right)=19 \) gives \( 6r^2-13r+6=0 \), so \( r=\frac32 \) or \( \frac23 \).
3. Only \( \frac32 \) appears among the options.

Teacher's Note:
1. The middle term is the cube root of the product.

Answer: (a) \( 1 \)

Explanation:
1. \( a2^{n-1}=128 \), so \( a2^n=256 \).
2. \( a(2^n-1)=255 \) gives \( 256-a=255 \), so \( a=1 \).

Teacher's Note:
1. Use the nth term to eliminate 2 to the n.

Answer: (a) \( 12 \)

Explanation:
1. With \( \frac ar,a,ar \), \( a^3=1728 \) so \( a=12 \).
2. (The numbers are 8, 12, 18 with sum 38, as required.)

Teacher's Note:
1. The cube root of the product is the middle number.

Answer: (a) \( 2 \)

Explanation:
1. \( S=\frac{lr-a}{r-1}=\frac{1458-a}2=728 \).
2. So \( a=2 \).

Teacher's Note:
1. Use S = (l r - a) / (r - 1).

Answer: (c) \( 15 \)

Explanation:
1. The right side is \( \frac{1-r^{16}}{1-r}=1+r+\dots+r^{15} \).
2. So \( n=15 \).

Teacher's Note:
1. Multiply and divide by (1 - r).

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

Explanation:
1. The sum is \( (1+i)\sum_{n=1}^{13}i^n \). Every 4 consecutive powers add to 0, so \( \sum_{n=1}^{13}i^n=i^{13}=i \).
2. \( (1+i)i=i-1 \).

Teacher's Note:
1. Powers of i repeat with period 4.

Answer: (b) \( 3\left(1-\frac1{3^{20}}\right) \)

Explanation:
1. It is a G.P. with \( a=2,\ r=\frac13 \).
2. \( S_{20}=\frac{2(1-3^{-20})}{2/3}=3\left(1-\frac1{3^{20}}\right) \).

Teacher's Note:
1. Use S = a(1 - r^n) / (1 - r).

Answer: (c) \( n+2^{-n}-1 \)

Explanation:
1. The \( k^{th} \) term is \( 1-2^{-k} \).
2. The sum is \( n-\left(1-2^{-n}\right)=n+2^{-n}-1 \).

Teacher's Note:
1. Write each term as 1 minus a power of 1/2.

Answer: (b) 2, 6, 18

Explanation:
1. Take \( \frac ar,a,ar \): \( a^3=216 \) so \( a=6 \).
2. Pair products: \( 36\left(\frac1r+1+r\right)=156 \) gives \( 3r^2-10r+3=0 \), so \( r=3 \) or \( \frac13 \).
3. The numbers are 2, 6, 18.

Teacher's Note:
1. Use a/r, a, ar.

Answer: (d) \( 2 \)

Explanation:
1. \( \frac{448r-7}{r-1}=889 \) gives \( 441r=882 \), so \( r=2 \).

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

Answer: (a) \( 6 \)

Explanation:
1. \( \frac{243\cdot3-a}2=364 \) gives \( a=1 \).
2. \( 3^{n-1}=243=3^5 \), so \( n=6 \).

Teacher's Note:
1. Find a first, then n.

Answer: (a) \( r \)

Explanation:
1. The even-place terms are the odd-place terms each multiplied by r.
2. So \( S_2=rS_1 \) and \( \frac{S_2}{S_1}=r \) (independent of a).

Teacher's Note:
1. Each even term is r times the previous odd term.