JEE Mathematics Sequence and Series MCQs Set 18

Answer: (b) \( \frac{a^2}{1-a^2}+\frac{ab}{1-ab} \)

Explanation:
1. The general term is \( a^{2k}+(ab)^k \).
2. The sum is \( \frac{a^2}{1-a^2}+\frac{ab}{1-ab} \).

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

Answer: (b) \( S\left[1-\left(1-\frac aS\right)^n\right] \)

Explanation:
1. \( r=1-\frac aS \).
2. \( S_n=\frac{a(1-r^n)}{1-r}=S\left[1-\left(1-\frac aS\right)^n\right] \).

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

Answer: (b) \( \frac12(\sqrt3-1) \)

Explanation:
1. The series sums to \( \frac{\sin^2x}{\cos^2x}=\tan^2x \), so the value is \( 2^{\tan^2x} \).
2. The roots are 1 and 8. \( 2^{\tan^2x}=1 \) needs \( \tan x=0 \), not allowed, so \( \tan^2x=3 \) and \( x=\frac\pi3 \).
3. \( \frac{\cos x}{\cos x+\sin x}=\frac1{1+\sqrt3}=\frac{\sqrt3-1}2 \).

Teacher's Note:
1. Apply exp and ln to simplify to 2 to the tan squared.

Answer: (b) \( \frac1{G^2} \)

Explanation:
1. \( G^2=xy \), so the sum is \( \frac1{x(y-x)}+\frac1{y(x-y)}=\frac1{y-x}\cdot\frac{y-x}{xy}=\frac1{xy} \).

Teacher's Note:
1. Substitute G squared = xy.

Answer: (c) \( a\left(\frac ba\right)^{\frac n{n+1}} \)

Explanation:
1. The ratio of the resulting G.P. is \( r=\left(\frac ba\right)^{1/(n+1)} \).
2. The \( n^{th} \) mean is \( ar^n=a\left(\frac ba\right)^{n/(n+1)} \).

Teacher's Note:
1. There are n + 2 terms in all.

Answer: (a) \( 3 \)

Explanation:
1. The product of the roots is 9, and the roots are positive.
2. The G.M. is \( \sqrt9=3 \).

Teacher's Note:
1. G.M. is the square root of the product.

Answer: (b) \( 6 \)

Explanation:
1. There are 7 terms, so \( r^6=\frac{2/3}{486}=\frac1{729} \), \( r^2=\frac19 \).
2. The fourth mean is \( 486r^4=486\cdot\frac1{81}=6 \) (the same for \( r=\pm\frac13 \)).

Teacher's Note:
1. The fourth mean is the fifth term.

Answer: (c) \( 20 \)

Explanation:
1. \( r^5=\frac5{160}=\frac1{32} \), so \( r=\frac12 \).
2. The means are 80, 40, 20, 10, so the third is 20.

Teacher's Note:
1. Find the ratio from the 6 terms.

Answer: (d) \( 1 \)

Explanation:
1. The product of n G.M.'s between a and b is \( (ab)^{n/2} \).
2. \( (4\cdot\frac14)^{3/2}=1 \).

Teacher's Note:
1. Pair the means equidistant from the centre.

Answer: (c) \( G_1G_2\cdots G_n=G^n \)

Explanation:
1. \( G_1G_n=G_2G_{n-1}=\dots=ab=G^2 \).
2. There are n means, giving n/2 pairs, so the product is \( (G^2)^{n/2}=G^n \).

Teacher's Note:
1. Pair the means from both ends.

Answer: (a) \( k^{\frac{n-1}{n+1}}:1 \)

Explanation:
1. First case: \( x\left(\frac yx\right)^{n/(n+1)} \). Second case: \( kx\left(\frac{y}{k^2x}\right)^{n/(n+1)}=k^{\frac{1-n}{n+1}}x\left(\frac yx\right)^{n/(n+1)} \).
2. The ratio first : second is \( k^{\frac{n-1}{n+1}}:1 \).

Teacher's Note:
1. Compute both n-th means with the formula.

Answer: (b) \( a(b^2+c^2)=c(a^2+b^2) \)

Explanation:
1. With \( b^2=ac \): \( a(b^2+c^2)=a^2c+ac^2 \) and \( c(a^2+b^2)=a^2c+ac^2 \).
2. They are equal. (Test 1, 2, 4 shows a and c fail.)

Teacher's Note:
1. Replace b squared by ac.

Answer: (c) \( \log_a(\log_ea)-\log_a(\log_eb) \)

Explanation:
1. \( a^x=\log_xa\cdot\log_bx=\log_ba \).
2. \( x=\log_a(\log_ba)=\log_a\left(\frac{\ln a}{\ln b}\right)=\log_a(\ln a)-\log_a(\ln b) \).

Teacher's Note:
1. Use log_x a times log_b x = log_b a.

Answer: (c) \( 4 \)

Explanation:
1. \( \frac{10}{9}\cdot\frac{n^2(n+1)^2(2n+1)^2}{36}=\frac{n(n+1)}2\cdot\frac{n^2(n+1)^2}4 \).
2. This gives \( 80(2n+1)^2=324n(n+1) \), that is \( n^2+n-20=0 \), so \( n=4 \).

Teacher's Note:
1. Middle squared equals product of the outer terms.

Answer: (a) \( a^2,b^2,c^2 \) are in G.P.

Explanation:
1. \( (b^2)^2=a^2c^2 \) follows from \( b^2=ac \).
2. The other two fail for 1, 2, 4.

Teacher's Note:
1. Square the G.P. relation.

Answer: (c) \( 17:15 \)

Explanation:
1. As before the terms are 1, 2, 4, 8, so \( p=ab=2 \) and \( q=cd=32 \).
2. \( (q+p):(q-p)=34:30=17:15 \).

Teacher's Note:
1. Find the ratio from the two sum conditions.

Answer: (a) \( c^3a=b^3d \)

Explanation:
1. Let the roots be \( \frac\alpha r,\alpha,\alpha r \). Then \( \alpha^3=-\frac da \), \( \frac ba=-\alpha K \) and \( \frac ca=\alpha^2K \) where \( K=1+r+\frac1r \).
2. Dividing \( \left(\frac ca\right)^3 \) by \( \left(\frac ba\right)^3 \) gives \( -\alpha^3=\frac da \), so \( c^3a=b^3d \).

Teacher's Note:
1. Use the product, sum and pairwise sums of roots.

Answer: (c) \( 0 \)

Explanation:
1. \( (4x+1)^2=(2x+1)(8x+1) \) gives \( 16x^2+8x+1=16x^2+10x+1 \), so \( x=0 \).
2. At \( x=0 \) all three numbers equal 1, so they are not unequal. So there are 0 values.

Teacher's Note:
1. Check the unequal condition on the solution.

Answer: (b) G.P.

Explanation:
1. These are \( S_n,\ r^nS_n,\ r^{2n}S_n \) (blocks of n terms).
2. They form a G.P. with ratio \( r^n \).

Teacher's Note:
1. Each block is the previous one times r to the n.

Answer: (a) A.P.

Explanation:
1. Let the common value be k. Then \( a=k^x,\ b=k^y,\ c=k^z \).
2. \( b^2=ac \) gives \( 2y=x+z \), so they are in A.P.

Teacher's Note:
1. Write a, b, c as powers of k.

Answer: (b) \( 17\frac17 \)

Explanation:
1. The reciprocals \( \frac1{30},\frac1{24},\frac1{20} \) are \( \frac4{120},\frac5{120},\frac6{120} \), an A.P., so the sequence is an H.P.
2. The next reciprocal is \( \frac7{120} \), so the term is \( \frac{120}7=17\frac17 \).

Teacher's Note:
1. Test the reciprocals for A.P.

Answer: (d) \( 10 \)

Explanation:
1. The reciprocals \( \frac12,\frac25,\frac3{10} \) form an A.P. with \( d=-\frac1{10} \).
2. The 5th reciprocal is \( \frac12-\frac4{10}=\frac1{10} \), so the term is 10.

Teacher's Note:
1. Work with the reciprocals.

Answer: (a) \( \frac1{89} \)

Explanation:
1. The reciprocals are in A.P.: \( 45 \) at 5th and \( 69 \) at 11th, so \( d=4 \).
2. The 16th is \( 45+11\cdot4=89 \), so the term is \( \frac1{89} \).

Teacher's Note:
1. Use the A.P. of reciprocals.

Answer: (c) \( \frac16 \)

Explanation:
1. Reciprocals: 61 and 105, so \( d=11 \).
2. The first reciprocal is \( 61-5\cdot11=6 \), so the first term is \( \frac16 \).

Teacher's Note:
1. Use the A.P. of reciprocals.

Answer: (a) \( 1\frac{10}{17} \)

Explanation:
1. The terms are \( 27,9,\frac{27}5,\frac{27}7 \), whose reciprocals \( \frac1{27},\frac3{27},\frac5{27},\frac7{27} \) are in A.P. with \( d=\frac2{27} \).
2. The 9th reciprocal is \( \frac{1+16}{27}=\frac{17}{27} \), so the term is \( \frac{27}{17}=1\frac{10}{17} \).

Teacher's Note:
1. Write all terms over 27.