JEE Mathematics Sequence and Series MCQs Set 39

Answer: (b) \( 0 \lt r \lt 1 \) or \( r \gt 3 \)

Explanation:
1. \( b_3 \gt 4b_2-3b_1 \) gives \( r^2 \gt 4r-3 \) (dividing by \( b_1 \gt 0 \)), i.e. \( r^2-4r+3 \gt 0 \).
2. \( (r-1)(r-3) \gt 0 \), so \( r \lt 1 \) or \( r \gt 3 \). With \( r \gt 0 \): \( 0 \lt r \lt 1 \) or \( r \gt 3 \).

Teacher's Note:
1. Divide by the positive first term and solve the quadratic inequality, remembering that \( r \) must be positive here.

Answer: (a) \( 3 \)

Explanation:
1. For \( k\ge2 \): \( S_k=\frac{(k-1)/k!}{1-1/k}=\frac{1}{(k-1)!} \). For \( k=1 \) the first term is \( 0 \), so \( S_1=0 \).
2. The term for \( k=2 \) is \( |4-6+1|\cdot1=1 \); for \( k\ge3 \) the factor \( k^2-3k+1 \gt 0 \). Write \( k^2-3k+1=(k-1)(k-2)-1 \), so the term is \( \frac{(k-1)(k-2)-1}{(k-1)!}=\frac{1}{(k-3)!}-\frac{1}{(k-1)!} \).
3. Telescoping for \( k=3 \) to \( 100 \): \( \sum\frac{1}{(k-3)!}-\sum\frac1{(k-1)!}=\left(1+1\right)-\left(\frac1{98!}+\frac1{99!}\right)=2-\frac{100}{99!} \). Adding the \( k=2 \) term \( 1 \) and \( \frac{100^2}{100!}=\frac{100}{99!} \) gives \( 1+2-\frac{100}{99!}+\frac{100}{99!}=3 \).

Teacher's Note:
1. The extra term \( \frac{100^2}{100!} \) is exactly what cancels the tail of the telescoping sum. A numerical check with exact fractions gives exactly \( 3 \).

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

Explanation:
1. In an A.P. the second difference is zero: \( a_1-2a_2+a_3=0 \), so (ii) holds.
2. The fourth difference \( a_1-4a_2+6a_3-4a_4+a_5 \) is also zero, so (iv) holds.
3. For \( a_n=n \): (i) gives \( 1+4+3=8\ne0 \) and (iii) gives \( 2+6-9-4=-5\ne0 \). So (ii) and (iv).

Teacher's Note:
1. The coefficients of a vanishing difference of an A.P. are binomial coefficients with alternating signs. Test the others with \( a_n=n \).

Answer: (d) \( 4\ell m^2n \)

Explanation:
1. Let the common ratio be \( R \), with \( R^4=\frac n\ell \). Then \( G_1=\ell R \), \( G_2=\ell R^2 \), \( G_3=\ell R^3 \).
2. \( G_1^4+2G_2^4+G_3^4=\ell^4R^4\left(1+2R^4+R^8\right)=\ell^4R^4\left(1+R^4\right)^2=\ell^3n\left(\frac{\ell+n}{\ell}\right)^2=\ell n(\ell+n)^2 \).
3. With \( \ell+n=2m \): \( \ell n\cdot4m^2=4\ell m^2n \).

Teacher's Note:
1. Insert three G.M.s as a G.P. of five terms so that \( R^4=n/\ell \). The expression factors as a perfect square.

Answer: (a) \( 150 \)

Explanation:
1. Since \( a_{2n+1}=r\,a_{2n} \), \( \sum a_{2n+1}=r\sum a_{2n} \), so \( 200=100r \) and \( r=2 \).
2. Also \( a_{2n-1}=\frac{a_{2n}}{r} \), so \( \sum_{n=1}^{100}a_{2n-1}=\frac{100}{2}=50 \).
3. \( \sum_{n=1}^{200}a_n=50+100=150 \).

Teacher's Note:
1. Each odd-numbered term is \( r \) times the preceding even-numbered one. Use this to compare the two given sums.

Answer: (a) \( 6 \)

Explanation:
1. Right triangle with sides in A.P.: the sides are \( 3k,4k,5k \).
2. Area \( =\frac12\cdot3k\cdot4k=6k^2=24 \), so \( k=2 \).
3. The smallest side is \( 3k=6 \).

Teacher's Note:
1. The only right triangle with sides in A.P. is similar to the \( 3,4,5 \) triangle. Use the area to find the scale factor.

Answer: (d) \( 248 \)

Explanation:
1. The \( k \)-th term is \( k^2 \) for odd \( k \) and \( 2k^2 \) for even \( k \). With \( k \) odd giving \( k^2 \) and \( k=2j \) giving \( 4j^2 \): \( A=\sum_{k=1}^{20}k^2+\sum_{j=1}^{10}4j^2=2870+1540=4410 \).
2. Similarly \( B=\sum_{k=1}^{40}k^2+\sum_{j=1}^{20}4j^2=22140+11480=33620 \).
3. \( B-2A=33620-8820=24800=100\lambda \), so \( \lambda=248 \).

Teacher's Note:
1. Express the series using \( k^2 \) plus an extra \( k^2 \) for even \( k \). The difference \( B-2A \) then simplifies to a closed-form number.

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

Explanation:
1. A.P.: \( b=2a-2 \). Sum: \( c=25-a-b=27-3a \). G.P.: \( c^2=18b \).
2. \( (27-3a)^2=36a-36 \) gives \( a^2-22a+85=0 \), so \( a=5 \) or \( a=17 \). The root \( a=17 \) gives \( b=32 \gt 18 \), which is rejected.
3. Hence \( a=5 \), \( b=8 \), \( c=12 \): \( b+c=20 \) and \( a+b+c=25 \), so all four statements are true.

Teacher's Note:
1. Reduce everything to one variable using the A.P. and G.P. conditions, and reject roots violating the stated range.

Answer: (c) \( 34 \) minutes

Explanation:
1. In the first \( 10 \) minutes he counts \( 1500 \) notes. The remaining \( 3000 \) notes are counted at \( a_{11}=148,\,146,\,\dots \) per minute (an A.P.).
2. Let \( m \) more minutes be needed: \( \frac m2\left[2\cdot148+(m-1)(-2)\right]=3000 \), so \( m^2-149m+3000=0 \), i.e. \( (m-24)(m-125)=0 \).
3. \( m=125 \) is impossible because the terms become negative; \( m=24 \). The total time is \( 10+24=34 \) minutes.

Teacher's Note:
1. Reject the larger root because the number of notes per minute cannot become negative. The ten equal minutes are counted separately.

Answer: (d) \( y(1-x)=1 \)

Explanation:
1. \( x=\frac{1}{1+\tan^2\theta}=\cos^2\theta \) (G.P. with ratio \( -\tan^2\theta \)).
2. \( y=\frac{1}{1-\cos^2\theta}=\frac1{\sin^2\theta} \).
3. So \( 1-x=\sin^2\theta=\frac1y \), i.e. \( y(1-x)=1 \).

Teacher's Note:
1. Both are infinite G.P.s that converge for \( 0 \lt \theta \lt \frac\pi4 \). Sum them with \( \frac{a}{1-r} \) and relate \( x \) and \( y \) through \( \cos^2\theta \).

Answer: (a) \( 9 \)

Explanation:
1. \( \frac{S_7}{S_{11}}=\frac{7(a+3d)}{11(a+5d)}=\frac{6}{11} \), so \( 7a+21d=6a+30d \) and \( a=9d \).
2. \( a_7=a+6d=15d \) must lie between \( 130 \) and \( 140 \): \( \frac{130}{15} \lt d \lt \frac{140}{15} \), so \( d=9 \).
3. Check: \( a=81 \) and \( a_7=135 \).

Teacher's Note:
1. The sum ratio gives \( a \) in terms of \( d \), and then the range for \( a_7 \) pins down the integer \( d \).

Answer: (c) \( 8 \)

Explanation:
1. For a G.P. \( \Delta^kT_n=T_n(r-1)^k \), so \( T_1(r-1)^2=36 \) with \( T_1 \) a natural number.
2. So \( (r-1)^2\in\{1,4,9,36\} \): \( r-1\in\{\pm1,\pm2,\pm3,\pm6\} \), i.e. \( r\in\{2,0,3,-1,4,-2,7,-5\} \).
3. The value \( r=0 \) does not give a G.P. and contributes \( 0 \) to the sum anyway: \( 2+3-1+4-2+7-5=8 \).

Teacher's Note:
1. Each difference operator multiplies a G.P. term by \( (r-1) \). Divisor conditions then restrict the ratio to a finite set.

Answer: (b) \( 9 \)

Explanation:
1. The sum of interior angles is \( (n-2)180^\circ=\frac n2\left[240+5(n-1)\right] \), which simplifies to \( 5n^2-125n+720=0 \), i.e. \( n^2-25n+144=0 \).
2. So \( n=9 \) or \( n=16 \). For a convex polygon every angle is below \( 180^\circ \): the largest angle is \( 120+5(n-1) \), which is \( 160^\circ \) for \( n=9 \) but \( 195^\circ \) for \( n=16 \).
3. Hence \( n=9 \).

Teacher's Note:
1. Two roots arise from the angle-sum equation, and convexity (largest angle less than \( 180^\circ \)) selects one of them.

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

Explanation:
1. Let the numbers be \( a,ar,ar^2 \). Doubling the middle: \( a,2ar,ar^2 \) in A.P. gives \( 4ar=a+ar^2 \).
2. \( r^2-4r+1=0 \), so \( r=2\pm\sqrt3 \).
3. The G.P. is increasing, so \( r \gt 1 \): \( r=2+\sqrt3 \).

Teacher's Note:
1. Form the A.P. condition \( 2\times\text{middle}=\text{sum of extremes} \) and solve the quadratic. The increasing condition decides the root.

Answer: (b) \( 3121 \)

Explanation:
1. \( 91=7\times13 \), so H.C.F.\( (91,n) \gt 1 \) means \( n \) is a multiple of \( 7 \) or of \( 13 \).
2. Multiples of \( 7 \) in \( (100,200) \): \( 105,\dots,196 \) (\( 14 \) terms, sum \( 2107 \)). Multiples of \( 13 \): \( 104,\dots,195 \) (\( 8 \) terms, sum \( 1196 \)). Multiples of \( 91 \): \( 182 \).
3. Sum \( =2107+1196-182=3121 \).

Teacher's Note:
1. Use inclusion-exclusion: subtract the multiples of \( 91 \) that were counted twice.

Answer: (b) (iii) and (iv) only

Explanation:
1. If \( a\ne1 \): \( S=\frac{a^{2017}-1}{a-1}=\frac{2a-2}{a-1}=2 \).
2. If \( a=1 \): it satisfies \( 1-2+1=0 \), and then \( S=2017 \) (2017 terms each equal to \( 1 \)).
3. So \( S \) can be \( 2017 \) or \( 2 \).

Teacher's Note:
1. Do not forget the special value \( a=1 \) when using the G.P. sum formula. It gives a different sum.

Answer: (c) \( 6 \)

Explanation:
1. It is a G.P., so \( a_2a_{n-1}=a_1a_n=128 \). With \( a_1+a_n=66 \): \( a_1,a_n \) are the roots of \( t^2-66t+128=0 \), i.e. \( 2 \) and \( 64 \). Increasing: \( a_1=2 \), \( a_n=64 \).
2. Sum: \( \frac{ra_n-a_1}{r-1}=\frac{64r-2}{r-1}=126 \), so \( 62r=124 \) and \( r=2 \).
3. \( 2\cdot2^{n-1}=64 \) gives \( n=6 \) (and the ratio is \( 2 \)).

Teacher's Note:
1. The product of terms equidistant from the ends of a G.P. is constant. Use sum and product to find the first and last terms.

Answer: (c) \( 2 \)

Explanation:
1. \( g(n)-f(n)=\frac{n(n+1)(2n+1)}6-\frac{n(n+1)}2=\frac{n(n+1)(n-1)}{3} \).
2. Values: \( n=2 \): \( 2 \); \( n=3 \): \( 8 \); \( n=4 \): \( 20 \); \( n=5 \): \( 40 \). Their greatest common divisor is \( 2 \).
3. The even natural number dividing all of them is \( 2 \).

Teacher's Note:
1. Simplify the difference to a product of three consecutive integers over \( 3 \), then evaluate a few values and take the common divisor.

Answer: (a) \( 2 \)

Explanation:
1. \( \Delta^kT_n=T_n(r-1)^k \), so \( (r-1)^7=(r-1)^3 \).
2. As \( r\ne1 \): \( (r-1)^4=1 \), so \( r-1=\pm1 \), i.e. \( r=2 \) or \( r=0 \). The value \( r=0 \) is excluded.
3. Hence \( r=2 \).

Teacher's Note:
1. Reduce to \( (r-1)^4=1 \) and discard the invalid roots \( r=0 \) and \( r=1 \).

Answer: (c) \( 21 \) months

Explanation:
1. The first three months give Rs. \( 600 \). From the fourth month the savings are \( 240,280,320,\dots \) (an A.P. with \( d=40 \)): the \( j \)-th later month is \( 200+40j \).
2. \( 600+\sum_{j=1}^{m}(200+40j)=11040 \) gives \( 200m+20m(m+1)=10440 \), so \( m^2+11m-522=0 \) and \( m=18 \).
3. The total number of months is \( 3+18=21 \).

Teacher's Note:
1. Treat the first three equal months separately and the rest as an A.P. Solve the resulting quadratic and keep the positive root.

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

Explanation:
1. The signs of \( (-1)^{k(k+1)/2} \) follow the pattern \( -,-,+,+ \) repeating. Each block of four gives \( (4j+3)^2+(4j+4)^2-(4j+1)^2-(4j+2)^2=32j+20 \) for \( j=0,1,\dots \).
2. So \( S_n=\sum_{j=0}^{n-1}(32j+20)=16n(n-1)+20n=16n^2+4n \).
3. \( 16n^2+4n=1056 \) gives \( n=8 \) and \( 16n^2+4n=1332 \) gives \( n=9 \); \( 1088 \) and \( 1120 \) give no integer \( n \).

Teacher's Note:
1. Group the terms in fours to find the closed form \( S_n=16n^2+4n=4n(4n+1) \). Then test each value for an integer \( n \).