JEE Mathematics Sequence and Series MCQs Set 15

Answer: (a) \( \tfrac13 \)

Explanation:
1. \( (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\ge3(ab+bc+ca) \), because \( a^2+b^2+c^2\ge ab+bc+ca \).
2. So \( ab+bc+ca\le\tfrac13 \), with equality at \( a=b=c=\tfrac13 \).

Teacher's Note:
1. The square of a sum can be compared with the pairwise products using a^2+b^2+c^2 >= ab+bc+ca.
2. Equality holds when all numbers are equal.

Answer: (c) \( \tfrac5{36} \)

Explanation:
1. For \( |t|<1 \): \( \sum_{n\ge1}n t^{n}(-1)^{n+1}=\frac{t}{(1+t)^2} \) (differentiate the geometric series).
2. With \( t=\tfrac15 \): \( \frac{1/5}{(6/5)^2}=\frac{1}{5}\cdot\frac{25}{36}=\frac5{36} \).

Teacher's Note:
1. Series of the type n x^n come from differentiating a geometric series.
2. Be careful with the alternating sign: use -t instead of t in the formula for the sum of n x^n.

Answer: (b) 20

Explanation:
1. Common terms form an A.P. whose common difference is the L.C.M. of 4 and 5, i.e. 20; the first common term is 21.
2. The common terms are \( 21+20k \) and must be at most 417 (the smaller last term): \( k\le19.8 \), so \( k=0,1,\dots,19 \).
3. That is 20 terms.

Teacher's Note:
1. Common terms of two A.P.s form another A.P. with difference equal to the L.C.M. of the differences.
2. Use the smaller of the two last terms as the upper limit.

Answer: (b) 4

Explanation:
1. Expand: \( \frac{1+x^2+x^6+x^8}{x^4}=x^{-4}+x^{-2}+x^{2}+x^{4} \).
2. By A.M.-G.M., \( x^4+x^{-4}\ge2 \) and \( x^2+x^{-2}\ge2 \).
3. So the minimum is \( 4 \), attained at \( x=\pm1 \).

Teacher's Note:
1. Expand and pair terms that multiply to 1 so A.M.-G.M. gives a constant bound.
2. Check that equality is reachable (x = 1 here).

Answer: (a) \( a\in[\tfrac34,\infty) \)

Explanation:
1. Let the common ratio be r: \( y=xr,\ z=xr^2 \) (x is non-zero).
2. Then \( a=1+r+r^2=\left(r+\tfrac12\right)^2+\tfrac34\ge\tfrac34 \).
3. All values from \( \tfrac34 \) upward are reached by real r, so \( a\in[\tfrac34,\infty) \).

Teacher's Note:
1. Write the three terms as x, xr, xr^2 and divide by x.
2. Complete the square to find the range.

Answer: (b) \( \tfrac1{60} \)

Explanation:
1. The k-th series has first term \( 2k-1 \) and ratio \( \tfrac2{2k+1} \), so \( s_k=\frac{2k-1}{1-\frac2{2k+1}}=\frac{(2k-1)(2k+1)}{2k-1}=2k+1 \).
2. The sum is \( \sum_{k\ge1}\frac1{(2k+1)(2k+3)(2k+5)} \).
3. Use \( \frac1{(2k+1)(2k+3)(2k+5)}=\frac14\left[\frac1{(2k+1)(2k+3)}-\frac1{(2k+3)(2k+5)}\right] \) (telescoping): the sum is \( \frac14\cdot\frac1{3\cdot5}=\frac1{60} \).

Teacher's Note:
1. Find s_k in closed form first; it simplifies to 2k+1.
2. Products of three consecutive terms of an A.P. in the denominator telescope.

Answer: (b) 1

Explanation:
1. The k-th term is \( \frac{k(k+1)/2}{[k(k+1)/2]^2}=\frac2{k(k+1)} \).
2. So \( S_n=2\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)=2\left(1-\frac1{n+1}\right) \).
3. This lies in \( [1,2) \) for every \( n\ge1 \), so the integral part is 1.

Teacher's Note:
1. Use the formulas for the sum of first k natural numbers and the sum of their cubes.
2. Telescoping gives a closed form that you can bound.

Answer: (b) 5050

Explanation:
1. The highest power in the k-th bracket is \( x^k \) (\( k=1,\dots,100 \)).
2. The highest exponent of the product is \( 1+2+\cdots+100=5050 \).

Teacher's Note:
1. The degree of a product equals the sum of the degrees of the factors.
2. Sum of first n natural numbers is n(n+1)/2.

Answer: (d) (11)! - 1

Explanation:
1. Write \( k\cdot k!=(k+1)!-k! \).
2. The sum telescopes: \( \sum_{k=1}^{10}[(k+1)!-k!]=11!-1!=11!-1 \).

Teacher's Note:
1. Rewrite k times k factorial as a difference of consecutive factorials.
2. Telescoping leaves only the first and last terms.

Answer: (c) 31

Explanation:
1. Let there be \( m \) stones on each side of the middle stone, so \( 2m+1 \) stones in all.
2. Bringing the k-th stone on one side to the middle and returning covers \( 2\times10k=20k \) metres, and there are two sides: total \( 2\sum_{k=1}^m20k=20m(m+1) \) metres.
3. Set \( 20m(m+1)=4800 \): \( m(m+1)=240 \), so \( m=15 \) and the number of stones is \( 31 \).

Teacher's Note:
1. The man goes out to a stone and comes back, so each stone costs a round trip.
2. Set up the total distance on both sides of the middle stone.

Answer: (a) 6479

Explanation:
1. Row n contains n numbers, and the pattern lists the odd numbers in order.
2. Odd numbers used up to row 80: \( 1+2+\cdots+80=\tfrac{80\cdot81}{2}=3240 \).
3. The m-th odd number is \( 2m-1 \), so the last entry is \( 2\cdot3240-1=6479 \).

Teacher's Note:
1. Count how many numbers have been written up to the end of row 80.
2. The m-th odd number is 2m-1.

Answer: (b) G.P.

Explanation:
1. \( a_1=S_1=2^p-1 \).
2. \( a_n=S_n-S_{n-1}=2^{np}-2^{(n-1)p}=2^{(n-1)p}(2^p-1) \) for \( n\ge2 \), and this formula also gives \( a_1=2^p-1 \) for \( n=1 \).
3. Consecutive ratio \( \frac{a_{n+1}}{a_n}=2^p \) is constant, so it is a G.P.

Teacher's Note:
1. Terms of a sequence come from differences of consecutive partial sums.
2. Check the first term separately.

Answer: (c) 360

Explanation:
1. Natural numbers from 1 to 399: 399 numbers.
2. Divisible by 17: \( \lfloor399/17\rfloor=23 \). Divisible by 23: \( \lfloor399/23\rfloor=17 \). Divisible by both (by 391): 1.
3. Divisible by 17 or 23: \( 23+17-1=39 \). Required count \( 399-39=360 \).

Teacher's Note:
1. Use inclusion-exclusion for two divisibility conditions.
2. Numbers divisible by both are multiples of 17 x 23 = 391.

Answer: (d) [12, infinity)

Explanation:
1. Let \( u=5^x+5^{-x}\ge2 \). Then \( 5^{1+x}+5^{1-x}=5u \) and \( 25^x+25^{-x}=u^2-2 \).
2. A.P. condition: \( a=5u+u^2-2 \).
3. For \( u\ge2 \) this is increasing, with minimum \( 10+4-2=12 \) at \( u=2 \). So \( a\ge12 \).

Teacher's Note:
1. Introduce u = 5^x + 5^(-x) and use A.M.-G.M. to get u >= 2.
2. Square u to express 25^x + 25^(-x).

Answer: (b) \( 5050\pi \)

Explanation:
1. The innermost disc is red, so the ring between radii 1 and 2 is green, 2 to 3 red, and so on: green rings lie between \( k \) and \( k+1 \) for odd k.
2. Area of such a ring \( \pi[(k+1)^2-k^2]=\pi(2k+1) \).
3. Sum over \( k=1,3,\dots,99 \): \( \pi\sum(2k+1)=\pi(2\cdot2500+50)=5050\pi \).

Teacher's Note:
1. Decide which rings are green by alternation, starting from a red centre.
2. Sum of the first n odd numbers is n squared.

Answer: (a) equal to 0

Explanation:
1. \( S_{11}=S_{19} \): \( \tfrac{11}{2}(2a+10d)=\tfrac{19}{2}(2a+18d) \).
2. Simplify: \( 22a+110d=38a+342d \), so \( 16a=-232d \), \( 2a=-29d \).
3. \( S_{30}=15(2a+29d)=15(-29d+29d)=0 \).

Teacher's Note:
1. Use S_n = n/2 (2a + (n-1)d) for both sums.
2. The relation between a and d makes the 30-term sum vanish.

Answer: (d) 11 : 12

Explanation:
1. Let the radius of C be R. Then \( P_0P_1=\tfrac R2,\ P_1P_2=\tfrac R4,\dots \), so the radii of the shaded circles are \( \tfrac R4,\tfrac R8,\dots \).
2. Shaded area \( \pi R^2\left(\tfrac1{16}+\tfrac1{64}+\cdots\right)=\pi R^2\cdot\dfrac{1/16}{1-1/4}=\tfrac{\pi R^2}{12} \).
3. Unshaded fraction \( 1-\tfrac1{12}=\tfrac{11}{12} \), ratio 11 : 12.

Teacher's Note:
1. The diameters halve each time, so the areas shrink by a factor of 4.
2. Use the sum of an infinite G.P. for the shaded area.

Answer: (c) \( g_1=1 \)

Explanation:
1. Let \( g_1=a \) and ratio r. \( a(1+r+r^2)=13 \) and \( a^2(1+r^2+r^4)=91 \).
2. Use \( 1+r^2+r^4=(1+r+r^2)(1-r+r^2) \): dividing gives \( a(1-r+r^2)=7 \). So \( \frac{1+r+r^2}{1-r+r^2}=\frac{13}7 \), i.e. \( 3r^2-10r+3=0 \), and for a decreasing G.P. \( r=\tfrac13 \).
3. Then \( a=9 \): the G.P. is 9, 3, 1, 1/3, ... So the greatest term is 9, \( 3g_4=1=g_3 \), \( g_2=3 \) hold, but \( g_1=1 \) does not.

Teacher's Note:
1. Factorise 1 + r^2 + r^4 to relate the two given sums.
2. Decreasing positive G.P.: the ratio is between 0 and 1.

Answer: (c) \( \tfrac{25}{54} \)

Explanation:
1. The series is \( \sum_{k\ge1}k^2x^{k-1} \) with \( x=-\tfrac15 \).
2. Known sum: \( \sum k^2x^{k-1}=\frac{1+x}{(1-x)^3} \).
3. So \( S=\frac{4/5}{(6/5)^3}=\frac45\cdot\frac{125}{216}=\frac{25}{54} \).

Teacher's Note:
1. Recognise the pattern k squared times a power of a fixed ratio.
2. The sum of k^2 x^(k-1) comes from differentiating x/(1-x)^2.

Answer: (a) \( \left[2^{\frac{1-n}2}\right]r \)

Explanation:
1. The first square has side \( 2r \) (the circle is inscribed), so its inradius is r.
2. Joining mid-points gives a square whose side is the original side divided by \( \sqrt2 \), so the inradius also shrinks by \( \tfrac1{\sqrt2} \) each time.
3. The inradius forms a G.P. \( r,\ \frac r{\sqrt2},\ \frac r2,\dots \), so the n-th is \( r\cdot2^{-(n-1)/2}=2^{\frac{1-n}2}r \).

Teacher's Note:
1. Joining mid-points of a square's sides produces a square rotated by 45 degrees with side divided by root 2.
2. Inradius is half the side, so it scales in the same way.