JEE Mathematics Progressions Related Inequalities and Series MCQs Set 13

Answer: (b) \(\frac12a(a-1)(4a-1)\)

Explanation:
1. Take \(A=(0,0),B=(a,0),C=(a,a),D=(0,a)\). Then \(L_n=(a,n)\) and \(M_n=(a-n,a)\).
2. \(AL_n^2=a^2+n^2\) and \(L_nM_n^2=n^2+(a-n)^2\).
3. Sum over \(n=1\) to \(a-1\) equals \(\frac12a(a-1)(4a-1)\). Check \(a=3\): \(33=\frac12\cdot3\cdot2\cdot11\).

Teacher's Note:
1. Place the square on coordinates and use the sum of squares formula.

Answer: (d) 12 cm

Explanation:
1. Edges \(\frac6r,6,6r\) (the middle edge cubed is 216, so it is 6).
2. Surface area: \(2\left(\frac{36}{r}+36+36r\right)=252\), so \(r+\frac1r=\frac52\), giving \(r=2\) or \(\frac12\).
3. The edges are 3, 6 and 12 cm; the longest is 12 cm.

Teacher's Note:
1. Write the edges as a/r, a, ar; the volume gives a directly.

Answer: (d) \(\frac{an}{2}\)

Explanation:
1. By similar triangles each segment is \(a\cdot\frac{i}{n+1}\) for \(i=1,\ldots,n\).
2. Sum \(=\frac{a}{n+1}\cdot\frac{n(n+1)}{2}=\frac{an}{2}\).

Teacher's Note:
1. Parallel segments in a triangle form an arithmetic progression.

Answer: (d) \(\{1,-1\}\)

Explanation:
1. \(5(5^x+5^{-x})=26\), so with \(t=5^x\): \(t+\frac1t=\frac{26}{5}\).
2. \(5t^2-26t+5=0\) gives \(t=5\) or \(\frac15\), so \(x=1\) or \(x=-1\).

Teacher's Note:
1. Put t = 5 to the power x and use t + 1/t.

Answer: (d) (iii) and (iv)

Explanation:
1. \(S_{3n}=\frac{3n}{2}\left[2a+(3n-1)d\right]\).
2. \(S_{2n}-S_n=\frac n2\left[2a+(3n-1)d\right]\).
3. The ratio is 3, a constant: (iii) and (iv) are true.

Teacher's Note:
1. Write each sum with the standard AP formula; the same bracket appears.

Answer: (b) (i) and (iii)

Explanation:
1. Let \(a^x=b^y=c^z=k\): \(x=\frac{\ln k}{\ln a}\) etc. \(y^2=xz\) gives \((\ln b)^2=\ln a\ln c\).
2. So \(\frac{\ln a}{\ln b}=\frac{\ln b}{\ln c}\), i.e. \(\log_b a=\log_c b\): (i) true.
3. Also \(\log_c b=\frac{\ln b}{\ln c}=\frac zy\) (from \(\ln b=\frac{\ln k}{y}\), \(\ln c=\frac{\ln k}{z}\)): (iii) true; (ii) is \(\frac{\ln b}{\ln a}\), different.

Teacher's Note:
1. Convert each exponent to a ratio of logarithms of k.

Answer: (c) All of (i), (ii) and (iii)

Explanation:
1. \(2n,\ n(n-1),\ n(n-1)(n-2)\) in AP: \(2n(n-1)=2n+n(n-1)(n-2)\), giving \(n^2-5n+6=0\), so \(n=2\) or \(3\).
2. \(n\ge3\) is needed for \({}^nP_3\), so \(n=3\) and the terms are \(6,6,6\).
3. Equal terms are in GP, in HP and equal: all three are true.

Teacher's Note:
1. Permutations nP3 need n at least 3, which leaves only n = 3.

Answer: (a) All of (i), (ii), (iii) and (iv)

Explanation:
1. \(a\cdot ar\cdot ar^2\cdot ar^3=a^4r^6=4\) and \(ar\cdot ar^3=a^2r^4=1\).
2. Dividing the square of the second by the first: \(\frac{1}{r^2}=4\), so \(r=\pm\frac12\), and \(a^2=16\), so \(a=\pm4\).
3. \(\frac{a}{1-r}\): \(\frac{4}{1/2}=8\), \(\frac{-4}{1/2}=-8\), \(\frac{4}{3/2}=\frac83\), \(\frac{-4}{3/2}=-\frac83\). All four values occur.

Teacher's Note:
1. Both the first term and the ratio have two possible signs, so four sums are possible.

Answer: (a) (i) and (ii)

Explanation:
1. \(a_n=n!\le n^n\) because each factor is at most \(n\): (i) true.
2. By AM-GM on \(1,2,\ldots,n\): \(\frac{n+1}{2}\ge(n!)^{1/n}\): (ii) true.
3. For \(n=1\): \(1\ge2\) is false, so (iii) is not true for all \(n\).

Teacher's Note:
1. AM-GM on the numbers 1 to n gives the second inequality.

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

Explanation:
1. \(x=\frac{a+b}2,\ y=\sqrt{ab},\ z=\frac{2ab}{a+b}\), so \(y^2=ab=xz\): (i) true.
2. \(a=5z\) gives \(10b=a+b\), so \(a=9b\); then \(x=5b,\ y=3b\) and \(x-y=2b=2\): \(b=1,\ a=9\): (iii) true.
3. The values are \(x=5,\ y=3,\ z=\frac95\), so \(x>y>z\): (ii) true. (iv) is false.

Teacher's Note:
1. Use a = 5z to get a = 9b, then x - y = 2 fixes b.

Answer: (b) (ii), (iii) and (iv)

Explanation:
1. Side of \(S_{n+1}\) is the side of \(S_n\) divided by \(\sqrt2\), so the area halves each time: \(A_n=100\cdot2^{-(n-1)}\).
2. \(A_n<1\) needs \(2^{n-1}>100\), so \(n-1\ge7\), \(n\ge8\).
3. \(A_7=\frac{100}{64}\approx1.56>1\); \(A_8=0.78,\ A_9,\ A_{10}\) are below 1. So 8, 9, 10.

Teacher's Note:
1. Areas form a GP with ratio one half.

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

Explanation:
1. Let the AP have difference \(d\) and the GP ratio \(r\): \(b=ar,\ c=ar^2,\ p=a+d,\ q=a+2d\).
2. \((b+p)^2=2a(c+q)\) simplifies to \(d^2+2ad(r-1)-a^2(r-1)^2=0\), with \(a(r-1)=b-a\).
3. So \(d=(b-a)(-1\pm\sqrt2)\): \(d=(\sqrt2-1)(b-a)\) or \(d=(\sqrt2+1)(a-b)\). These are (iv) and (ii).

Teacher's Note:
1. Reduce to a quadratic in d with b - a as a parameter.

Answer: (a) (ii) and (iv)

Explanation:
1. For numbers \(p,q\): \(g_1g_2=pq\).
2. \(a_1=\frac{2p+q}{3},\ a_2=\frac{p+2q}{3}\) and \(h_1=\frac{3pq}{p+2q},\ h_2=\frac{3pq}{2p+q}\).
3. \(a_1h_2=pq\) and \(a_2h_1=pq\), while \(a_1h_1\) and \(a_2h_2\) are not constant multiples of \(pq\). So (ii) and (iv).

Teacher's Note:
1. Write the means explicitly for the two numbers p and q.

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

Explanation:
1. AP: terms with common difference \(\frac3m\) reach 4 after \(m\) steps and 16 after \(5m\) steps, for every positive integer \(m\): infinitely many APs.
2. GP: ratio \(4^{1/k}\) for any positive integer \(k\) passes through 1, 4, 16: infinitely many GPs.
3. So (iii) and (iv) are true.

Teacher's Note:
1. Choose a smaller common difference or ratio that still hits all three numbers.