JEE Mathematics Sequence and Series MCQs Set 11

Answer: (d) 650

Explanation:
1. The n-th term is \( 50-2(n-1)=52-2n \), which is positive for n ≤ 25 and zero for n = 26.
2. The maximum sum \( S_{26}=\frac{26}{2}(50+0)=650 \).

Teacher's Note:
1. The sum is largest when only non-negative terms are added.
2. The 26th term is zero.

Answer: (b) \( \gt2b^3 \)

Explanation:
1. Let \( a\ln x=b\ln y=c\ln z=k \). Since \( 2\ln y=\ln x+\ln z \), \( \frac2b=\frac1a+\frac1c \): a, b, c are in HP.
2. So \( b=\frac{2ac}{a+c}\le\sqrt{ac} \) (HM ≤ GM, strict for a ≠ c), and \( b^3<(ac)^{3/2}\le\frac{a^3+c^3}2 \).
3. Hence \( a^3+c^3>2b^3 \).

Teacher's Note:
1. x, y, z in GP means ln x, ln y, ln z are in AP, so a, b, c are in HP.
2. Use HM < GM < power mean.

Answer: (d) \( \frac12+\frac1{\sqrt2} \)

Explanation:
1. AP: \( a+c=2b \), so \( 3b=\frac32\Rightarrow b=\frac12 \) and \( a+c=1 \).
2. GP: \( b^4=a^2c^2\Rightarrow ac=\pm\frac14 \). The case ac = 1/4 gives a = c = 1/2, which violates a > b > c.
3. So \( ac=-\frac14 \): a, c are roots of \( t^2-t-\frac14=0 \), t = (1 ± √2)/2. As a > c, \( a=\frac12+\frac1{\sqrt2} \).

Teacher's Note:
1. From a + b + c = 3/2 and AP, b = 1/2.
2. The GP condition gives b⁴ = (ac)², so ac may be negative.

Answer: (b) 363

Explanation:
1. Sum of 21 terms of an AP = 21 × (middle term): \( 21a_{11}=693\Rightarrow a_{11}=33 \).
2. The terms a₁, a₃, ..., a₂₁ form an AP of 11 terms with middle term a₁₁, so the sum is \( 11\times33=363 \).

Teacher's Note:
1. The sum of an odd number of AP terms is the number of terms times the middle term.
2. Alternate terms of an AP again form an AP.

Answer: (a) 76

Explanation:
1. GP gives \( x_1x_{20}=4 \). Let \( \frac1{x_r} \) be an AP with common difference d.
2. \( x_rx_{r+1}=\frac{x_r-x_{r+1}}{d} \), so the sum telescopes to \( \frac{x_1-x_{20}}{d} \).
3. Also \( 19d=\frac1{x_{20}}-\frac1{x_1}=\frac{x_1-x_{20}}{x_1x_{20}}=\frac{x_1-x_{20}}4 \). Hence the sum is 4 × 19 = 76.

Teacher's Note:
1. Reciprocals of an HP form an AP.
2. Use 1/x_{r+1} − 1/x_r = d to write x_r x_{r+1} as a difference.

Answer: (c) 8

Explanation:
1. \( (a+b)(c+d)\ge16 \).
2. By AM-GM, \( (a+b)+(c+d)\ge2\sqrt{16}=8 \).

Teacher's Note:
1. Combine the logarithms.
2. AM-GM on the two brackets.

Answer: (c) \( \frac35(6^{10}-1) \)

Explanation:
1. The terms are 1, 2, 6, 12, 36, 72, ... (a₂ = 2a₁, a₃ = 3a₂ = 6, a₄ = 2a₃ = 12, ...).
2. Pair them: \( (1+2)+(6+12)+(36+72)+\cdots=3(1+6+36+\cdots) \): a GP with ratio 6 and 10 pairs.
3. Sum \( =3\cdot\frac{6^{10}-1}{5}=\frac35(6^{10}-1) \).

Teacher's Note:
1. Write the first few terms and group them in pairs.
2. Each pair is 6 times the previous pair.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The given relation is \( x^2+9y^2+25z^2-15yz-5zx-3xy=0 \), i.e. \( \frac12\left[(x-3y)^2+(3y-5z)^2+(5z-x)^2\right]=0 \).
2. By R, \( x=3y=5z \), so \( \frac1x:\frac1y:\frac1z=1:3:5 \): the reciprocals are in AP, so x, y, z are in HP. R is the reason.

Teacher's Note:
1. Write the equation as a sum of three squares.
2. Then use the fact that a sum of squares of real numbers is zero only if each is zero.