Class 11 Mathematics Sequences and Series MCQs Set 13

Answer: (a) 2

Explanation:
1. The numbers of pots 2, 4, 8, … form a GP.
2. r = 4/2 = 8/4 = 2

Teacher's Note:
1. In a GP the ratio of any term to its preceding term is the common ratio.
2. Here each row has twice as many pots as the row before.

Answer: (b) 256

Explanation:
1. a = 2, r = 2
2. a₈ = 2 × 2⁷ = 2⁸ = 256

Teacher's Note:
1. aₙ = arⁿ⁻¹; here it simplifies to 2ⁿ.
2. Check with row 3: 2³ = 8.

Answer: (d) 96

Explanation:
1. a₇ = 2⁷ = 128 and a₅ = 2⁵ = 32
2. Difference = 128 − 32 = 96

Teacher's Note:
1. The nth row has 2ⁿ pots in this arrangement.
2. In a GP the difference of terms is not constant, unlike an AP.

Answer: (b) 2046

Explanation:
1. The rows contain 2, 4, 8, …, a GP with a = 2, r = 2, n = 10.
2. \( S_{10} = \frac{2(2^{10} - 1)}{2 - 1} = 2 \times 1023 = 2046 \)

Teacher's Note:
1. Use the GP sum formula with r > 1.
2. 1024 is the number of pots in the 10th row alone, not the total.

Answer: (b) 8

Explanation:
1. 2(2ⁿ − 1) = 510 ⇒ 2ⁿ − 1 = 255
2. 2ⁿ = 256 = 2⁸ ⇒ n = 8

Teacher's Note:
1. Set the sum formula equal to the given total and solve for n.
2. Check: 2 + 4 + … + 256 = 510.

Answer: (d) None of these

Explanation:
1. Joining mid-points of a square of side s gives a square of side \( \frac{s}{\sqrt2} \), so the sides form a GP with a = 10, \( r = \frac{1}{\sqrt2} \).
2. Side of 4th square = \( 10 \left(\frac{1}{\sqrt2}\right)^3 = \frac{10}{2\sqrt2} = \frac{5}{\sqrt2} = \frac{5\sqrt2}{2} \) cm
3. This value is not among (a), (b), (c), so the answer is (d).

Teacher's Note:
1. By Pythagoras, the new side is √((s/2)² + (s/2)²) = s/√2.
2. The sides are 10, 5√2, 5, 5/√2, 5/2, …

Answer: (d) \( \frac{25}{4} \)

Explanation:
1. Side of 5th square = \( 10\left(\frac{1}{\sqrt2}\right)^4 = \frac{10}{4} = \frac{5}{2} \) cm
2. Area = \( \left(\frac{5}{2}\right)^2 = \frac{25}{4} \) sq cm

Teacher's Note:
1. The areas form a GP with ratio 1/2: 100, 50, 25, 25/2, 25/4, …
2. Each new square has exactly half the area of the previous one.

Answer: (c) 5

Explanation:
1. Side of 7th square = \( 10\left(\frac{1}{\sqrt2}\right)^6 = \frac{10}{8} = \frac{5}{4} \) cm
2. Perimeter = \( 4 \times \frac{5}{4} = 5 \) cm

Teacher's Note:
1. Perimeters form a GP with the same ratio 1/√2 as the sides.
2. (1/√2)⁶ = 1/8.

Answer: (b) 200

Explanation:
1. Areas: 100, 50, 25, … form an infinite GP with a = 100, r = 1/2.
2. \( S_\infty = \frac{100}{1 - \frac{1}{2}} = 200 \) sq cm

Teacher's Note:
1. An infinite GP has a finite sum a/(1 − r) when |r| < 1.
2. The ratio of areas is the square of the ratio of sides: (1/√2)² = 1/2.

Answer: (a) \( 80 + 40\sqrt2 \)

Explanation:
1. Perimeters: 40, \( \frac{40}{\sqrt2} \), … form an infinite GP with a = 40, \( r = \frac{1}{\sqrt2} \).
2. \( S_\infty = \frac{40}{1 - \frac{1}{\sqrt2}} = \frac{40\sqrt2}{\sqrt2 - 1} = 40\sqrt2(\sqrt2 + 1) = 80 + 40\sqrt2 \) cm

Teacher's Note:
1. Rationalise by multiplying numerator and denominator by √2 + 1.
2. The sum of perimeters is a finite number even though infinitely many squares are drawn.

Answer: (c) 1.5

Explanation:
1. The mid-point triangle has half the side, so the sides form a GP: 24, 12, 6, 3, 1.5, … with r = 1/2.
2. Side of 5th triangle = \( 24 \times \left(\frac{1}{2}\right)^4 = 1.5 \) cm

Teacher's Note:
1. By the mid-point theorem, the segment joining mid-points of two sides is half the third side.
2. Count carefully: the original triangle is the 1st one.

Answer: (b) \( \frac{567}{4} \)

Explanation:
1. Perimeters: 72, 36, 18, … form a GP with a = 72, r = 1/2.
2. \( S_6 = \frac{72\left(1 - \frac{1}{64}\right)}{1 - \frac{1}{2}} = 144 \times \frac{63}{64} = \frac{567}{4} \) cm

Teacher's Note:
1. Use S = a(1 − rⁿ)/(1 − r) when r < 1.
2. The perimeter of an equilateral triangle is 3 × side.

Answer: (b) \( 192\sqrt3 \)

Explanation:
1. Area of the first triangle = \( \frac{\sqrt3}{4} \times 24^2 = 144\sqrt3 \) sq cm.
2. Each new triangle has 1/4 of the previous area, so the areas form a GP with r = 1/4.
3. \( S_\infty = \frac{144\sqrt3}{1 - \frac{1}{4}} = 192\sqrt3 \) sq cm

Teacher's Note:
1. When lengths are halved, areas become one-fourth.
2. Area of an equilateral triangle of side a is \( \frac{\sqrt3}{4}a^2 \).

Answer: (a) 144

Explanation:
1. Perimeters: 72, 36, 18, … form an infinite GP with a = 72, r = 1/2.
2. \( S_\infty = \frac{72}{1 - \frac{1}{2}} = 144 \) cm

Teacher's Note:
1. An infinite GP with ratio 1/2 sums to twice its first term.
2. Compare with the sum of the first 6 perimeters (567/4 ≈ 141.75), which is just below 144.

Answer: (b) \( \frac{9}{8} \)

Explanation:
1. Perimeters form a GP with a = 3 × 24 = 72 and r = 1/2.
2. Perimeter of 7th triangle = \( 72 \times \left(\frac{1}{2}\right)^6 = \frac{72}{64} = \frac{9}{8} \) cm

Teacher's Note:
1. The nth term of the GP is arⁿ⁻¹, so the 7th uses r⁶.
2. Equivalently, side of 7th triangle = 24/64 = 3/8 cm, and 3 × 3/8 = 9/8.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Putting n = 1, 2, 3, … in 2n gives 2, 4, 6, …, so A is true.
2. Putting n = 1, 2, 3, … in 2n − 1 gives 1, 3, 5, …, so R is true.
3. R describes odd numbers and does not explain A.

Teacher's Note:
1. Test a general-term formula with the first few values of n.
2. Each odd number is one less than the corresponding even number.

Answer: (d) A is false, but R is true.

Explanation:
1. ar³ = (ar)² ⇒ r = a = −3.
2. a₇ = ar⁶ = −3 × 729 = −2187, not 2187. So A is false.
3. For the AP, \( S_{10} = \frac{10}{2}[2 \times 6 + 9 \times 2] = 5 \times 30 = 150 \), so R is true.

Teacher's Note:
1. From ar³ = a²r², dividing by ar² (non-zero) gives r = a.
2. An even power of a negative ratio is positive, but the negative first term keeps the sign negative.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. ar⁴ = 48 and ar⁷ = 384 ⇒ r³ = 8 ⇒ r = 2, so A is true.
2. x = (18 + 14)/2 = 16, so R is true.
3. R is about an AP and does not explain A.

Teacher's Note:
1. Dividing two terms of a GP eliminates a.
2. The middle term of three numbers in AP is their arithmetic mean.

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

Explanation:
1. By R, \( \frac{K}{-2/7} = \frac{-7/2}{K} \Rightarrow K^2 = \left(-\frac{2}{7}\right)\left(-\frac{7}{2}\right) = 1 \)
2. So K = ±1; A is true and follows from R.

Teacher's Note:
1. Three numbers a, b, c are in GP exactly when b² = ac.
2. Both signs of K are valid because only K² is fixed.

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

Explanation:
1. a = 4, r = 4: by R, \( S_6 = \frac{4(4^6 - 1)}{4 - 1} = \frac{4 \times 4095}{3} = 5460 \)
2. So A is true and it is obtained from the formula in R.

Teacher's Note:
1. The formula a(rⁿ − 1)/(r − 1) is valid for every r ≠ 1; it is just more convenient when r > 1.
2. 4⁶ = 4096.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Each term = 3 × (sum of terms after it): \( a = 3 \cdot \frac{ar}{1 - r} \Rightarrow 1 - r = 3r \Rightarrow r = \frac{1}{4} \). A is true (then a + ar = 5 gives a = 4).
2. 10th term of the AP = 3 + 9 × 3 = 30, so R is true.
3. R does not explain A.

Teacher's Note:
1. The sum of the terms after the first term of an infinite GP is ar/(1 − r).
2. The condition on each term gives r; the sum of the first two terms then gives a.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( S_n = \frac{6}{9}[0.9 + 0.99 + 0.999 + \ldots] = \frac{2}{3}\left[(1 - 0.1) + (1 - 0.01) + \ldots\right] \)
2. \( = \frac{2}{3}\left[n - \frac{0.1(1 - 0.1^n)}{1 - 0.1}\right] = \frac{2}{3}\left[n - \frac{1}{9}\left(1 - \frac{1}{10^n}\right)\right] \). A is true.
3. R is the correct formula for the general term, but A uses the sum of a GP, so R is not the explanation.

Teacher's Note:
1. Write 0.6, 0.66, … as (6/9)(1 − 0.1), (6/9)(1 − 0.01), … to reveal a GP.
2. The sum 0.1 + 0.01 + … (n terms) is a GP with ratio 1/10.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( S_n = \frac{5}{9}[9 + 99 + 999 + \ldots] = \frac{5}{9}[(10 - 1) + (100 - 1) + \ldots] \)
2. \( = \frac{5}{9}\left[\frac{10(10^n - 1)}{9} - n\right] \), using the GP sum 10 + 100 + … . A is true.
3. R is a true statement about APs but is not used in A.

Teacher's Note:
1. Multiply and divide by 9 to change 5, 55, 555, … into 9, 99, 999, …
2. Each 99…9 = 10ᵏ − 1, which splits the series into a GP and a constant series.