CBSE Class 10 Mathematics Arithmetic Progressions MCQs Set 24

Answer: (b) 17

Explanation:
1. \( a_{2} = S_{2} - S_{1} \).
2. \( S_{2} = 20 + 4 = 24 \) and \( S_{1} = 5 + 2 = 7 \).
3. \( a_{2} = 24 - 7 = 17 \).

Teacher's Note:
1. The nth term is \( S_{n} \) - Sₙ₋₁.
2. Check that the series is an AP with \( d = 10 \).

Answer: (b) 1

Explanation:
1. The terms 1, 4, 7, ..., 28 form an AP with 10 terms (\( 28 = 1 + 3(n - 1) \) gives \( n = 10 \)).
2. Their sum is \( 10(1 + 28)/2 = 145 \).
3. So \( 10x + 145 = 155 \) and \( x = 1 \).

Teacher's Note:
1. Count the terms first.
2. Separate the \( x's \) from the AP of constants.

Answer: (b) 2

Explanation:
1. \( a_{1} = S_{1} = 1 \) and \( S_{2} = 4 \), so \( a_{2} = 3 \).
2. \( d = a_{2} - a_{1} = 2 \).

Teacher's Note:
1. Use \( a_{1} = S_{1} \) and \( a_{2} = S_{2} - S_{1} \).
2. The AP is 1, 3, 5, ...

Answer: (c) 1

Explanation:
1. The arithmetic mean is \( (a + b)/2 \).
2. Put \( n = 1 \): \( (a + b)/(1 + 1) = (a + b)/2 \), which matches.
3. For other \( n \) the equality does not hold for all a, \( b \), so \( n = 1 \).

Teacher's Note:
1. Test the options in the given expression.
2. \( n = 1 \) gives denominators \( a^{0} + b^{0} = 2 \).

Answer: (d) \( \frac{n - 1}{\sqrt{a_1} + \sqrt{a_n}} \)

Explanation:
1. Rationalise: 1/(√aₖ + √aₖ₊₁) = (√aₖ₊₁ - √aₖ)/d, where \( d \) is the common difference.
2. The sum telescopes to \( (\sqrt{a}_{n} - \sqrt{a}_{1})/d \).
3. Since \( a_{n} - a_{1} = (n - 1)d \), this equals \( (n - 1)/(\sqrt{a}_{n} + \sqrt{a}_{1}) \).

Teacher's Note:
1. Multiply numerator and denominator by the conjugate.
2. aₖ₊₁ \( - a_{k} = d \) for every \( k \).

Answer: (a) \( -5\sqrt{2} \)

Explanation:
1. \( a = 2\sqrt{2} \) and \( d = \sqrt{2} - 2\sqrt{2} = -\sqrt{2} \).
2. \( a_{8} = a + 7d = 2\sqrt{2} - 7\sqrt{2} = -5\sqrt{2} \).

Teacher's Note:
1. Find \( d \) from the first two terms.
2. Use \( a_{n} = a + (n - 1)d \).

Answer: (c) 1 : 2

Explanation:
1. The ratio of the mth terms equals the ratio of the sums of \( (2m - 1) \) terms.
2. For \( m = 24, 2m - 1 = 47 \): \( (3 \times 47 - 13)/(5 \times 47 + 21) = 128/256 \).
3. The ratio is 1 : 2.

Teacher's Note:
1. Replace \( n \) by \( 2m - 1 \) in the ratio of sums.
2. \( S_{n}/S'_{n} = (2a + (n - 1)d)/(2a' + (n - 1)d') \).

Answer: (d) 17696

Explanation:
1. The numbers are 56, 63, ..., 497.
2. \( 497 = 56 + 7(n - 1) \) gives \( n = 64 \).
3. Sum = \( 64 \times (56 + 497)/2 = 32 \times 553 = 17696 \).

Teacher's Note:
1. Find the first and last multiples of 7 in the range.
2. Use \( S = n(a + l)/2 \).

Answer: (a) 2475

Explanation:
1. The numbers are 11, 13, ..., 99.
2. \( 99 = 11 + 2(n - 1) \) gives \( n = 45 \).
3. Sum = \( 45 \times (11 + 99)/2 = 45 \times 55 = 2475 \).

Teacher's Note:
1. List the first and last term.
2. Count the terms using the nth-term formula.

Answer: (c) 0

Explanation:
1. Let the AP have first term \( A \) and common difference \( D \): \( a = A + (p - 1)D \), and so on.
2. Σ \( A(q - r) = A\cdot 0 = 0 \).
3. \( D[(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)] = D[0 - 0] = 0 \) as each bracket sum cancels.

Teacher's Note:
1. Write a, \( b, c \) using \( A \) and \( D \).
2. Terms with \( A \) and with \( D \) both vanish separately.

Answer: (b) \( a^2, b^2, c^2 \) are in AP

Explanation:
1. The condition is \( 2/(c + a) = 1/(b + c) + 1/(a + b) \).
2. \( 2(b + c)(a + b) = (c + a)(a + 2b + c) \), which simplifies to 2(ab \( + b^{2} \) + ac + bc) = ac + 2bc \( + c^{2} + a^{2} \) + 2ab + ac.
3. This gives \( 2b^{2} = a^{2} + c^{2} \), so \( a^{2}, b^{2}, c^{2} \) are in AP.

Teacher's Note:
1. Use 2 × middle = sum of the others.
2. Expand and cancel carefully.

Answer: (a) 12

Explanation:
1. Let there be \( 2m \) means. Their sum = \( 2m \) × (the mean of the two numbers) = \( 2m \times 13/12 = 13m/6 \).
2. \( 13m/6 = 2m + 1 \), so \( 13m = 12m + 6 \) and \( m = 6 \).
3. The number of means is \( 2m = 12 \).

Teacher's Note:
1. The sum of \( n \) AMs between a and \( b \) is \( n(a + b)/2 \).
2. The sum here is 13/6, so \( (a + b)/2 = 13/12 \).

Answer: (c) \( -\frac{47}{16} \)

Explanation:
1. \( n = 2S/(a + l) = (407/8)/(37/8) = 11 \).
2. \( d = (l - a)/(n - 1) = (-99/8 - 17)/10 = (-235/8)/10 = -47/16 \).

Teacher's Note:
1. Find \( n \) from \( S = n(a + l)/2 \).
2. Then use \( l = a + (n - 1)d \).

Answer: (b) \( 2^{2/3} \)

Explanation:
1. Let the numbers be \( b - d, b, b + d \): \( b(b^{2} - d^{2}) = 4 \).
2. So \( b^{3} = 4 \) + bd² ≥ 4, i.e. \( b \ge 4^{1/3} = 2^{2/3} \).
3. The minimum occurs at \( d = 0 \).

Teacher's Note:
1. Write the numbers as \( b - d, b, b + d \).
2. bd² cannot be negative.

Answer: (a) 1020

Explanation:
1. \( a = -6 \) and \( d = 6 \).
2. \( S_{20} = 20/2 \times [2(-6) + 19 \times 6] = 10 \times 102 = 1020 \).

Teacher's Note:
1. Use \( S = n/2 [2a + (n - 1)d] \).
2. Check \( a_{20} = 108 \) and \( S = 20(-6 + 108)/2 \).

Answer: (b) 20700

Explanation:
1. \( a_{1} = 9 \) and \( a_{100} = 405 \).
2. \( S_{100} = 100 \times (9 + 405)/2 = 50 \times 414 = 20700 \).

Teacher's Note:
1. Find the first and last term.
2. Use \( S = n(a + l)/2 \).

Answer: (a) 7500

Explanation:
1. The numbers are 101, 103, ..., 199, which are 50 in number.
2. Sum = \( 50 \times (101 + 199)/2 = 50 \times 150 = 7500 \).

Teacher's Note:
1. First term 101, last term 199.
2. Count: \( (199 - 101)/2 + 1 = 50 \).

Answer: (d) 3050

Explanation:
1. Multiples of 2: \( 2 + 4 + ... + 100 = 2550 \). Multiples of 5: \( 5 + 10 + ... + 100 = 1050 \).
2. Multiples of 10 are counted twice: \( 10 + 20 + ... + 100 = 550 \).
3. Sum = \( 2550 + 1050 - 550 = 3050 \).

Teacher's Note:
1. Use inclusion and exclusion.
2. Each sum is an AP.

Answer: (c) 11

Explanation:
1. \( S_{2n} = 2n/2 \times [4 + (2n - 1)3] = n(6n + 1) \).
2. \( S_{n} = n/2 \times [114 + (n - 1)2] = n(n + 56) \).
3. \( 6n + 1 = n + 56 \), so \( n = 11 \).

Teacher's Note:
1. Write both sums with the AP formula.
2. Cancel the common factor \( n \).

Answer: (c) 1

Explanation:
1. T_m - T_n = (m - n)d = 1/n - 1/m = (m - n)/(mn), so \( d \) = 1/(mn).
2. T_m = a + (m - 1)/(mn) = \( 1/n \) gives \( a \) = 1/(mn).
3. \( T_{mn} = a \) + (mn - 1)d = 1/(mn) + (mn - 1)/(mn) = 1.

Teacher's Note:
1. Subtract the two given terms to find \( d \).
2. Then find a.

Answer: (d) 336 m

Explanation:
1. The distances form an AP with \( a = 16 \) and \( d = 32 \).
2. \( a_{11} = 16 + 10 \times 32 = 336 \).

Teacher's Note:
1. Identify the AP.
2. Use \( a_{n} = a + (n - 1)d \).

Answer: (c) 0

Explanation:
1. Write the terms as \( c - 2k, c - k, c, c + k, c + 2k \).
2. \( a - 4b + 6c - 4d + e = (c - 2k) - 4(c - k) + 6c - 4(c + k) + (c + 2k) \).
3. The \( c \) terms: \( 1 - 4 + 6 - 4 + 1 = 0 \); the \( k \) terms: \( -2 + 4 - 4 + 2 = 0 \). So the value is 0.

Teacher's Note:
1. Centre the AP at \( c \).
2. The coefficients 1, -4, 6, -4, 1 come from (1 - 1)⁴.

Answer: (a) \( \frac{n}{2n - 1}\left(a_1^2 - a_{2n}^2\right) \)

Explanation:
1. Each pair: a₂ₖ₋₁² \( - a_{2k}^{2} \) = -d(a₂ₖ₋₁ \( + a_{2k} \)).
2. Sum = \( -d \) × (sum of all \( 2n \) terms) = \( -d \times n(a_{1} + a_{2n}) \).
3. Using \( d = (a_{2n} - a_{1})/(2n - 1) \): sum = \( n(a_{1} - a_{2n})(a_{1} + a_{2n})/(2n - 1) = n(a_{1}^{2} - a_{2n}^{2})/(2n - 1) \).

Teacher's Note:
1. Factorise each pair as a difference of squares.
2. The common difference \( d \) is the same in every pair.

Answer: (b) 42

Explanation:
1. The sum of \( n \) AMs between \( p \) and \( q \) is \( n(p + q)/2 \).
2. Sum = \( 6 \times (2 + 12)/2 = 42 \).

Teacher's Note:
1. The mean of the AMs equals the mean of the two end numbers.
2. \( 6 \times 7 = 42 \).

Answer: (a) 0

Explanation:
1. \( S_{i}/n_{i} = a + (n_{i} - 1)d/2 = A \) + Bnᵢ with constants \( A \) and \( B \).
2. The sum is \( A[(n_{2} - n_{3}) + (n_{3} - n_{1}) + (n_{1} - n_{2})] + B[n_{1}(n_{2} - n_{3}) + n_{2}(n_{3} - n_{1}) + n_{3}(n_{1} - n_{2})] \).
3. Both brackets are 0, so the value is 0.

Teacher's Note:
1. \( S_{i}/n_{i} \) is a linear function of \( n_{i} \).
2. Expand the second bracket to see it cancels.

Answer: (b) 2(d - c)

Explanation:
1. Let the common difference be δ. Then \( e - c \) = 2δ.
2. \( d - c \) = δ, so \( 2(d - c) \) = 2δ.
3. The other options give 4δ, 4δ and δ.

Teacher's Note:
1. Express every difference using δ.
2. \( e - c \) spans two steps.

Answer: (c) ±3

Explanation:
1. Let the roots be \( a - d, a, a + d \): sum \( 3a = 12 \), so \( a = 4 \).
2. Product \( (4 - d)(4)(4 + d) = 28 \) gives \( 16 - d^{2} = 7 \), so \( d^{2} = 9 \).
3. \( d = \pm 3 \) (the roots are 1, 4, 7).

Teacher's Note:
1. Use the sum of roots for the middle term.
2. Then use the product.