CBSE Class 10 Mathematics Arithmetic Progression MCQs Set 18

Answer: (c) 20

Explanation:
1. \( S_n = \frac{n}{2}(a + a_n) \), so \( 860 = \frac{n}{2}(5 + 81) = 43n \).
2. Hence \( n = 20 \).

Teacher's Note:
1. Use the sum formula with the first and last terms.
2. Divide 860 by 43.

Answer: (d) \( k = 3 \)

Explanation:
1. Middle term condition: \( 2(4k - 6) = (k + 2) + (3k - 2) \).
2. \( 8k - 12 = 4k \), so \( k = 3 \).

Teacher's Note:
1. For three terms in A.P., twice the middle term equals the sum of the other two.
2. Check: the terms are 5, 6, 7.

Answer: (b) −142

Explanation:
1. \( a_{100} = 5 + 99d = -292 \), so \( d = -3 \).
2. \( a_{50} = 5 + 49(-3) = -142 \).

Teacher's Note:
1. Find d first.
2. Then use a + (n − 1)d.

Answer: (a) \( 4abc \)

Explanation:
1. Since \( a, b, c \) are in A.P., \( 2b = a + c \).
2. \( a + 2b - c = 2a \), \( 2b + c - a = 2c \), \( c + a - b = 2b - b = b \).
3. Product = \( 2a \cdot 2c \cdot b = 4abc \).

Teacher's Note:
1. Replace 2b by a + c in the first two brackets.
2. In the third bracket use c + a = 2b.

Answer: (c) \( \frac{n(n + 1)}{\sqrt{2}} \)

Explanation:
1. The terms are \( \sqrt{2},\ 2\sqrt{2},\ 3\sqrt{2},\ 4\sqrt{2} \), an A.P. with \( a = d = \sqrt{2} \).
2. \( S_n = \frac{n}{2}[2\sqrt{2} + (n - 1)\sqrt{2}] = \frac{n(n + 1)\sqrt{2}}{2} = \frac{n(n + 1)}{\sqrt{2}} \).

Teacher's Note:
1. Simplify each surd first.
2. Then use the A.P. sum formula.

Answer: (c) 0

Explanation:
1. \( 8(a + 7d) = 12(a + 11d) \), so \( -4a - 76d = 0 \), i.e. \( a + 19d = 0 \).
2. \( a_{20} = a + 19d = 0 \).

Teacher's Note:
1. Expand and collect terms; the result is exactly a + 19d.
2. That expression is the 20th term.

Answer: (a) −43

Explanation:
1. Subtract: \( 9d = 81 \), so \( d = 9 \).
2. \( a = 20 - 9 \times 9 = -61 \).
3. \( a_3 = -61 + 2 \times 9 = -43 \).

Teacher's Note:
1. Subtract the two equations to eliminate a.
2. Then find a_3.

Answer: (a) \( 7^2 \)

Explanation:
1. \( a_{12} = S_{12} - S_{11} \).
2. \( S_{12} = 288 + 36 = 324 \), \( S_{11} = 242 + 33 = 275 \).
3. \( a_{12} = 49 = 7^2 \).

Teacher's Note:
1. The nth term is S_n − S_(n−1).
2. √625 = 25, so option (c) is not 49.

Answer: (c) −1

Explanation:
1. \( d = \frac{1 - p}{p} - \frac{1}{p} = \frac{-p}{p} = -1 \).

Teacher's Note:
1. Subtract the first term from the second.
2. Cancel p.

Answer: (b) \( (2n - 1)a \)

Explanation:
1. \( d = 3a - a = 2a \).
2. \( a_n = a + (n - 1)2a = (2n - 1)a \).

Teacher's Note:
1. Find d first.
2. Use a_n = a + (n − 1)d.

Answer: (b) 200

Explanation:
1. \( a = 2,\ d = 3 \). \( \frac{n}{2}[4 + 3(n - 1)] = 60100 \), so \( 3n^2 + n - 120200 = 0 \).
2. Putting \( n = 200 \): \( 3(40000) + 200 = 120200 \), which is true.
3. So \( n = 200 \).

Teacher's Note:
1. Form the quadratic in n; the negative root is rejected.
2. Verify by substitution.

Answer: (d) 2

Explanation:
1. Take the terms \( a - 3x,\ a - x,\ a + x,\ a + 3x \). Sum \( 4a = 20 \), so \( a = 5 \).
2. \( \frac{a^2 - 9x^2}{a^2 - x^2} = \frac{2}{3} \) gives \( 3(25 - 9x^2) = 2(25 - x^2) \), so \( x^2 = 1 \).
3. The common difference is \( 2x = 2 \) (taking x = 1).

Teacher's Note:
1. Choose terms symmetric about the middle for four terms in A.P.
2. The common difference is 2x, not x.

Answer: (d) 98

Explanation:
1. \( 185 = 8 + 59d \), so \( d = 3 \).
2. \( a_{31} = 8 + 30 \times 3 = 98 \).

Teacher's Note:
1. Use the last term to get d.
2. Then the 31st term.

Answer: (d) −2, −6, −10, −14

Explanation:
1. Six terms: 2, ..., −18, so \( -18 = 2 + 5d \) and \( d = -4 \).
2. The means are \( -2,\ -6,\ -10,\ -14 \).

Teacher's Note:
1. Inserting four means makes 6 terms in all.
2. The last term is the 6th.

Answer: (c) \( \frac{(a + c)(b + c - 2a)}{2(b - a)} \)

Explanation:
1. Common difference \( d = b - a \). From \( c = a + (n - 1)(b - a) \), \( n = \frac{b + c - 2a}{b - a} \).
2. \( S = \frac{n}{2}(a + c) = \frac{(a + c)(b + c - 2a)}{2(b - a)} \).

Teacher's Note:
1. Find n in terms of a, b, c first.
2. Then substitute into S = (n/2)(a + l).

Answer: (b) 330

Explanation:
1. The middle term of 11 terms is the 6th term, \( a_6 = 30 \).
2. \( S_{11} = \frac{11}{2}(a_1 + a_{11}) = 11 \times a_6 = 330 \).

Teacher's Note:
1. The first and last terms add up to twice the middle term.
2. So S = 11 × middle term.

Answer: (b) \( 2S \)

Explanation:
1. \( S = \frac{n}{2}(a + \ell) \), so \( n = \frac{2S}{a + \ell} \).
2. \( d = \frac{\ell - a}{n - 1} = \frac{\ell - a}{\frac{2S}{a + \ell} - 1} = \frac{\ell^2 - a^2}{2S - (\ell + a)} \).
3. Comparing, \( k = 2S \).

Teacher's Note:
1. Express n through S, then d through n.
2. Compare with the given form.

Answer: (a) 5, 10, 15, 20

Explanation:
1. Let the numbers be \( a, a + d, a + 2d, a + 3d \). Greatest = 4 × least gives \( a + 3d = 4a \), so \( d = a \).
2. Sum: \( 4a + 6d = 10a = 50 \), so \( a = 5 \).
3. The numbers are 5, 10, 15, 20.

Teacher's Note:
1. Use the two conditions to find a and d.
2. Check: the sum is 50 and 20 = 4 × 5.

Answer: (c) 11

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

Teacher's Note:
1. Write both sums with the A.P. sum formula.
2. Cancel n from both sides.

Answer: (b) 30

Explanation:
1. \( \frac{n}{2}[10 + 2(n - 1)] = n(n + 4) = 1020 \), so \( n^2 + 4n - 1020 = 0 \).
2. \( (n - 30)(n + 34) = 0 \), so \( n = 30 \).

Teacher's Note:
1. Form and factorise the quadratic in n.
2. Reject the negative root.

Answer: (b) 4

Explanation:
1. \( a_{n+1} - a_n = [4(n + 1) + 1] - (4n + 1) = 4 \).

Teacher's Note:
1. The coefficient of n is the common difference.
2. Or subtract a₁ from a₂.

Answer: (b) \( 2(d - c) \)

Explanation:
1. Let the common difference be \( \delta \). Then \( e - c = 2\delta \) and \( d - c = \delta \).
2. So \( e - c = 2(d - c) \).

Teacher's Note:
1. e is two steps after c.
2. Compare each option with 2δ.

Answer: (c) 30

Explanation:
1. They are 12, 15, ..., 99 with \( a = 12,\ d = 3 \).
2. \( 99 = 12 + (n - 1)3 \), so \( n = 30 \).

Teacher's Note:
1. Find the first and last two-digit multiples of 3.
2. Then use the nth term formula.

Answer: (b) 245

Explanation:
1. \( a_1 = 2,\ a_{10} = 47 \).
2. \( S_{10} = \frac{10}{2}(2 + 47) = 245 \).

Teacher's Note:
1. Find the first and 10th terms.
2. Use S = (n/2)(first + last).

Answer: (d) −2550

Explanation:
1. Next five terms sum to \( S_{10} - S_5 \), so \( S_5 = \frac{1}{4}(S_{10} - S_5) \), i.e. \( 5S_5 = S_{10} \).
2. \( S_5 = 10 + 10d \) and \( S_{10} = 20 + 45d \), so \( 50 + 50d = 20 + 45d \) and \( d = -6 \).
3. \( S_{30} = 15[4 + 29(-6)] = 15(-170) = -2550 \).

Teacher's Note:
1. Convert the condition into 5S₅ = S₁₀.
2. Find d, then S₃₀.

Answer: (c) \( -(m + n) \)

Explanation:
1. From the two conditions, \( 2a + (m - 1)d = \frac{2n}{m} \) and \( 2a + (n - 1)d = \frac{2m}{n} \).
2. Subtracting: \( (m - n)d = \frac{2(n^2 - m^2)}{mn} \), so \( d = -\frac{2(m + n)}{mn} \).
3. \( 2a + (m + n - 1)d = \frac{2n}{m} + nd = \frac{2n}{m} - \frac{2(m + n)}{m} = -2 \).
4. \( S_{m + n} = \frac{m + n}{2}(-2) = -(m + n) \).

Teacher's Note:
1. Write each condition as an expression for 2a + (k − 1)d.
2. Note that (m + n − 1) − (m − 1) = n.

Answer: (a) 0.2, 0.4, 0.6, ....

Explanation:
1. Option (a) has a constant difference \( 0.4 - 0.2 = 0.6 - 0.4 = 0.2 \).
2. (b) and (c) have constant ratios 2 and 3, not constant differences, and (d) has differences 0.5, 1, 4.

Teacher's Note:
1. Check the difference between consecutive terms.
2. A constant ratio gives a G.P., not an A.P.

Answer: (c) \( 6n + 11 \)

Explanation:
1. Replace \( n \) by \( n + 1 \): \( t_{n+1} = 6(n + 1) + 5 = 6n + 11 \).

Teacher's Note:
1. Substitute n + 1 in the formula for t_n.
2. Compare the options after expanding.

Answer: (b) \( \frac{n}{2}[2a + (n - 1)d] \)

Explanation:
1. The sum of n terms of an A.P. with first term a and common difference d is \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
2. Option (a) is wrong (it should be \( \frac{n}{2}(a + l) \)), and (c), (d) are G.P. sums.

Teacher's Note:
1. Learn both forms of the sum formula.
2. Terms with r^n belong to G.P.

Answer: (a) 18

Explanation:
1. \( \frac{n}{2}[108 + (n - 1)(-3)] = 513 \), so \( n(111 - 3n) = 1026 \), i.e. \( n^2 - 37n + 342 = 0 \).
2. \( (n - 18)(n - 19) = 0 \), so \( n = 18 \) or \( 19 \).
3. The least value is 18.

Teacher's Note:
1. Both roots are valid because the 19th term is 0.
2. Pick the smaller value as asked.

Answer: (d) \( n^2 \)

Explanation:
1. \( a_1 = 1,\ a_n = 2n - 1 \).
2. \( S_n = \frac{n}{2}(1 + 2n - 1) = n^2 \).

Teacher's Note:
1. This is the sum of the first n odd numbers.
2. Use S = (n/2)(first + last).

Answer: (c) \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \)

Explanation:
1. Add 2 to each term: \( \frac{a + b + c}{a},\ \frac{a + b + c}{b},\ \frac{a + b + c}{c} \) are in A.P.
2. Dividing each by \( a + b + c \) keeps an A.P.: \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in A.P.

Teacher's Note:
1. Adding a constant to every term keeps an A.P.
2. Dividing every term by a constant also keeps an A.P.

Answer: (a) 2

Explanation:
1. \( a_{n+1} - a_n = (2n + 5) - (2n + 3) = 2 \).

Teacher's Note:
1. The coefficient of n is the common difference.
2. Subtract two consecutive terms to check.

Answer: (a) 6

Explanation:
1. \( 2(x + 10) = 2x + (3x + 2) \), so \( 2x + 20 = 5x + 2 \).
2. \( 3x = 18 \), so \( x = 6 \).

Teacher's Note:
1. Use the middle term condition.
2. Check: the terms are 12, 16, 20.

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

Explanation:
1. Dividing the terms \( a, b, c \) by \( abc \) gives \( \frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab} \). The Assertion is true.
2. Dividing every term of an A.P. by a non-zero constant keeps the common difference constant, so the Reason is true and explains the Assertion: option (a).

Teacher's Note:
1. Divide each of a, b, c by abc.
2. The Reason is exactly the step used.

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

Explanation:
1. Here \( a = -0.5,\ d = -0.5 \), so \( S_{10} = 5[-1 + 9(-0.5)] = 5(-5.5) = -27.5 \).
2. The sum is −27.5, not 27.5, so the Assertion is false; the Reason is true: option (d).

Teacher's Note:
1. All terms are negative, so the sum must be negative.
2. Use the sum formula to confirm.

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

Explanation:
1. Using the Reason, \( d = [7 - 4(n + 1)] - (7 - 4n) = -4 \), so the Assertion is true.
2. The Reason is true and gives the method: option (a).

Teacher's Note:
1. Subtract consecutive terms.
2. The Reason is the definition used.

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

Explanation:
1. The Assertion is true, because \( S_n - S_{n-1} \) is exactly the nth term.
2. For the Reason, \( a_{10} = 5 + 9(3) = 32 \), not 35, so the Reason is false: option (c).

Teacher's Note:
1. Check the Reason by direct calculation.
2. A true Assertion with a false Reason gives option (c).

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

Explanation:
1. By the Reason, the mean of 8 and 12 is \( \frac{8 + 12}{2} = 10 \).
2. Both statements are true and the Reason explains the Assertion: option (a).

Teacher's Note:
1. Apply the formula to 8 and 12.
2. The Reason is the formula used.