CBSE Class 10 Mathematics Arithmetic Progression MCQs Set 19

Answer: (c) 5

Explanation:
1. Three terms are in A.P. when the middle term is the mean of the other two: \( 2(3y+5)=(3y-1)+(5y+1) \).
2. \( 6y+10=8y \), so \( y=5 \).

Teacher's Note:
1. For three numbers in A.P., 2 × middle = sum of the other two.
2. Check: the terms are 14, 20, 26 with d = 6.

Answer: (c) 400

Explanation:
1. The odd numbers 1, 3, 5, ... form an A.P. with \( a=1 \), \( d=2 \).
2. \( S_{20}=\frac{20}{2}[2(1)+19(2)]=10\times 40=400 \).

Teacher's Note:
1. The sum of the first n odd numbers is n².
2. So the sum of the first 20 is 20² = 400.

Answer: (c) 50

Explanation:
1. \( d=5 \).
2. \( t_{30}-t_{20}=10d=50 \).

Teacher's Note:
1. The difference of two terms 10 places apart is 10d.
2. Here d = 7 − 2 = 5.

Answer: (c) 15

Explanation:
1. \( t_{20}-t_{15}=(20-15)d=5\times3=15 \).

Teacher's Note:
1. Two terms k places apart differ by kd.
2. Here k = 5.

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

Explanation:
1. First term \( =a \), common difference \( =3a-a=2a \).
2. \( t_n=a+(n-1)2a=(2n-1)a \).

Teacher's Note:
1. Find d first, then use a + (n − 1)d.
2. Check with n = 2: (2·2 − 1)a = 3a.

Answer: (d) 28

Explanation:
1. \( a_7=a+6d \Rightarrow 4=a-24 \), so \( a=28 \).

Teacher's Note:
1. Substitute into a_n = a + (n − 1)d.
2. Solve the one-step equation for a.

Answer: (c) \( 4n-9 \)

Explanation:
1. \( a_n=S_n-S_{n-1}=(2n^2-7n)-[2(n-1)^2-7(n-1)] \).
2. \( =2n^2-7n-2n^2+4n-2+7n-7=4n-9 \).

Teacher's Note:
1. The nth term equals S_n − S_(n−1).
2. Check: a_1 = S_1 = −5 and 4(1) − 9 = −5.

Answer: (c) 21

Explanation:
1. Let the terms be \( a-d,\ a,\ a+d \). Sum \( =3a=51 \), so \( a=17 \).
2. \( (17-d)(17+d)=273\Rightarrow 289-d^2=273\Rightarrow d=\pm4 \); increasing means \( d=4 \).
3. Third term \( =17+4=21 \).

Teacher's Note:
1. Use a − d, a, a + d for three terms.
2. Increasing A.P. means d is positive.

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

Explanation:
1. Let them be \( a-3d,a-d,a+d,a+3d \). Sum \( =4a=50 \), \( a=\frac{25}{2} \).
2. \( a+3d=4(a-3d)\Rightarrow 15d=3a\Rightarrow a=5d \), so \( d=\frac52 \).
3. Numbers: 5, 10, 15, 20.

Teacher's Note:
1. Use symmetric terms a ± d, a ± 3d for four numbers.
2. Verify: 5 + 10 + 15 + 20 = 50 and 20 = 4 × 5.

Answer: (d) none of these

Explanation:
1. \( a+8d=449 \) and \( a+448d=9 \). Subtracting, \( 440d=-440 \), so \( d=-1 \) and \( a=457 \).
2. \( a+(n-1)d=0\Rightarrow 457-(n-1)=0\Rightarrow n=458 \).

Teacher's Note:
1. Subtract the two equations to find d.
2. 458th is not among the first three options.

Answer: (b) 3 : 1

Explanation:
1. \( \frac{a+17d}{a+10d}=\frac32\Rightarrow 2a+34d=3a+30d\Rightarrow a=4d \).
2. \( \frac{a_{21}}{a_5}=\frac{a+20d}{a+4d}=\frac{24d}{8d}=3:1 \).

Teacher's Note:
1. Cross-multiply to relate a and d.
2. Then substitute a = 4d.

Answer: (b) an A.P. with \( d=4 \)

Explanation:
1. \( -6-(-10)=4,\ -2-(-6)=4,\ 2-(-2)=4 \). Constant difference 4.

Teacher's Note:
1. Check every consecutive difference.
2. Subtracting a negative number adds.

Answer: (c) 88

Explanation:
1. \( d=3 \), last term \( l=130 \).
2. 15th from the end \( =l-(15-1)d=130-42=88 \).

Teacher's Note:
1. Count backwards using the last term and −d.
2. Or reverse the A.P. with first term 130 and d = −3.

Answer: (a) 8

Explanation:
1. \( a_{18}-a_{14}=4d=32\Rightarrow d=8 \).

Teacher's Note:
1. Terms 4 places apart differ by 4d.
2. Divide 32 by 4.

Answer: (b) 3.5

Explanation:
1. \( a_{101}=3.5+100\times0=3.5 \).

Teacher's Note:
1. With d = 0 every term equals the first term.

Answer: (b) 10th

Explanation:
1. \( 21+(n-1)21=210\Rightarrow 21n=210\Rightarrow n=10 \).

Teacher's Note:
1. The terms are 21 × 1, 21 × 2, ...
2. So 210 = 21 × 10.

Answer: (c) 24 terms

Explanation:
1. \( d=-2 \). \( 5+(n-1)(-2)=-41\Rightarrow n-1=23 \), so \( n=24 \).

Teacher's Note:
1. Use the last term as the nth term.
2. Do not forget the +1 at the end.

Answer: (a) 0

Explanation:
1. \( S_6=\frac62[2(-5)+5\times2]=3\times0=0 \).

Teacher's Note:
1. Substitute a = −5, d = 2, n = 6 in the sum formula.

Answer: (b) 30

Explanation:
1. \( t_5=40 \), \( t_2=10 \). Difference \( =30 \).

Teacher's Note:
1. Substitute n = 5 and n = 2 separately.

Answer: (b) \( \frac{mn-m+1}{m} \)

Explanation:
1. \( a=\frac1m \), \( d=\frac{m+1}{m}-\frac1m=1 \).
2. \( t_n=\frac1m+(n-1)=\frac{1+mn-m}{m} \).

Teacher's Note:
1. Find d first (it is 1).
2. Combine over the common denominator m.

Answer: (c) 40

Explanation:
1. \( 25+(k-1)25=1000\Rightarrow 25k=1000\Rightarrow k=40 \).

Teacher's Note:
1. The kth term is 25k.

Answer: (b) 25

Explanation:
1. \( a=320,\ d=40 \). \( \frac n2[640+40(n-1)]=20000 \).
2. \( n(20n+300)=20000\Rightarrow n^2+15n-1000=0\Rightarrow (n-25)(n+40)=0 \), so \( n=25 \).

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

Answer: (b) \( nd \)

Explanation:
1. \( t_{2n}-t_n=[a+(2n-1)d]-[a+(n-1)d]=nd \).

Teacher's Note:
1. The two terms are n places apart.

Answer: (c) 3 times the previous common difference

Explanation:
1. Let the A.P. be \( a,\ a+d,\ a+2d,\ldots \). Multiplying by 3 gives \( 3a,\ 3a+3d,\ldots \).
2. The new common difference is \( 3d \).

Teacher's Note:
1. Multiplying every term by k multiplies d by k.
2. Adding a constant to every term leaves d unchanged.

Answer: (a) 3, 17, 31

Explanation:
1. \( 10=x+d \) and \( 24=x+3d \), so \( 2d=14 \), \( d=7 \).
2. \( x=3,\ y=17,\ z=31 \).

Teacher's Note:
1. Terms 2 and 4 are two steps apart, so the difference is 2d.
2. Then fill in the other terms by adding or subtracting d.

Answer: (a) 14.7

Explanation:
1. \( a=0.40,\ d=0.03,\ l=1 \). \( n=\frac{1-0.40}{0.03}+1=21 \).
2. \( S_{21}=\frac{21}{2}(0.40+1)=14.7 \).

Teacher's Note:
1. Find n from the last term first.
2. Then use S = n/2 (a + l).

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

Explanation:
1. For \( S_n=2n^2+3n \): \( t_n=S_n-S_{n-1}=4n+1 \), valid for every \( n \) (also \( t_1=S_1=5 \)). So the series is an A.P. with \( d=4 \): Assertion is true.
2. The sum of an A.P. is \( \frac d2n^2+\left(a-\frac d2\right)n \), so the coefficient of \( n^2 \) is \( \frac d2 \) and \( d=2\times \) that coefficient: Reason is true and explains the Assertion.

Teacher's Note:
1. S_n of an A.P. has no constant term.
2. The coefficient of n² is half the common difference.

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

Explanation:
1. Reason is true: \( t_n=S_n-S_{n-1} \) for every sequence.
2. For the Assertion, the ratio of the \( n^{th} \) terms equals \( \frac{S_{2n-1}}{S'_{2n-1}}=\frac{3(2n-1)+8}{7(2n-1)+15}=\frac{6n+5}{14n+8} \), which depends on \( n \) and is not a fixed 3 : 16. (For \( n=12 \) it is 7 : 16.) Assertion is false.

Teacher's Note:
1. Use t_n = S_n − S_(n−1) for both A.P.s.
2. The ratio of nth terms is 6n + 5 : 14n + 8, not a constant.