JEE Mathematics Sequence and Series MCQs Set 12

Answer: (b) \( 3-\frac pn \)

Explanation:
1. The first term is \( 3-\frac1n \) and the common difference is \( -\frac1n \).
2. \( T_p=3-\frac1n-\frac{p-1}n=3-\frac pn \).

Teacher's Note:
1. Use T_p = a + (p - 1)d.

Answer: (b) \( 2:1 \)

Explanation:
1. \( a+8d=0 \), so \( a=-8d \).
2. \( T_{29}=a+28d=20d \) and \( T_{19}=a+18d=10d \).
3. The ratio is \( 20d:10d=2:1 \).

Teacher's Note:
1. Express every term in d.

Answer: (b) \( p+q-r \)

Explanation:
1. Subtracting \( a+(p-1)d=q \) and \( a+(q-1)d=p \): \( (p-q)d=q-p \), so \( d=-1 \).
2. Then \( a=q+p-1 \), and \( T_r=a+(r-1)d=p+q-1-(r-1)=p+q-r \).

Teacher's Note:
1. Subtract the two conditions to find d.

Answer: (b) \( 79 \)

Explanation:
1. Subtracting, \( 10d=40 \), so \( d=4 \).
2. \( T_{20}=T_{19}+d=75+4=79 \).

Teacher's Note:
1. From T19 to T20 add one common difference.

Answer: (b) \( \sqrt{y^2-xz} \)

Explanation:
1. \( y=\frac{x+z}2 \), so \( y^2-xz=\frac{(x+z)^2}4-xz=\frac{(x-z)^2}4 \).
2. The common difference is \( d=\frac{z-x}2 \), so \( d^2=y^2-xz \) and \( |d|=\sqrt{y^2-xz} \).

Teacher's Note:
1. Use y = (x + z) / 2.

Answer: (b) \( \sqrt{300} \)

Explanation:
1. The terms are \( \sqrt3,2\sqrt3,3\sqrt3,\dots \), an A.P. with \( a=d=\sqrt3 \).
2. \( T_{10}=10\sqrt3=\sqrt{300} \).

Teacher's Note:
1. Pull out the common factor root 3.

Answer: (a) 5th

Explanation:
1. The real parts form an A.P. \( -8,-6,-4,\dots \) with \( T_n=-8+2(n-1) \).
2. The real part is zero when \( n=5 \), where the imaginary part is \( 18-3\cdot4=6\ne0 \).
3. So the 5th term is purely imaginary.

Teacher's Note:
1. Set the real part equal to zero.

Answer: (c) \( -1 \)

Explanation:
1. The first condition gives \( a+3d=10 \). The second gives \( 2a+9d=17 \).
2. Substituting \( a=10-3d \): \( 20+3d=17 \), so \( d=-1 \).

Teacher's Note:
1. Write each term with a and d.

Answer: (a) A.P.

Explanation:
1. \( n\theta=k\pi+m\theta \), so \( \theta=\frac{k\pi}{n-m} \) for integer k.
2. These values differ by the constant \( \frac\pi{n-m} \), so they form an A.P.

Teacher's Note:
1. Use the general solution of tan A = tan B.

Answer: (a) \( \frac49 \)

Explanation:
1. The 12th terms are \( 3(12)+8=44 \) and \( 7(12)+15=99 \).
2. The ratio is \( \frac{44}{99}=\frac49 \).

Teacher's Note:
1. Substitute n = 12 in both formulas.

Answer: (c) \( \frac23 \)

Explanation:
1. \( a_1=2-5d,\ a_4=2-2d,\ a_5=2-d \). The product is \( f(d)=(2-5d)(2-2d)(2-d) \).
2. \( f'(d)=0 \) gives \( 15d^2-34d+16=0 \), so \( d=\frac23 \) or \( \frac85 \).
3. \( f''(d)=4(17-15d) \) is positive at \( d=\frac23 \), so that is the minimum.

Teacher's Note:
1. Write the product in d and differentiate.

Answer: (a) \( 0 \)

Explanation:
1. \( p\{a+(p-1)d\}=q\{a+(q-1)d\} \) gives \( (p-q)a+\{p(p-1)-q(q-1)\}d=0 \).
2. Divide by \( p-q \): \( a+(p+q-1)d=0 \), which is \( T_{p+q}=0 \).

Teacher's Note:
1. Factor p(p - 1) - q(q - 1) as (p - q)(p + q - 1).

Answer: (d) \( x-\frac53y \)

Explanation:
1. \( 3d=(x+y)-(x-y)=2y \), so \( d=\frac{2y}3 \).
2. \( a=(x-y)-\frac{2y}3=x-\frac{5y}3 \).

Teacher's Note:
1. Subtract T2 from T5 to get 3d.

Answer: (c) \( 20 \)

Explanation:
1. The first sequence is \( 17+4k \) and the second \( 16+5m \). A common term satisfies \( x\equiv1\pmod4 \) and \( x\equiv1\pmod5 \), that is, \( x\equiv1\pmod{20} \).
2. The common terms are 21, 41, ..., 401 (the next, 421, exceeds 417).
3. Their number is \( \frac{401-21}{20}+1=20 \).

Teacher's Note:
1. The common terms form an A.P. with difference the L.C.M. of 4 and 5.

Answer: (c) \( n(A\cap B)=66 \) and \( n(A\cup B)=384 \)

Explanation:
1. A has multiples of 2 up to 400 and B has multiples of 3 up to 750.
2. The common elements are multiples of 6 up to 400: \( \lfloor400/6\rfloor=66 \).
3. \( n(A\cup B)=200+250-66=384 \).

Teacher's Note:
1. Common elements are multiples of the L.C.M. 6.

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

Explanation:
1. The numbers 1, 2, ..., n form an A.P. with \( a=1 \) and \( l=n \).
2. \( S_n=\frac n2(1+n) \).

Teacher's Note:
1. Use S = (n / 2)(first + last).

Answer: (d) \( -\frac32 \)

Explanation:
1. The common difference is \( \frac13-\frac12=-\frac16 \) and \( \frac16-\frac13=-\frac16 \), so it is an A.P.
2. \( S_9=\frac92\left(2\cdot\frac12+8\cdot\left(-\frac16\right)\right)=\frac92\cdot\left(-\frac13\right)=-\frac32 \).

Teacher's Note:
1. Check the common difference before using the formula.

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

Explanation:
1. \( a=1,\ d=2 \), so \( S_n=\frac n2(2+2(n-1))=n^2 \).

Teacher's Note:
1. The sum of the first n odd numbers is n squared.

Answer: (b) \( 200 \)

Explanation:
1. \( a=2,\ d=3 \): \( S_n=\frac n2(4+3(n-1))=\frac{n(3n+1)}2 \).
2. \( 3n^2+n=120200 \) gives \( n=200 \) (check: \( 120000+200 \)).

Teacher's Note:
1. Form the quadratic in n.

Answer: (a) \( 55350 \)

Explanation:
1. The numbers are 108, 117, ..., 999. Number of terms \( =\frac{999-108}9+1=100 \).
2. \( S=\frac{100}2(108+999)=55350 \).

Teacher's Note:
1. Find the first and last multiples of 9 in the range.

Answer: (b) 3, 5, 7

Explanation:
1. Take \( a-d,a,a+d \). The sum \( 3a=15 \) gives \( a=5 \).
2. The squares: \( 3a^2+2d^2=83 \), so \( 75+2d^2=83 \) and \( d=\pm2 \).
3. The numbers are 3, 5, 7.

Teacher's Note:
1. Three terms in A.P. are best taken as a - d, a, a + d.

Answer: (a) 21, 17, 13

Explanation:
1. The middle term is \( 51/3=17 \). With difference d, \( (17-d)(17+d)=273 \), so \( d^2=16 \).
2. The terms are 13, 17, 21 (or 21, 17, 13).

Teacher's Note:
1. The middle term is the average.

Answer: (d) \( Q \)

Explanation:
1. Comparing with \( S_n=na+\frac{n(n-1)}2d \), the common difference is \( Q \).

Teacher's Note:
1. Compare with the standard formula.

Answer: (d) \( 156375 \)

Explanation:
1. The numbers are 252, 255, ..., 999, which are \( \frac{999-252}3+1=250 \) terms.
2. \( S=\frac{250}2(252+999)=156375 \).

Teacher's Note:
1. First multiple of 3 above 250 is 252.

Answer: (d) \( 1 \)

Explanation:
1. Take \( a-3d,a-d,a+d,a+3d \). The extremes give \( 2a=8 \), so \( a=4 \).
2. \( (4-d)(4+d)=15 \) gives \( d^2=1 \).
3. The terms are 1, 3, 5, 7 (or reversed), so the least is 1.

Teacher's Note:
1. Four terms: a - 3d, a - d, a + d, a + 3d.