JEE Mathematics Sequence and Series MCQs Set 25

Answer: (a) 324

Explanation:
1. Let the first term be a and the ratio r: \( \frac a{1-r}=162 \) and \( a\frac{1-r^n}{1-r}=160 \), so \( 1-r^n=\frac{160}{162}\Rightarrow r^n=\frac1{81} \).
2. With 1/r an integer: r = 1/3 with n = 4 gives a = 162(2/3) = 108; r = −1/3 with n = 4 gives a = 162(4/3) = 216. Other integers (±9, ±81, ...) do not give a whole n.
3. Sum of possible first terms = 108 + 216 = 324.

Teacher's Note:
1. Use rⁿ = 1/81 and test r = 1/m.
2. a = 162(1 − r).

Answer: (d) \( a=3,\ r=\frac14 \)

Explanation:
1. Sum \( \frac a{1-r}=4 \) and second term ar = 3/4.
2. So \( a=4(1-r) \) and \( 4r(1-r)=\frac34\Rightarrow16r^2-16r+3=0 \), giving r = 1/4 or 3/4.
3. r = 1/4 gives a = 3; r = 3/4 gives a = 1, which is not among the options.

Teacher's Note:
1. Use a = 4(1 − r) in ar = 3/4.
2. Check which option satisfies both conditions.

Answer: (d) A = 3, B = 8

Explanation:
1. \( \frac1\alpha+\frac1\gamma=\frac{\alpha+\gamma}{\alpha\gamma}=4 \) and \( \frac1\beta+\frac1\delta=6 \).
2. Reciprocals \( t_1,t_2,t_3,t_4 \) form an A.P.: \( t_1+t_3=2t_2=4 \) so \( t_2=2 \); \( t_2+t_4=2t_3=6 \) so \( t_3=3 \). Then d = 1, \( t_1=1,\ t_4=4 \).
3. α = 1, γ = 1/3 give \( A=\frac1{\alpha\gamma}=3 \); β = 1/2, δ = 1/4 give \( B=\frac1{\beta\delta}=8 \).

Teacher's Note:
1. Take reciprocals so that the H.P. becomes an A.P.
2. 1/α + 1/γ = sum/product of roots.

Answer: (c) A.P.

Explanation:
1. Sum of roots \( -\frac ba \); sum of squares of reciprocals \( \frac{\alpha^2+\beta^2}{\alpha^2\beta^2}=\frac{b^2-2ac}{c^2} \).
2. So \( -bc^2=ab^2-2a^2c \), i.e. \( 2a^2c=ab^2+bc^2 \).
3. This says \( 2(ca^2)=ab^2+bc^2 \), so bc², ca², ab² are in A.P.

Teacher's Note:
1. Express both sides using a, b, c.
2. Compare with the A.P. condition 2y = x + z.

Answer: (b) \( x=2\sqrt2,\ y=3 \)

Explanation:
1. The first sum: \( \log_2x\left(1+\frac12+\frac14+\cdots\right)=2\log_2x=y \).
2. The second: numerator \( \sum_{k=1}^{y}(4k+1)=2y^2+3y \), denominator \( y^2 \), ratio \( \frac{2y+3}{y} \). Also \( 4\log_4x=2\log_2x=y \).
3. So \( \frac{2y+3}y=y\Rightarrow y^2-2y-3=0\Rightarrow y=3 \) and \( \log_2x=\frac32\Rightarrow x=2\sqrt2 \).

Teacher's Note:
1. The logs form a G.P. with ratio 1/2.
2. Both sums are sums of A.P.s.

Answer: (a) 11

Explanation:
1. Left side: \( n(6n+1) \). Right side: \( \frac n2(112+2n)=n(n+56) \).
2. Equating: 6n + 1 = n + 56, so 5n = 55 and n = 11.

Teacher's Note:
1. Write each sum with S = (m/2)[2a + (m − 1)d].
2. Cancel the common factor n.

Answer: (d) in H.P.

Explanation:
1. Divide each product by abcd: abc/abcd = 1/d, abd/abcd = 1/c, acd/abcd = 1/b, bcd/abcd = 1/a.
2. a, b, c, d in A.P. means 1/d, 1/c, 1/b, 1/a are in H.P. (reciprocals of an A.P.), and multiplying all terms by abcd keeps an H.P.

Teacher's Note:
1. Divide all four numbers by abcd.
2. The reciprocals of an A.P. form an H.P.

Answer: (d) \( \frac12-\frac1{\sqrt2} \)

Explanation:
1. A.P. gives 3b = 3/2, so b = 1/2. Write a = 1/2 − d and c = 1/2 + d with d > 0.
2. a², b², c² in G.P. gives \( b^4=a^2c^2\Rightarrow b^2=\pm ac \). The case \( b^2=ac \) forces a = b = c, which contradicts a < b < c, so \( b^2=-ac \).
3. Then \( \frac14=-\left(\frac14-d^2\right)\Rightarrow d^2=\frac12 \) and \( a=\frac12-\frac1{\sqrt2} \).

Teacher's Note:
1. Find b from the sum first.
2. a²c² = b⁴ allows ac = ±b²; ac = b² would force a = b = c.

Answer: (d) \( 0\lt x\lt 10 \)

Explanation:
1. \( \frac x{1-r}=5\Rightarrow r=1-\frac x5 \).
2. Convergence needs |r| < 1 (and r ≠ 0 for a genuine progression): \( -1\lt 1-\frac x5\lt 1 \) gives 0 < x < 10.

Teacher's Note:
1. Express r in terms of x.
2. Convergence requires |r| < 1.

Answer: (c) \( c\Delta=0 \)

Explanation:
1. Let s = α + β and p = αβ. G.P. gives \( (s^2-2p)^2=s(s^3-3sp) \).
2. Expanding: \( s^4-4s^2p+4p^2=s^4-3s^2p\Rightarrow p(4p-s^2)=0 \).
3. So p = 0 (which means c = 0) or \( s^2=4p \) (which means Δ = 0). Hence cΔ = 0.

Teacher's Note:
1. Write the G.P. condition in s and p.
2. Translate p = 0 and s² = 4p back to c and Δ.

Answer: (b) 5

Explanation:
1. \( A_n=\frac{3/4\left[1-(-3/4)^n\right]}{1+3/4}=\frac37\left[1-\left(-\frac34\right)^n\right] \).
2. \( B_n\gt A_n\iff A_n\lt \frac12\iff1-\left(-\frac34\right)^n\lt \frac76 \), i.e. \( \left(-\frac34\right)^n\gt -\frac16 \).
3. This fails for odd n when \( (3/4)^n\ge\frac16 \) i.e. n = 1, 3, 5 (since (3/4)⁵ ≈ 0.237 > 1/6) and holds for n = 7, 9, ... (0.133 < 1/6). Also all even n hold. So the condition holds for every n > 5 but fails at n = 5.

Teacher's Note:
1. Sum the G.P. with ratio −3/4.
2. Test odd n separately since the sign alternates.

Answer: (a) \( \frac1{12}n(n+1)(3n^2+n+2) \)

Explanation:
1. \( V_r=\frac r2[2r+(r-1)(2r-1)]=\frac r2(2r^2-r+1)=r^3-\frac{r^2}2+\frac r2 \).
2. \( \sum V_r=\frac{n^2(n+1)^2}4-\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}4=\frac{n(n+1)(3n^2+n+2)}{12} \).

Teacher's Note:
1. Find V_r with the A.P. sum formula.
2. Use the standard sums of r, r² and r³.

Answer: (d) \( Q_1,Q_2,Q_3,\ldots \) are in A.P. with common difference 6

Explanation:
1. \( T_r=3r^2+2r-1 \), so \( Q_r=T_{r+1}-T_r=6r+5 \).
2. Qᵣ = 11, 17, 23, … are in A.P. with common difference 6.

Teacher's Note:
1. Use Tᵣ from the previous part.
2. Qᵣ is linear in r.

Answer: (c) \( G_1=G_2=G_3=\cdots \)

Explanation:
1. Always \( A_kH_k=G_k^2 \) for two numbers. So \( G_n^2=A_{n-1}H_{n-1}=G_{n-1}^2 \).
2. Hence all the geometric means are equal: G₁ = G₂ = G₃ = …

Teacher's Note:
1. Recall A·H = G² for two positive numbers.
2. Gₙ is the geometric mean of Aₙ₋₁ and Hₙ₋₁.

Answer: (b) \( A_1\gt A_2\gt A_3\gt \cdots \)

Explanation:
1. For distinct positive numbers, \( A_{n-1}\gt G\gt H_{n-1} \).
2. So \( A_n=\frac{A_{n-1}+H_{n-1}}2\lt A_{n-1} \), and the sequence Aₙ is strictly decreasing.

Teacher's Note:
1. Compare Aₙ₋₁ with Hₙ₋₁.
2. Aₙ is the mean of a larger and a smaller number.

Answer: (c) B and D only

Explanation:
1. By the intersecting chords theorem, \( PS\cdot ST=QS\cdot SR \).
2. AM–HM: \( \frac1{PS}+\frac1{ST}\ge\frac2{\sqrt{PS\cdot ST}}=\frac2{\sqrt{QS\cdot SR}} \), with strict inequality unless PS = ST (which would make S the centre for chord QR... excluded). So (B) is true.
3. Also \( \sqrt{QS\cdot SR}\le\frac{QS+SR}2=\frac{QR}2 \), so \( \frac2{\sqrt{QS\cdot SR}}\ge\frac4{QR} \) and hence (D) is true.

Teacher's Note:
1. Use the power of the point S.
2. Apply AM–GM to QS and SR.

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

Explanation:
1. Take a, ar, ar², ar³: \( b_1=a,\ b_2=a(1+r),\ b_3=a(1+r+r^2),\ b_4=a(1+r+r^2+r^3) \).
2. The differences \( ar,\ ar^2,\ ar^3 \) are not equal and the ratios are not equal for r ≠ 1, so Statement 1 is true.
3. Using r = 2: b = 1, 3, 7, 15 and reciprocals 1, 1/3, 1/7, 1/15 are not in A.P., so Statement 2 is false.

Teacher's Note:
1. Test with a specific ratio such as r = 2.
2. Check the three conditions (A.P., G.P., H.P.) separately.

Answer: (a) 0

Explanation:
1. The recurrence makes the terms an A.P. with a₁ = 15 and common difference d. The mean of the 11 terms is a₆ = 15 + 5d.
2. For i = 0 to 10: \( \sum(15+id)^2=11\cdot225+30d\cdot55+d^2\cdot385 \), so the mean of squares is \( 225+150d+35d^2=90 \), i.e. \( 7d^2+30d+27=0 \), giving d = −3 or d = −9/7.
3. 27 − 2a₂ > 0 means a₂ < 13.5, i.e. 15 + d < 13.5, so d < −1.5. This rejects d = −9/7 and leaves d = −3.
4. The mean of the terms is 15 + 5(−3) = 0.

Teacher's Note:
1. The recurrence defines an A.P.
2. Use the condition on a₂ to reject one root.

Answer: (c) 3 or 9

Explanation:
1. \( \frac{S_{5n}}{S_n}=\frac{5n[6+(5n-1)d]}{n[6+(n-1)d]}=5\cdot\frac{(6-d)+5nd}{(6-d)+nd} \).
2. This is independent of n if 6 − d = 0 (ratio 25) or d = 0 (ratio 5).
3. d = 6 gives a₂ = 9; d = 0 gives a₂ = 3.

Teacher's Note:
1. Write S₅ₙ/Sₙ using the sum formula.
2. A ratio of two linear functions of n is constant in only two ways.

Answer: (d) 8

Explanation:
1. Apply AM–GM to the six numbers (with 3a⁻³ written as three equal terms a⁻³, giving 8 terms in all): \( \frac{a^{-5}+a^{-4}+3a^{-3}+1+a^8+a^{10}}{8}\ge\left(a^{-5-4-9+8+10}\right)^{1/8}=1 \).
2. Equality holds at a = 1, so the minimum is 8.

Teacher's Note:
1. Count 3a⁻³ as three terms so that there are 8 terms.
2. The exponents add up to zero.

Answer: (c) \( \frac7{81}(179+10^{-20}) \)

Explanation:
1. The kth term is \( 0.77\ldots7=\frac79\left(1-10^{-k}\right) \).
2. Sum of 20 terms: \( \frac79\left[20-\frac{1-10^{-20}}9\right]=\frac7{81}\left(179+10^{-20}\right) \).

Teacher's Note:
1. Write 0.77…7 as 7(1 − 10^−k)/9.
2. Sum the G.P. of 10^−k.

Answer: (b) 1056 and 1332

Explanation:
1. The signs follow the pattern −, −, +, +, … so each block of four terms gives \( -(4m+1)^2-(4m+2)^2+(4m+3)^2+(4m+4)^2=32m+20 \).
2. Therefore \( S_n=\sum_{m=0}^{n-1}(32m+20)=16n^2+4n \).
3. n = 8 gives 1056 and n = 9 gives 1332. The values 1088 and 1120 do not make 16n² + 4n an integer solution.

Teacher's Note:
1. Group the terms in fours.
2. Check which options equal 16n² + 4n for a positive integer n.

Answer: (b) \( 2+\sqrt3 \)

Explanation:
1. Let the numbers be a, ar, ar². After doubling the middle term, a, 2ar, ar² are in A.P.: \( 4ar=a+ar^2\Rightarrow r^2-4r+1=0 \).
2. r = 2 ± √3; for an increasing G.P. r > 1, so r = 2 + √3.

Teacher's Note:
1. Use 2 × middle = sum of the extremes.
2. An increasing G.P. of positive terms has r > 1.

Answer: (a) 100

Explanation:
1. Divide by \( 10^9 \): \( 1+2x+3x^2+\cdots+10x^9 \) with x = 11/10.
2. The sum \( \sum_{j=1}^{10}jx^{j-1}=\frac{1-11x^{10}+10x^{11}}{(1-x)^2} \) at x = 1.1 evaluates to 100.

Teacher's Note:
1. Factor out 10⁹ to get an arithmetico-geometric series.
2. Use the A.G.P. sum formula.

Answer: (c) 4

Explanation:
1. Let b = ar with r an integer (r ≥ 2) and c = ar². The A.M. condition: \( a(1+r+r^2)=3(ar+2) \) gives \( a(r-1)^2=6 \).
2. Integer solutions with r ≥ 2: r = 2, a = 6 (r = 3 gives a = 3/2, r = 4 gives 2/3, ...).
3. Then \( \frac{36+6-14}{7}=4 \).

Teacher's Note:
1. Write the A.M. condition using b = ar and c = ar².
2. a(r − 1)² = 6 has one integer solution with r ≥ 2.