Question 1

Let p, q and r be three unequal numbers such that p, q and r are in AP. If (q-p), (r-q) and p are in GP, then (p+q):(q+r):(r+p) equals

Accepted Answer

Since p, q, and r are in AP, let the common difference be d. Then q - p = d and r - q = d. It is given that (q - p), (r - q), and p are in GP. Substituting the values, we get d, d, and p are in GP. Therefore, d^2 = d p. Since p, q, and r are unequal, d 0. Thus, p = d. Now, q = p + d = 2d and r = p + 2d = 3d. We need to find the ratio (p+q) : (q+r) : (r+p). Substituting the values of p, q, and r in terms of d: p + q = d + 2d = 3d q + r = 2d + 3d = 5d r + p = 3d + d = 4d The required ratio is 3d :

Question 2

If p^ th term of an AP is k, then what is the sum of p^ th term, (p+q)^ th term and (p-q)^ th term of the AP ?

Accepted Answer

Let the first term of the AP be a and the common difference be d. The p^ th term is given by T_p = a + (p-1)d = k. The (p+q)^ th term is T_ p+q = a + (p+q-1)d. The (p-q)^ th term is T_ p-q = a + (p-q-1)d. The required sum is T_p + T_ p+q + T_ p-q . Substituting the values, we get: [a + (p-1)d] + [a + (p+q-1)d] + [a + (p-q-1)d] = 3a + d[(p-1) + (p+q-1) + (p-q-1)] = 3a + d(3p - 3) = 3[a + (p-1)d] Since a + (p-1)d = k, the sum is 3k. Answer: 3k

Question 3

If p, g_1, g_2 and q are in GP and m is the arithmetic mean of p and q, then g_1^2 g_2 + g_2^2 g_1 is equal to

Accepted Answer

Given p, g_1, g_2, q are in GP. Let the common ratio of the GP be r. Then g_1 = pr, g_2 = pr^2, and q = pr^3. The arithmetic mean of p and q is m, so m = p + q 2 p + q = 2m. Consider the expression g_1^2 g_2 + g_2^2 g_1 . Substituting the values of g_1 and g_2: g_1^2 g_2 + g_2^2 g_1 = (pr)^2 pr^2 + (pr^2)^2 pr = p^2r^2 pr^2 + p^2r^4 pr = p + pr^3 Since q = pr^3, we get: = p + q Substituting p + q = 2m: g_1^2 g_2 + g_2^2 g_1 = 2m Answer: 2m

Question 4

The sum of the first n terms of an AP is 3n^2+5n. If the m^ th term of the AP is 68, then what is the value of m ?

Accepted Answer

Given S_n = 3n^2 + 5n The n^ th term of the AP is given by T_n = S_n - S_ n-1 T_n = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)] T_n = 3n^2 + 5n - (3n^2 - 6n + 3 + 5n - 5) T_n = 6n + 2 Given T_m = 68 6m + 2 = 68 6m = 66 m = 11 Answer: 11

Question 5

Passage: Let X(a,p), Y(b,q) and Z(c,r) be the points such that a, b and c are in AP. Question: If p, q and r are not in AP and b=c, then the line joining the points X, Y and Z is parallel to

Accepted Answer

Given that a, b, and c are in AP, we have 2b = a + c. It is also given that b = c. Substituting this into the first equation gives 2b = a + b, which implies a = b. Therefore, a = b = c. The coordinates of the points become X(a, p), Y(a, q), and Z(a, r). Since the x-coordinates of all three points are equal, they lie on the vertical line x = a. This line is parallel to the y-axis. Answer: y-axis

Question 6

If (10 + _ 10 x), (10 + _ 10 y) and (10 + _ 10 z) are in AP, then consider the following statements: I. The GM of x and z is y^2. II. The AM of _ 10 x and _ 10 z is _ 10 y. Which of the statements giv

Accepted Answer

Given (10 + _ 10 x), (10 + _ 10 y), and (10 + _ 10 z) are in AP. 2(10 + _ 10 y) = (10 + _ 10 x) + (10 + _ 10 z) 20 + 2 _ 10 y = 20 + _ 10 x + _ 10 z 2 _ 10 y = _ 10 x + _ 10 z _ 10 y = _ 10 x + _ 10 z 2 This shows that the AM of _ 10 x and _ 10 z is _ 10 y. Thus, Statement II is correct. From 2 _ 10 y = _ 10 x + _ 10 z, we get: _ 10 y^2 = _ 10 (xz) y^2 = xz The GM of x and z is xz = y^2 = y. Thus, Statement I is incorrect. Only Statement II is correct. Answer: II only

Question 7

How many terms of the series 1 + 3 + 5 + 7 + amount to a sum equal to 12345678987654321?

Accepted Answer

The given series is 1 + 3 + 5 + 7 + This is an arithmetic progression with first term a = 1 and common difference d = 2. The sum of the first n terms of this series is given by: S_n = n 2 [2a + (n-1)d] = n 2 [2(1) + (n-1)2] = n^2 We are given that S_n = 12345678987654321. Therefore, n^2 = 12345678987654321. Using the pattern of squares of numbers consisting entirely of ones: 1^2 = 1 11^2 = 121 111^2 = 12321 11 1 _ k times ^2 = 12 k 21 (for k 9) Since the maximum digit in the given sum is 9, the

Question 8

For the following two (02) items: Let (6 + 10 + 14 + up to m terms ) = (1 + 3 + 5 + 7 + up to n terms ), where m How many values of m are possible?

Accepted Answer

The sum of the first m terms of the arithmetic progression 6, 10, 14, is given by: S_m = m 2 [2(6) + (m - 1)4] = m 2 [12 + 4m - 4] = 2m^2 + 4m The sum of the first n terms of the arithmetic progression 1, 3, 5, 7, is given by: S_n = n 2 [2(1) + (n - 1)2] = n^2 Equating the two sums, we get: 2m^2 + 4m = n^2 2(m^2 + 2m + 1) - 2 = n^2 2(m + 1)^2 - n^2 = 2 Let u = m + 1. The equation becomes 2u^2 - n^2 = 2. This implies that n must be an even number. Let n = 2v. Substituting this into the equation y

Question 9

For the following two (02) items: Let (6 + 10 + 14 + up to m terms ) = (1 + 3 + 5 + 7 + up to n terms ), where m What is the relation between m and n?

Accepted Answer

The first series is an arithmetic progression with first term a = 6 and common difference d = 4. The sum up to m terms is given by: S_m = m 2 [2(6) + (m-1)4] S_m = m 2 [12 + 4m - 4] = m 2 [4m + 8] = 2m(m + 2) The second series is an arithmetic progression with first term a = 1 and common difference d = 2 (which is the sum of the first n odd natural numbers). The sum up to n terms is given by: S_n = n 2 [2(1) + (n-1)2] = n^2 Equating the two sums, we get: n^2 = 2m(m + 2)

Question 10

How many terms are identical in the two APs 19, 21, 23, up to 110 terms and 19, 22, 25, 28, up to 75 terms?

Accepted Answer

For the first AP: 19, 21, 23, First term a_1 = 19, common difference d_1 = 2, number of terms n_1 = 110. Last term L_1 = a_1 + (n_1 - 1)d_1 = 19 + (110 - 1) 2 = 19 + 218 = 237. For the second AP: 19, 22, 25, 28, First term a_2 = 19, common difference d_2 = 3, number of terms n_2 = 75. Last term L_2 = a_2 + (n_2 - 1)d_2 = 19 + (75 - 1) 3 = 19 + 222 = 241. The identical terms of both APs form a new AP. The first common term is a = 19. The common difference of the new AP is D = LCM (d_1, d_2) = LCM

Question 11

If p, 1, q are in AP and p, 2, q are in GP, then which of the following statements is/are correct? I. p, 4, q are in HP. II. 1 p , 1 4 , 1 q are in AP. Select the answer using the code given below.

Accepted Answer

Since p, 1, q are in AP, we have p + q = 2(1) = 2. Since p, 2, q are in GP, we have pq = 2^2 = 4. For p, 4, q to be in HP, the middle term must be the harmonic mean of p and q, which is given by 2pq p+q . Substituting the values of p+q and pq, we get 2(4) 2 = 4. Thus, p, 4, q are in HP. Statement I is correct. By the definition of harmonic progression, if p, 4, q are in HP, their reciprocals 1 p , 1 4 , 1 q must be in AP. Thus, Statement II is also correct. Both statements I and II are correct.

Question 12

The sum of the first 8 terms of a GP is five times the sum of its first 4 terms. If r 1 is the common ratio, then what is the number of possible real values of r?

Accepted Answer

Let the first term of the GP be a and the common ratio be r. The sum of the first n terms is given by S_n = a(r^n - 1) r - 1 . According to the given condition, S_8 = 5S_4. a(r^8 - 1) r - 1 = 5 a(r^4 - 1) r - 1 For a non-zero first term a and given r 1: r^8 - 1 = 5(r^4 - 1) (r^4 - 1)(r^4 + 1) = 5(r^4 - 1) Assuming r^4 - 1 0 to avoid the trivial zero-sum case: r^4 + 1 = 5 r^4 = 4 Since r is a real number, r^2 must be positive. r^2 = 2 r = 2 Therefore, there are two possible real values of r.

Question 13

If 5th, 7th and 13th terms of an AP are in GP, then what is the ratio of its first term to its common difference?

Accepted Answer

Let the first term of the AP be a and the common difference be d. The 5th, 7th, and 13th terms of the AP are a + 4d, a + 6d, and a + 12d respectively. Since these terms are in GP, the square of the middle term is equal to the product of the other two terms: (a + 6d)^2 = (a + 4d)(a + 12d) Expanding both sides: a^2 + 12ad + 36d^2 = a^2 + 16ad + 48d^2 Simplifying the equation: 4ad + 12d^2 = 0 4d(a + 3d) = 0 Assuming d 0 for a non-constant AP, we get: a + 3d = 0 a = -3d The ratio of the first term t

Question 14

The sum of the first k terms of a series S is 3k^2 + 5k. Which one of the following is correct?

Accepted Answer

The sum of the first k terms of the series is given by S_k = 3k^2 + 5k. The k-th term of the series, T_k, is obtained by subtracting the sum of the first k-1 terms from the sum of the first k terms: T_k = S_k - S_ k-1 Substituting the given expression for S_k and S_ k-1 : T_k = (3k^2 + 5k) - [3(k-1)^2 + 5(k-1)] T_k = 3k^2 + 5k - [3(k^2 - 2k + 1) + 5k - 5] T_k = 3k^2 + 5k - (3k^2 - 6k + 3 + 5k - 5) T_k = 3k^2 + 5k - (3k^2 - k - 2) T_k = 6k + 2 To find the common difference, we calculate the diffe

Question 15

What is the harmonic mean of the numbers C(10, 3), C(10, 4), C(10, 5), C(10, 6) and C(10, 7)?

Accepted Answer

The given numbers are ^ 10 C_ 3 , ^ 10 C_ 4 , ^ 10 C_ 5 , ^ 10 C_ 6 , and ^ 10 C_ 7 . Calculating their values: ^ 10 C_ 3 = 10 9 8 3 2 1 = 120 ^ 10 C_ 4 = 10 9 8 7 4 3 2 1 = 210 ^ 10 C_ 5 = 10 9 8 7 6 5 4 3 2 1 = 252 ^ 10 C_ 6 = ^ 10 C_ 4 = 210 ^ 10 C_ 7 = ^ 10 C_ 3 = 120 The harmonic mean H of n numbers x_1, x_2, , x_n is given by: H = n 1 x_1 + 1 x_2 + + 1 x_n Here n = 5. The sum of the reciprocals is: S = 1 120 + 1 210 + 1 252 + 1 210 + 1 120 S = 2 120 + 2 210 + 1 252 S = 1 60 + 1 105 + 1 252

Sequences and Series — Previous Year Questions & Solutions