JEE Main Mathematics Sequences and Series Sequences Online Test

Let AP series is a, a + d, a + 2d, a + 3d, ……

Now, given a = 2

According to the given condition,

Sum of first five terms = 1/4 (Sum of next five terms)

⇒ a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d)

= 1/4 [a + 5d + a + 6d + a + 7d + a + 8d + a + 9d]

⇒ 5a + 10d = 1/4 (5a + 35d)

⇒ 4(5a + 10d) = 5a + 35d

⇒ 20a + 40d = 5a + 35d ⇒ 20a – 5a = 35d – 40d

⇒ 15a = -5d

⇒ 15 × 2 = -5d

⇒ 30 -5d                             (∵ a = 2)

Now, T_n = a + (n – 1)d

⇒ T₂₀ = 2 + (20 – 1)(-6) = 2 + 19 (-6)

= 2 – 19 × 6 = 2 – 114 = -112

Common terms are 5, 11, 17, ….

∴ n^th term of common sequence

t_n = 5 + (n – 1)6 = (6n – 1)

Also 100^th terms of the first sequence

= 2 + (100 – 1)3 = 299

and 100^th term of the second sequence

= 3 + (100 – 1)2 = 201

⇒ t_n ≤ 201

⇒ 6n – 1 ≤ 201

⇒ n ≤ 33

∴ n = 33                     (∵ n ∈ N)

Let the first term of GP is A and common ratio is R.

Given, pth term = T_p = a ⇒ AR^p-1 = a         …….(i)

qth term = T_q = b ⇒ AR^{q – 1} = b                …….(ii)

rth term = T_r = c ⇒ AR^r-1 = c                   …….(iii)

∴ a^q-r b^r-p c^p-a

= (AR^p-1)^q-r (AP^q-1)r-p (AR^r-1)^p-q

= A^q-rR^(p-1)(q-r)A^r-p R^(q-1)(r-p) A^p-q R^(r-1)(p-q)

= A^q-r+r-p+p-q R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}

= A⁰ R^{pq-pr-q+r+qr-pq+r+p+rp-rq-p+q}

= A⁰ R⁰ = 1

H = 4, G² = AH

⇒ 2A + G² = 27

⇒ 2A + AH = 27

⇒ 6A = 2A ⇒ A = 9/2, G² = 18

a + b = 9, ab = 18

∴ a = 3, b = 6 or a = 6, b = 3

∵ a, b, c, d are in G.P.

⇒ b = ar, c = ar², d = ar²                       ……..(i)

Now, (aⁿ + bⁿ) (cⁿ + dⁿ)

= (aⁿ + aⁿrⁿ) × (aⁿr²ⁿ + aⁿr³ⁿ)               [from Eq. (i)]

= aⁿ (1 + rn) aⁿr²ⁿ (1 + rⁿ)

= a²ⁿr²ⁿ(1 + rⁿ)² = (aⁿrⁿ(1 + rⁿ))2 = [aⁿrⁿ + aⁿr²ⁿ]²

= [(ar)ⁿ + (ar²)ⁿ]² = (bⁿ + cⁿ)²             [from Eq. (i)]

Hence, they are in GP.

Let a^1/x = b^1/y = c^1/z

= k ⇒ a = k^x, b = k^y, c = k^z

Now, a,b,c are in G.P.

⇒ b² = ac ⇒ k^2y = k^x.k^z = k^{x + z}

⇒ 2y = x + z

∴ x,y,z are in A.P.

Here, number of factors  = 50

∴The coefficient of x⁴⁹ = -1 – 3 – 5 – … – 99

= -2500

Suppose that the added number be x, then x + 2, x + 14, x + 62 are in GP.

∴ (x + 14)² = (x + 2)(x + 62)

⇒ x² + 196 + 28x = x² + 64x + 124

⇒ 36x = 72 ⇒ x = 2

t_m + t_n

1 + (m – 1), 10 = 31 + (n – 1).5

10m – 9 = 5n + 26

10m = 5n + 34 = 5(n + 7)

m = λ and n = 2λ – 7

m ≤ 100 and n ≤ 100

λ ≤ 100 and 2λ – 7 ≤ 100 ⇒ λ ≤ 53 1/2

so 1 ≤ λ ≤ 53 [No. of common term = 53]

t₅₃ = 1 + 52.10 = 521

Suppose that ∠A = x, then ∠B = x + 10^o,

∠C = x + 20^o and ∠D = x + 30^o

So, we know that ∠A + ∠B + ∠C + ∠D = 360^o

On putting these values, we get

(x) + (x + 10°) + (x + 20°) + (x + 30°) = 360°

⇒                    x = 75°

Hence, the angles of the quadrilateral are 75°, 85°, 95°, 105°.

28, A1, A2 … …. A10, A11, 10

A₁ = 28 + d ⇒ clearly not integral

The A.M.’s will be integral only when 2d, 4d ………types of terms occur So, A₂, A₄, A₆, A₈, A₁₀ are the A.M.

Total number = 5

Since, x, 2y and 3z are in AP.

∴         4y  = x + 3z                                         …..(i)

And     x, y, z are in GP.

∴         y = rx and z  = xr²

On putting the values of y and z in Eq. (i), we get 4xr = x + 3xr²

⇒ 3r² – 4r + 1 = 0

⇒ (3r – 1)(r – 1) = 0

⇒ r = 1/3, 1

∴ r = 1/3                 (∵ r ≠ 1)

Let a, ar, ar² are in GP and a, ar, ar² – 64 are in AP, we get

a(r² – 2r + 1) = 64                              ….(i)

Again, a, ar – 8, ar² – 64 are in GP.

∴ (ar – 8)² = a(ar² – 64)

⇒ a(16r – 64) = 64                           ….(ii)

On solving Eqs. (i) and (ii), we get r = 5, a = 4

Thus, required numbers are 4, 20, 100.

∵ 1029 = 3.7³ → number of divisors = (1 + 1) (1 +3) = 8

1547 = 7.11² → number of divisors = (1 + 1) (1 +3) = 6

122 = 2.61 → number of divisors = (1+ 1 ) (1 + 1) = 4

Also 8, 6, 4 are in AP