Maths Quiz on Sequences and Series for JEE Mains/Advance

a) 1

Solution:

x + y + z = 15, if x = (z⁻³)⁻¹ = z³ are in A.P.

Sum =9+15+a=52(9+a)

⇒ 24 + a = 5 / 2 (9 + a)

⇒ 48 + 2a = 45 + 5a

⇒ 3a = 3

⇒ a = 1 ..(i) and

[1 / x] + [1 / y] + [1 / z] = 5 / 3, if 9, x, y, z, a are in H.P.

Sum = 1 / 9 + 5 / 3 + 1 / a

= 5 / 2 [1 / 9 + 1 / a]

⇒ a = 1

b) H.P.

Solution:

x, 1, z are in A.P., then

2 = x + z ……(i) and

4 = xz ……(ii)

Divide (ii) by (i), we get

[x z] / [x + z] = 4 / 2 or

2xz / [x + z] = 4

Hence, x, 4, z will be in H.P.

a) 9

Solution:

Let the number of sides of the polygon be n.

Then the sum of interior angles of the polygon = (2n − 4) [π / 2] = (n − 2)π

Since the angles are in A.P. and a = 120^o, d = 5, therefore

[n / 2] * [2 * 120 + (n − 1)5] = (n − 2) * 180

n² − 25n + 144 = 0

(n − 9) (n − 16) = 0

n = 9, 16

But n = 16 gives

T₁₆ = a + 15d = 120^o + 15.5^o = 195^o, which is impossible as interior angle cannot be greater than 180^o.

Hence, n = 9.

a) 55 / 155

Solution:

We have Sn₁ / Sn₂ = [2n + 3] / [6n + 5]

{[n / 2] * [2a₁ + (n − 1)d₁]} / {[n / 2] * [2a₂ + (n − 1)d₂]}

= [2n + 3] / [6n + 5]

{2 * [a₁ + ([n − 1] / 2)d1]} / {2 * [a₂ + ([n − 1] / 2)d2]} = [2n + 3] / [6n + 5]

{a₁ + ([n − 1] / 2)d₁} / {a₂+([n − 1] / 2)d₂} = [2n + 3] / [6n + 5]

Put n = 25 then,

[a₁ + 12d₁] / [a₂ + 12d₂]= [2 (25) + 3] / [6 (25) + 3]

T13₁ / T13₂ = 53 / 155

a) 0

Solution:

Suppose that the first term and the common difference of A.P.s are A and D respectively.

Now,

p^th term = A + (p − 1)D = a …… (i)

q^th term = A + (q − 1)D = b …… (ii) and

r^th term = A + (r − 1)D = c .….. (iii)

So, [a (q − r) + b (r − p) + c (p − q)] = a * {[b − c] / [D]} + b * {[c − a] / [D]} + c * {[a − b] / [D]}

= [1 / D] (ab − ac + bc − ab + ca −bc)

= 0

b) 3

Solution:

log₃ 2, log₃ (2^x − 5) and log₃ (2^x − [7 / 2]) are in A.P.

⇒ 2 log₃ (2^x − 5) = log₃ [(2) (2^x − 72)]

⇒ (2^x − 5)² = 2^x+1 − 7

⇒ [2^2x − 12] * [2^x − 32] = 0

⇒ x = 2, 3

But x = 2 does not hold, hence x = 3.

c) 13

Solution:

Given series 63 + 65 + 67 + 69 + . . . . . . . . . (i) and 3 + 10 + 17 + 24 + . . . . . . (ii)

Now from (i), m^th term = (2m + 61) and m^th term of (ii) series = (7m − 4)

Under condition, ⇒ 7m − 4 = 2m + 61

⇒ 5m = 65

⇒ m = 13

a) 3050

Solution:

The sum of integers from 1 to 100 that are divisible by 2 or 5

= sum of series divisible by 2 + sum of series divisible by 5 – sum of series divisible by 2 and 5.

= (2 + 4 + 6 + . . . . . . + 100) + (5 + 10 + 15 . . . . . . . + 100) − (10 + 20 + 30 + . . . . . . . . + 100)

= [50 / 2] {2 * 2 + (50 − 1) 2} + [20 / 2] {2 * 5 + (20 − 1)5} − [10 / 2] {10 * 2 + (10 − 1)10}

= 2550 + 1050 − 550

= 3050

b) 1

Solution:

log_e {1 / [1 − x − x² + x³]} = log_e {1 / [(1 − x) − x² (1 − x)]}

= log_e {1 / [(1 − x) (1 − x²)]}

= log_e {[1 / (1 − x)² (1 + x)]}

= log_e {(1 − x)⁻² (1 + x)⁻¹} ……… (i)

= log_e (1 − x)⁻² + log_e (1 + x)⁻¹

= −2 log_e (1 − x) − log_e (1 + x)

= −2 [(−x) − (x² / 2) − (x³ / 3) − (x⁴ / 4) − . . . . . . . ∞] − [x + (x² / 2) + (x³ / 3) + (x⁴ / 4) − . . . . ∞],

Hence, coefficient of x = 2 − 1 = 1

d) 6e − 1

Solution:

S = 4 + 11 + 22 + 37 + . . . . + T_n−1 + T_n or

S = 4 + 11 + 22 + 37 + . . + T_n−1 + T_n

Therefore, on subtracting we get 0 = 4 + [7 + 11 + 15 + 19 + . . . . + (T_n − T_n−1)] − T_n

0 = 4 + [n − 1] / [2] * [14 + (n − 2)4] − T_n

So, T_n = 2n² + n + 1

Thus, n^th term of the given series is T_n = [2n² + n + 1] / [(n)!]

= [2n] / [(n − 1)!] + [1] / [(n − 1)!] + [1]/ [n!]

= [2 (n − 1 + 1)] / [(n − 1)!] + 1 / [(n − 1)!] + [1] / [n!]

= 2 / [(n − 2)!] + 3 / [(n − 1)!] + 1 / n!

Sum = ∑_n=1^∞ T_n = 2e + 3e + e − 1

= 6e − 1