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y = x + x² + x³ + … ∞

⇒ y = x( 1 + x + x² + … ∞)

⇒ y = x (1/(1 - x))

⇒ y (1 – x) = x

⇒ y – xy – x = 0

⇒ y – x(y + 1) = 0

⇒ x(1 + y) = y

⇒ x = (y/(1 + y))

Answer is (a) 23

-4, -1, + 2, + 5, …

Here a = – 4

d = (- 1) – (- 4) = – 1 + 4 = 3

10^th term = T₁₀

= a + 9d

= -4 + 9 × 3

= – 4 + 27 = 23

series = 5 + 10 + 20 + 40 + ….

Common ratio r = T₂/T₁ = 10/5 = 2

n^th term = T_n = ar^{n – 1}

= 5 × (2)^{n – 1} = 5.2^{n – 1}

Answer is (b) 79

Given,

T₉ = 35

⇒ a + 8d = 35 …(i)

and T₁₉ = 75

⇒ a + 18d = 75 …(ii)

Subtracting equation (i) from equation (ii),

10d = 40 ⇒ d = 4

Put d = 4 in equation (i),

a + 8 × 4 = 35

⇒ a + 32 = 35

⇒ a = 35 – 32 = 3 ⇒ a = 3

Thus, 20^th term = T₂₀

= a + 19 d

= 3 + 19 × 4 = 3 + 76 = 79

Let a is first term and R is common ratio of G.P., then according to question

T₄ = aR^{4 – 1}

p = aR³           ….. (i)

T₇ = aR^{7 – 1}

q = aR⁶           ….. (ii)

T₁₀ = aR^{10 – 1}

r = aR⁹            ….. (iii)

Multiplying equation (i) and (iii), we get

a² R¹² = pr    ….. (iv)

Squaring of equation (ii),

q² = a² R¹² Hence, from equation (iv) and (v)

q² = pr         ….. (v)

Hence Proved.

∵ x²(y + z), y²(z + x), z²(x + y) are in A.P.

∴ Adding xyz in each terms x²(y + z) + xyz, y²(z + x) + xyz, z²(x + y) + xyz also will be in A.P.

or x(xy + yz + zx), y(xy +yz + zx), z(xy + yz + zx)

also will be in A.P.

∴ 2y(xy + yz + zx) = x(xy + yz + zx) + z(xy + yz + zx)

⇒ 2y(xy + yz + zx) = (xy + yz + zx) (x + z)

⇒ 2y(xy + yz + zx) – (xy + yz + zx) (x + z) = 0

⇒ (xy + yz + zx) (2y – x – z) = 0

If  2y – x – z = 0

Then 2y = x + z

⇒ x, y, z are in A.P.

or xy + yz + zx = 0

Hence Proved.

Answer is (d) n²

1,3,5, …

n^th term of the given series

T_n = 2n – 1

Sum of n terms

s_n = ∑(2n – 1)

= 2∑n – ∑1

= 2(n(n + 1))/2 – n

= n² + n – n = n²

First two digit number which is divisible by 3 is 12 and last two digit number is 99

∵ We have to find only the numbers which are divisible by 3, so common difference is 3.

a= 12, d = 3, l = 99

Hence, l = a + (n – 1 )d

⇒  12 + (n- 1) × 3 = 99

⇒  3(n – 1) = 99 – 12

⇒ n - 1 = 87/3 = 29

⇒  n= 29 + 1 = 30

Hence, two digit natural number which are divisible by 3 are 30.

Sum of terms

S_n= n² + 2n

and S, (n – 1)² + 2(n-1)

We know that nth term of A.P.

T_n= S_n – S_{n – 1}

⇒ T_n= (n² + 2n) – [(n – 1)² + 2(n − 1)]

= [n² + 2n] – [n² + 1 – 2n + 2n – 2]

= [n² + 2n] – [n² – 1]

= n² + 2n – n² + 1

= 2n + 1

First term T₁= 2 × 1 + 1 = 2 + 1 = 3

Second term, T₂ = 2 2 + 1 = 4 + 1 = 5

Common difference d = T₂ – T₁ = 5 – 3 = 2

Hence, first term is 3 and common difference is 2.

Let a be the first term and d be the common difference then 9^th

term = T₉ = 0 Then a + (9 – 1)d= 0

⇒ a + 8d = 0   ….. (i)

29^th term = T₂₉ = a + 28d  …(ii)

19^th term = T₁₉ = a + 18d  …(iii)

Putting the value of equation (i) into equation (ii) and (iii)

we get

T₂₉ = (a + 8 d) + 20 d

⇒ T₂₉ = 0 + 20 d

⇒ T₂₉ = 20d … (iv)

⇒ T₂₉ = (a + 8 d) + 10 d

⇒ T₂₉ = 0 + 10d

⇒ T₂₉ = 10d … (v)

From equation (iv) and (v), we have

T₂₉ = 20d = 2 × 10d =2 × T₂₉

T₂₉ = 2T₂₉

Hence Proved.

Answer is (c) 1250

First term, a = 20

Last term, l = 30

Number of terms = 52

Sum of 52 terms

S₅₂ =  52/2(20 + 30)

= 26 × 50 = 1300

Thus, sum of 50 A.M. between 20 and 30 is

= 1300 – 20 – 30

= 1300 – 50 = 1250

Answer is (C) a + c = 2b

H.M. of a, b, c

H =  (2ac)/(a + c)….(i)

G.M. of a,b,c

G = √ac ….(ii)

From equation (i) and (ii),

b = (2ac)/(a + c)

We know that GM. > H.M.

G > H

√ac > b

Given series = 2 + 5 + 8 + 11 + ….

Let its nth term is 65

Then T_n = 65 and d = 5 – 2 = 3

⇒ a + (n – 1)d – 65

⇒ 2 + (n – 1) × 3 = 65

⇒ 2 + 3n – 3 = 65

⇒ 3n – 1 = 65

⇒ n = (65 + 1)/3 = 66/3 = 22

Hence, 22nd term is 65

Answer is (A) 1/3

Common ratio (1/√3)/√3 = 1/3

Given series = 4 + 9 + 14 + 19 +…+ 124 a = 4, d = 9 – 4 = 5, l= 124

13^th term from last

= l – (n – 1)d

= 124 – (13 – 1) × 5

= 124 – 12 × 5

= 124 – 60 = 64

Hence, 13^th term from last is 64.

Answer is (b) 1/16

The given series is a H.P. because the corresponding A.P. will be 1, 4, 7, 10,…. 

In which a = 1, d = 4 – 1 = 3

∴ If 6^th term a₆ = a + 5d

= 1 + 5 × 3 = 1 + 15 = 16

Thus, 6th term of corresponding H.P. is 1/16.