Problems on Arithmetic Geometric Progressions

Answer:(a) 2

x, x + 2, 3x are in A.P

∴ (x + 2) - x = 3x - (x + 2)

∴ x + 2 - x = 3x - x - 2

∴ 2 = 2x - 2

∴ 2x = 4

∴ x = 2

Answer:(d) 5⁵

t₃ = 5

ar^3-1 = 5

ar² = 5

Product of 1st 5 terms is, a X ar X ar² X ar³ X ar⁴

= a⁵r¹⁰

= (ar²)⁵

= 5⁵

Answer:(c) 256

Here a = 4

r = 2

t₇ = ?

t_n = ar^7-1

=ar⁶

=4 X 2⁶

=4 X 64

=256

Answer:(b) 225

The first 15 odd number are

1, 3, 5, 7, 9, 11, 13, ..... 29

Now 1 + 3 + 5 + ...... + 29

Here a = 1, n = 15, d = 2

∴ S_n = n/2 { 2a + (n-1) d}

∴ = 15/2 { 2 X 1 + (15-1) X 2 }

∴ = 15/2 { 2 + 28 }

∴ = 15 X 15

∴ = 225

Answer:(a) 4, 8, 16

Let the numbers be a/r, a, ar

⇒ a³ = 512

⇒ a³ = 8³

⇒ a = 8

⇒ 2r² + 2r + 2 = 7r

⇒ 2r² - 5r + 2 = 0

⇒ 2r² - 4r - r + 2 = 0

⇒ 2r(r-2) - (r-2) = 0

⇒ (r-2)(2r-1) = 0

∴ r - 2 = 0 or 2r - 1 = 0

⇒ r = 2        ⇒ r = 1/2

∴ numbers are 4, 8, 16

Answer:(b) 9

Let the nth term is zero

Here a = 64, d = 56 - 64 = -8

Now t_n = a + (n-1) d

⇒ 0 = 64 + (n-1) X (-8)

⇒ 0 = 64 - 8n + 8

⇒ 8n = 72

⇒ n = 9

∴ 9^th term is zero

Answer:(d) 180

The 3-digit numbers which are divisible by 5 are

100, 105, 110, 115, ...... 995

Here a = 100

d = 105 - 100 = 5

t_n = 995

n = ?

We have t_n = a + (n-1) d

⇒ 995 = 100 + (n-1) 5

⇒ n-1 = (995-100)/5

⇒ n-1 = 179

⇒ n = 180

Answer:(c) 30

The numbers are .....

24, 30, 36, 42, .........., 198

Here a = 24

d = 30 - 24 = 6

t_n = 198

n = ?

We have, t_n = a + (n-1) d

⇒ 198 = 24 + (n-1) 6

⇒ n - 1 = (198 - 24) / 6

⇒ n - 1 = 29

⇒ n = 30

Answer:(c) 51

a = 6,      d = 5;      t₁₀ = ?

We have,

t₁₀ = a + (10-1) d

= 6 + 9 X 5

= 6 + 45

= 51

Answer:(a) 15

Here a = 7,      d = 3;      t₅ = ?

Now,

t₅ = a + (5-1) d

= 7 + 4 X 2

= 7 + 8

= 15