# Problems on Arithmetic Geometric Progressions

Q11. If x, x+2, 3x are in A.P. then x = ?

(a) 2

(b) 3

(c) 4

(d) 5

**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

Q12. The third term of a geometrical progression is 5. The product of the first 5 terms is

(a) 2^{2}

(b) 3^{3}

(c) 4^{4}

(d) 5^{5}

**Answer:**(d) 5^{5}

t_{3} = 5

ar^{3-1} = 5

ar^{2} = 5

Product of 1st 5 terms is, a X ar X ar^{2} X ar^{3} X ar^{4}

= a^{5}r^{10}

= (ar^{2})^{5}

= 5^{5}

Q13. In a G.P, the first term is 4 and the common ratio is 2. The 7th term is

(a) 198

(b) 206

(c) 256

(d) 312

**Answer:**(c) 256

Here a = 4

r = 2

t_{7} = ?

t_{n} = ar^{7-1}

=ar^{6}

=4 X 2^{6}

=4 X 64

=256

Q14. The sum of first 15 natural odd number is

(a) 175

(b) 225

(c) 275

(d) 315

**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

Q15. Three numbers are in G.P. Their sum is 28 and product is 512. The numbers are

(a) 4, 8, 16

(b) 6, 12, 24

(c) 3, 9, 27

(d) 2, 4, 16

**Answer:**(a) 4, 8, 16

Let the numbers be a/r, a, ar

Now | a | + a + ar = 28 ------> (1) |

r |

And | a | . a . ar = 512 |

r |

⇒ a^{3} = 512

⇒ a^{3} = 8^{3}

⇒ a = 8

(1) ⇒ a ( | 1 | + 1 + r ) = 28 |

r |

⇒ 8 ( | 1 | + 1 + r ) = 28 |

r |

⇒ | 1 + r + r^{2} |
= | 28 |

r | 8 |

⇒ 2r^{2} + 2r + 2 = 7r

⇒ 2r^{2} - 5r + 2 = 0

⇒ 2r^{2} - 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

Q16. Which term of the series 64 + 56 + 48 + ------ is zero

(a) 7

(b) 9

(c) 11

(d) 12

**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

Q17. How many 3-digit numbers are divisible by 5 ?

(a) 160

(b) 165

(c) 175

(d) 180

**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

Q18. How many numbers are there between 20 and 200 which are exactly divisible by 6 ?

(a) 20

(b) 25

(c) 30

(d) 35

**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

Q19. The first term of an A.P. is 6 and its common difference is 5. The 10th term is

(a) 39

(b) 43

(c) 51

(d) 57

**Answer:**(c) 51

a = 6, d = 5; t_{10} = ?

We have,

t_{10} = a + (10-1) d

= 6 + 9 X 5

= 6 + 45

= 51

Q20. What is the next number in A.P. 7, 9, 11, 13, ?

(a) 15

(b) 17

(c) 19

(d) 21

**Answer:**(a) 15

Here a = 7, d = 3; t_{5} = ?

Now,

t_{5} = a + (5-1) d

= 7 + 4 X 2

= 7 + 8

= 15

## Practice Test Exam