Numbers and Sequences

If the H.C.F of 65 and 117 is expressible in the form of 65m – 117 , then the value of m is

• 4

• 2

• 1

• 3

Answer explanation

H.C.F. of 65 and 117

117 = 65 × 1 + 52

65 = 52 × 1 + 13

52 = 13 × 4 + 0

∴ 13 is the H.C.F. of 65 and 117.

65m – 117 = 65 × 2 – 117

130 – 117 = 13

∴ m = 2

The least number that is divisible by all the numbers from 1 to 10

(both inclusive) is

• 2025

• 5220

• 5025

• 2520

Answer explanation

The first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P

• 4551

• 10091

• 7881

• 13531

Answer explanation

Tags

CCSS.HSF.BF.A.2

If 6 times of 6^th term of an A.P is equal to 7 times the 7^th term, then the 13^th term of the A.P. is ………..

• 0

• 6

• 7

• 13

Answer explanation

6 t₆ = 7 t₇

6(a + 5d) = 7 (a + 6d) ⇒ 6a + 30d = 7a + 42d

30 d – 42 d = 7a – 6a ⇒ -12d = a

t₁₃ = a + 12d (12d = -a)

= a – a = 0

Tags

CCSS.HSF.BF.A.2

An A.P consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P. is

• 16m

• 62m

• 31m

• 

Answer explanation

Tags

CCSS.HSF.BF.A.2

In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?

• 6

• 7

• 8

• 9

Answer explanation

A = 2⁶⁵ and

B = 2⁶⁴ + 2⁶³ + 2⁶² …. + 2⁰

 which of the following is true?

• B is 2⁶⁴ more than A

• A and B are equal

• B is larger than A by 1

• A is larger than B by 1

Answer explanation

A = 2⁶⁵

B = 2⁶⁴⁺⁶³ + 2⁶² + …….. + 2⁰

= 2

= 1 + 2² + 2² + ……. + 2⁶⁴

a = 1, r = 2, n = 65 it is in G.P.

S₆₅ = 1 (2⁶⁵ – 1) = 2⁶⁵ – 1

A = 2⁶⁵ is larger than B

Tags

CCSS.8.EE.A.3

The next term of the sequence  3/16, 1/8 , 1/12, 1/18,........is

• 1/24

• 1/27

• 2/3

• 1/81

Answer explanation

Tags

CCSS.HSF.BF.A.2

If the sequence t₁,t₂,t₃ … are in A.P. then the sequence t₆,t₁₂,t₁₈ … is

• a Geometric Progression

• an Arithmetic Progression

• neither an Arithmetic Progression nor a Geometric Progression

•  a constant sequence

Answer explanation

If t₁, t₂, t₃, … is 1, 2, 3, …

If t₆ = 6,

t₁₂ = 12,

t₁₈ = 18

then 6, 12, 18 … is an arithmetic progression

Tags

CCSS.HSF.BF.A.2

The value of (1³ + 2³ + 3³ + ……. + 15³) – (1 + 2 + 3 + …….. + 15) is …………….

• 14400

• 14200

• 14280

• 14520

Answer explanation

(15×16/2)²-⁽15x16)/2

=1202×120

=14280