Numerical Ability Test - 31

Explanation:

Let's represent the three consecutive terms as a, a + d, and a + 2d, where a is the first term, and d is the common difference between the terms.

The sum of the three consecutive terms is 15.

We can write the equation:

a + (a + d) + (a + 2d) = 15

3a + 3d = 15

a = (15 - 3d) / 3

Let  a = 5 because if we substitute d = 0, then a = 5, and the three terms become 5, 5, and 5, whose sum is 15.

While taking a + d = 15, 20, 10 then This will not form Arithmetic Progression.

Hence, the middle term (a + d) = 5 + 0 = 5

Hence, the correct option is: 2) 5

Given:

10^th term of an A.P = 21

17^th term = 8 + 13^th term

Formula used:

A.P = a + (n - 1)d

Where, a = first term, d = difference and n = nth term

Calculation:

According to question,

10^th term = a + (10 - 1)d = a + 9d = 21   ----(1)

17^th term = a + 16d = 8 + a + 12d

⇒ 4d = 8

⇒ d = 2

So, from (1) we get

a = 21 - 9 x 2 = 3 

A.P. series = 3, 5, 7, 9, 11, .......

∴ The answer is 3, 5, 7, 9 ......

Given:

An A.P. series 4, 11, 18, ........

Formula used:

Where, a = first term, d = difference between consecutive numbers

n = n^th term

Calculation:

Here, S_n = 795 ,a = 4, d = 18 - 11 = 7

Given:

Numbers 2, 7, 12, 17, .......

Formula Used:

Given:

a_n = 40

a = 3

Concept used:

Arithmetic Progression

a_n = a + (n - 1)d,

where a = first term,

d = common difference,

n = number of terms

Calculation:

⇒ a_n = 7 + (n - 1)3

⇒ 40 = 7 + (n - 1)3

Solving for n,

⇒ 40 - 7 = 3n - 3

⇒ 33 + 3 = 3n

⇒ n = 36/3

⇒ n = 12

∴ The term that has a value of 40 is the 12th term in the sequence.

Given:

numbers exist between 100 and 1000 which are divisible by 7.

Formula Used:

General term(n^th term) of an AP ⇒ a_n = a + (n-1)d.

a_n = the n^th term in the sequence

a = the first term in the sequence

d = the common difference between terms

Calculation:

The numbers between 100 and 1000 that are divisible by 7 are,

105, 112, 119, ......, 994 which form an A. P with

First term a = 105

Common Difference d = 7

last term a_n = 994

⇒ a + (n-1)d = 994

⇒ 105 + (n-1)7 = 994

⇒ 7(n-1) = 994-105

⇒ 7n - 7 = 889

⇒ 7n = 896

⇒ n = 128.

∴ There are 128 numbers are there between 100 and 1000 which are divisible by 7.

Shortcut Trick

Number between 100 to 1000 

⇒ 1000/7 - 100/7

Mention the quotient after dividing

⇒ 142 - 14

⇒ 128

∴ There are 128 numbers are there between 100 and 1000 which are divisible by 7.

Formula used:

A n =  a + (n - 1)d

Where a = first term, n = no of terms and d = common difference

Calculation:

Here,

a = √2

d = 3√2 - √2 = 2√2

A 18  = √2 + (18 - 1)2√2

A 18  = √2 + 17 × 2√2

A 18  = 35√2

∴ The correct answer is 35√2.

Given:

a = ­-10,  a8 = 11

Formula used:

a_n = a + (n - 1)d

Where, a = first term, d = difference between consecutive terms

n = n^th term

Calculation:

a = -10

a₈ = a + (8 - 1)d = a + 7d

⇒ 11 = -10 + 7d

∴ The answer is 3 .

Given:

The common difference of an AP is 5.

Concept:

For an AP such that its first term is ' a ' and the common difference is ' d ' then,

a_n = a + (n -1)d

Solution:

According to the question,

The common difference ' d ' = 5

a18 = a + 17d

a13 = a + 12d

a18 - a13  = a + 17d - ( a + 12d )

a18 - a13  = 5d = 25

Hence, option 3 is correct.

Concept:  

A.P having Common ratio d = a_n - a_{n -1} 

The first term is a 

a_n  = a + (n - 1)d

Solution:

For the given A.P 

a = 1/p d = ((1 - p)/p) -(1/p) = -1 

4th term for the A.P = a + (4 -1) d = a + 3d = 1/p - 3 

Hence, The correct option is 4