Explore topic-wise InterviewSolutions in .

Concept:

• Common ratio = r = \(\frac{{{{\rm{a}}_2}}}{{{{\rm{a}}_1}}} = \frac{{{{\rm{a}}_3}}}{{{{\rm{a}}_2}}} = \ldots = \frac{{{{\rm{a}}_{\rm{n}}}}}{{{{\rm{a}}_{{\rm{n}} - 1}}}}\)
• nth  term of the G.P. is an = arn−1
• Sum of n terms of GP = sn = \(\frac{{{\rm{a\;}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - {\rm{\;}}1}}\); where r >1
• Sum of n terms of GP = sn = \(\frac{{{\rm{a\;}}\left( {1 - {\rm{\;}}{{\rm{r}}^{\rm{n}}}} \right)}}{{1 - {\rm{\;r}}}}\); where r <1
• Sum of infinite GP = \({{\rm{s}}_\infty } = {\rm{\;}}\frac{{\rm{a}}}{{1{\rm{\;}} - {\rm{\;r}}}}{\rm{\;}}\) ; |r| < 1

Calculation:

Given series is 5, 10, 20, ...

Here, a = 5, r = 2

Sum of n numbers = sn = 1275

As we know that, Sum of n terms of GP = sn = \(\frac{{{\rm{a\;}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - {\rm{\;}}1}}\); where r >1

sn = \(\frac{{{\rm{5\;}}\left( {{{\rm{2}}^{\rm{n}}} - 1} \right)}}{{{\rm{2}} - {\rm{\;}}1}}\)

1275 = 5 × (2n - 1)

⇒ 255 = (2n - 1)

⇒ 2n = 256

⇒ 2n = 28

⇒ n = 8

Thus the correct answer is 8.

Calculation:

1 × 1 = 1

11 × 11 = 121

111 × 111 = 12321

1111 × 1111 = 1234321

∴ 11111 × 11111 = 123454321

Calculation:

(2n + 1) + (2n + 3) + (2n + 5) + ...... + (2n + 47) = 5280 

(2n + 5) – (2n + 3) = 2

(2n + 3) – (2n + 1) = 2

On the Left Hand Side, (2n + 1) + (2n + 3) + (2n + 5) + ...... + (2n + 47) is an arithmetic progression with common difference 2 

First term of the arithmetic progression is (2n + 1) and the last term is (2n + 47)

Let the number of terms be T 

(2n + 47) = (2n + 1) + (T – 1)2

⇒ 2T – 2 = 2n + 47 – 2n – 1

⇒ 2T = 46 + 2

⇒ 2T = 48

⇒ T = 24

Number of terms of the arithmetic progression = 24

Sum of arithmetic progression = 24/2[(2n + 1) + (2n + 1) + (24 – 1)2]

⇒ 24/2[(2n + 1) + (2n + 1) + (24 – 1)2] = 5280

⇒ 24/2[4n + 2 + 23 × 2] = 5280

⇒ 24/2[4n + 48] = 5280

⇒ 48[n + 12] = 5280

⇒ n + 12 = 110

⇒ n = 110 – 12

⇒ n = 98

The value of 1 + 2 + 3 + .... + n =  1 + 2 + 3 + .... + 98

⇒ 98(98 + 1)/2

⇒ 49 × 99

⇒ 4851

∴ The value of 1 + 2 + 3 + .... + n is 4851

Formula Used:

nth term of an arithmetic progression = a + (n – 1)d

Here, n = Number of terms , a = First term, d = Common difference (Difference between two consecutive terms)

Sum of an arithmetic progression = n/2[a + a + (n – 1)d]

Sum of first n natural numbers = n(n + 1)/2

Given:

The initial speed of the car = 60 km/h

Increase in speed per hour = 5 km/h

Distance to be covered = 435 km

Formula used:

Sum of n-terms of an A.P is given by

S_n = (n/2) [2a + (n - 1)d]

where n = number of terms, a = 1^st term, d = common difference

Calculation:

Let n be the time taken by the car to cover 435 km

The initial speed of the car = 60 km/h

Speed for next hour = 65 km/h and so on

According to the question,

The distance covered by the car in 1st hour is 60km

The distance covered by the car in 2nd hour is 65km and so on

60 + 65 + ............ = 435

Here S_n = 435, a = 60 and 

d = 65 - 60

⇒ 5

Now,

435 = (n/2) [2 × 60 + (n - 1) × 5]

⇒ 870 = 5n [24 + n - 1]

⇒ 174 = n [23 + n] 

⇒ n [n + 23] = 6 × 29

⇒ n [n + 23] = 6 × [6 + 23]

⇒ n = 6

∴ The time taken by the car to cover 435 km is 6 hour.

Given:

0, 6, 24, 60, 120, ?, 336

Calculations:

It is double line difference

→ 6 - 0 = 6

→ 24 - 6 = 18

→ 60 - 24 = 36 

→120 - 60 = 60

→ x - 120 = 

→ 336 - x = 

Next difference in series of answer we got in first difference

→ 18 - 6 = 6

→ 36 -18 = 18 

→ 60 - 36 = 24

In the above series difference is 6 = 30 - 24

→ y - 60 = 30

y = 30 + 60 = 90

→ 90 - z = 36

⇒ z = 90 + 36

⇒ z = 126

So series   

 0          6         24         60       120              ?          336

      + 6    +18       +36      +60        + 90          + 126

             +12       +18     +24        +30         + 36

                    + 6        + 6       + 6         +6

120 + 90 = 210 or

336 - 126 = 210

∴ The missing number is 210

Calculation

We can see the following trend in the series : 

40 × 3 = 120

120 ÷ 2 = 60

60 × 3 = 180

180 ÷ 2 = 90

So, the next term will be 90 × 3 = 270

270 ÷ 2 = 135

∴ The missing term is 270

Calculation

The series is 2, 7, 24, 77, ?, 723

Its terms can be written as :

2 × 3 + 1 = 7

7 × 3 + 3 = 24

24 × 3 + 5 = 77

So, the next term will be

77 × 3 + 7 = 238

And last term is 

238 × 3 + 9 = 723

∴ The required term is 238

Given:

x , 2x + 2 & 3x + 3 are in G.P

Concept used:

G.P is the equence of numbers in which the ratio between consecutive terms is constant.

Calculations:

x, 2x + 2 & 3x + 3 are in G.P

So, 2nd/1st = ( 2x + 2)/2

⇒ 2nd/3rd = (3x+3)/(2x+2)

Now,

For, 11x, 22x + 22 , 33x + 33

(22x+22)/11x = (2x+2)/x

(33x+33)/(22x+22) = (3x+3)/(2x+2)

The ratio for both the series is coming the same, so both the series are in G.P.

So, 11x, 22x+22 & 33x+33 series is in G.P.

∴ The correct choice is option 2.

Given :

a, b & c are in A.P

Concept used :

If the difference between the number of a group is the same then they are in A.P.

If the sequence of numbers in which the ratio between consecutive terms is constant then they are in G.P.

Calculations :

a, b & c are in A.P

Let a, b, c be 1, 2 & 3

So,

⇒ 3^a = 3 , 3^b = 3² = 9, 3^c = 3³ = 27

3^c/3^b = 3, 3^b/3^a = 3

So this is a G.P as ratio is constant.

∴ The correct choice is option 2.

Given: 

The 31st term of an AP remains the same even if the common difference decreases by 9.

Formula Used:

n-th term of an AP = a + (n - 1)d, where a is the first term, n is the total number of terms and d is the common difference.

Calculation:

Let the initial first term be a₁ and the required first term be a₂.

Let d be the common difference.

Now, a₁ + 30d = a₂ + {30 × (d - 9)}

 ⇒ a₁ + 30d = a₂ + 30d – 270

⇒ a₁ = a₂ – 270

∴ The first term will be increased by 270.

Calculation

1, 4, 14, 45, 139, .....

These terms can be written as :

4 = (1 × 3) + 1

14 = (4 × 3) + 2

45 = (14 × 3) + 3

139 = (45 × 3) + 4

∴ The next term will be

(139 × 3) + 5 = 422

Formula used:

sum of natural numbers = n(n + 1)/2

Calculations:

The sum of natural numbers 1 to 100 

n = 100 

100(100 + 1)/2 = 5050

∴ The sum of all natural numbers from 1 to 100 is 5050

Given

Numbers from 41 to 80

All number between 41 to 80 should be even number

Formula used

T_n = a + (n-1)d

S_n = \(\frac{n}{2}\left\{ {2A + \left( {N - 1} \right)D} \right\}\)

Where, T_n = last number in the series

S_n = Sum of series of AP

N = number of terms in series

A = first number of the series

D = difference of consecutive series

Calculation:

Even number between 41 to 80

⇒ 42, 44, 46, 48, 50 ..............80

⇒ This series is in AP

Where, a = 42

⇒ d = a₂ - a₁

_⇒ 44 - 42 = 2

T_n = a + (n-1)d

⇒ 80 = 42 + (n-1)2

⇒ 38 = (n-1)2

⇒ n = 20

⇒ S_n  = \(\frac{{20}}{2}\left\{ {2 × 42 + \left( {20 - 1} \right)2} \right\}\)

⇒ 20{42 + 19}

⇒ 20 × 61

∴ 1220

Given:

The first term as well as the common difference of the second AP each is one more than the 1^st term and the common difference of the 1^st AP. The fourth term of the second AP is 15 and 5^th term of the 1^st AP is 13.

Formula used:

nth term of an AP, T_n = a + (n - 1)d, where ‘a’ is the first term, ‘n’ is the total number of terms and ‘d’ is the common difference.

Calculation:

Let the first term and common difference of the first AP is a1 and d1 respectively.

Let the first term and common difference of the second AP is a2 and d2 respectively.

From the question

a2 – a1 = 1     ----(1)

d2 – d1 = 1     ----(2)

Fourth term of second AP = a2 + 3d2 = 15      ----(3)

Fifth term of first AP = a1 + 4d1 = 13      ----(4)

From equation (1) (2) and (3) by putting values we get,

a1 + 1 + 3(d1 + 1) = 15

⇒ a1 + 3d1 = 11      ----(5)

Subtract equation (5) from (4)

d1 = 2

And a1 = 5

By putting the value of a1 and d1 in equation (1) and (2) we get,

a2 = 6 and d2 = 3

Second term of first AP = a1 + d1 = 7

Second term of second AP = a2 + d2 = 9

Average of both the terms = (9 + 7)/2 = 8

Thus the correct answer is 8.

Given:

The difference between the 6th term and 3rd term of an AP is equal to the second term of the AP.

The sum of the first and the 10th term is 52.

Formula Used:

n-th term of an AP = a + (n - 1)d, where a is the first term, n is the total number of terms and d is the common difference.

Calculation:

Let the first term and common difference be a and d respectively.

(a + 5d) – (a + 2d) = a + d

⇒ 3d = a + d

⇒ a = 2d      ----(1)

Now, a + (a + 9d) = 52

2a + 9d = 52

⇒ 4d + 9d = 52 [Using (1)]

⇒ 13 × d  = 52

⇒ d = 4

So, a = 2 × 4 = 8

9th term = a + 8d = 8 + 8 × 4 = 8 + 32 = 40

∴ 9th term is 40

Given:

How many numbers are divisible by 13 from 100 to 300

Calculation:

⇒ Among the numbers 100 to 300 the first number is divisible by 13 is 104 and the last number is 299

⇒ It is an arithmetic progression and we can find the terms that are multiples of 13 as

⇒ 299 = a + (n - 1)d

⇒ 299 = 104 + (n - 1) × 13

⇒ n = 16

∴ The required result will be 16.

Given: 

The ratio of the second term to the third term is 8 : 11.

The common difference is 3.

Formula Used:
n-th term of an AP = a + (n - 1)d, where a is the first term, n is the total number of terms and d is the common difference.

Calculation:

Let the first term be a and the common difference be d

Now, (a + d)/(a + 2d) = 8/11

⇒ 11a + 11d = 8a + 16d

⇒ 3a = 5d

⇒ a = 5d/3

⇒ (5 × 3)/3

⇒ 5

∴ The first term is 5

Given :

Number be divisible by 8

Common difference (d) = 8

To find :

the sum of all numbers divisible by 8 between 100 - 300

Formula used :

S_n = (n/2) × (a + a_n)

a_n = a + (n - 1)d

Where,

S_n is the sum of all terms

n is the number of terms

a is the first term

a_n is last term

d is the common difference

Calculation :

The first number (a) divisible by 8 is 104

⇒ The last number (a_n) divisible by 8 is 296

⇒ 296 = 104 + (n - 1) × 8

⇒ (n - 1) × 8 = 296 - 104

⇒ n - 1 = 192/8

⇒ n = 24 + 1

⇒ n = 25

⇒ S_n = (25/2) × (104 + 296)

⇒ S_n = 5000

∴ the sum of all numbers divisible by 8 between 100 - 300 is 5000.

Given :

6 + 8 + 10 +12 + 14 .........       + 40 

Formula used :

Sum of first n even natural numbers = n (n + 1)

Calculations :

Add and subtract 6 in the series

6 + 6 + 8 + 10 + 12 + 14 .................. + 40 - 6 

2 + 4 + 6 + 8 + 10 + 12 ...................+ 40 - 6     (6 = 2 + 4)

Total number of terms from 2 to 40

n = (40 - 2)/2 + (1) = 20 (n is the total number of term, 40 is the last term and 2 is the common difference in the series)

Total number of terms(n) = 20

So,

2 + 4 + 6 + 8 + 10 ...................... 40 = 20 (20 + 1) 

⇒ 420 

2 + 4 + 6 + 8 + 10..................+ 40 - 6 = 420 - 6 

⇒ 414 

∴ The sum of the series will be 414 

GIVEN:

A number between 11 and 90 that is divisible by 7 are 14, 21, 28 .............84

FORMULA USED:

Last term(l) = a + (n - 1)d, where a = 1st term, n = Number of terms and d = Common difference(difference between two consecutive term)

CALCULATION:

All the numbers are in the Arithmetic series

⇒ a = 14, d = (14 - 7) = 7 and l = 84

⇒ Last term(l) = a + (n - 1)d

⇒ 84 = 14 + (n - 1)7

⇒ 84 - 14 = (n - 1)7

⇒ 70 = (n - 1)7

⇒ 10 = (n -1)

⇒ 10 + 1 = n

⇒ 11 = n

∴ There are total 11 numbers that is divisbile by 7 between 11 and 90

Given:

The average of three consecutive terms starting from the third term in an AP is 18.

Formula:

nth term of an AP, T_n = a + (n - 1)d, where ‘a’ is the first term, ‘n’ is the total number of terms and ‘d’ is the common difference.

Calculation:

Let the first term is ‘a’ and common difference is ‘d’.

From the question

(T₃ + T₄ + T₅)/3 = 18

⇒ a + 2d + a + 3d + a + 4d = 18 × 3

⇒ 3a + 9d = 54

⇒ a + 3d = 18     ----(1)

The average of five consecutive terms of the same AP starting from the second term is

= (T₂ + T₃ + T₄ + T₅ + T₆)/5

⇒ (a + d + a + 2d + a + 3d + a + 4d + a + 5d)/5

⇒ (5a + 15d)/5

⇒ a + 3d = 18 [from equation (1)]

∴ required answer is 18.

Given: The sum of the first term and the last term is 51. The second last term is 42.

Formula:

Nth term of an AP, T_n = a + (n - 1)d, where ‘a’ is the first term, ‘n’ is the total number of terms and ‘d’ is the common difference.

Calculation:

Let the first term is ‘a’ and common difference is ‘d’.

From the question

a + T_n = 51

⇒ a + a + (n – 1)d = 51

⇒ 2a + (n – 1)d = 51     -----(1)

Second last term = 42

T_{(n- 1 )} = 42

⇒ a + (n – 2)d = 42     ----(2)

Subtracting equation (2) from (1) we get

a + (n – 1)d – (n – 2)d = 9

⇒ a + nd –d –nd + 2d = 9

⇒ a + d = 9

Second term = a + d = 9

∴ required answer is 9.

Given:

The sum of the first 12 terms of an AP is 264. The sum of the next 12 terms of the same AP is 792.

Formula:

Sum of n terms = n × middle term = n(first term + last term)/2

Calculation:

Let the first term is ‘a’ and common difference is ‘d’.

From the question

Sum of first 12 terms = 264

⇒ [(T₁ + T₁₂)/2] × 12 = 264

⇒ (a + a + 11d) × 6 = 264

⇒ 2a + 11d = 44      ----(1)

Sum of next 12 terms = 792

⇒ [(T13 + T24)/2] × 12 = 792

⇒ (a + 12d + a + 23d) × 6 = 792

⇒ 2a + 35d = 132     ----(2)

Subtract equation (1) from (2)

24d = 88

⇒ d = 11/3

The difference between the third and the second term = (a + 2d) – (a + d)

⇒ d

⇒ d = 11/3

Thus the correct answer is 11/3.

Given:

The 7th term of an AP is 71. The third term is 43.

Formula used:

Nth term of an AP, T_n = a + (n - 1)d, where ‘a’ is the first term, ‘n’ is the total number of terms and ‘d’ is the common difference.

Calculation:

Let the first term is ‘a’ and common difference is‘d’.

T₇ = a + (7 – 1)d

⇒ 71 = a + 6d      ----(1)

T₃ = a + (3 – 1)d

⇒ 43 = a + 2d      ----(2)

From equation from (1) and (2)

71 – 6d = 43 – 2d

⇒ 4d = 28

⇒ d = 7

And a = 29

Fourth term, T₄ = a + 3d

⇒ 29 + 3 × 7 = 50

∴ required answer is 50.

Given:

Sum of three middle terms = 726

Sum of last three terms = 1320

Formula used:

a_n = a + (n – 1)d

Concept:

Using the concept of linear equation we can find the value of 17^th term.

Calculation:

Middle terms of the given AP must be 10^th, 11^th and 12^th

As the numbers will of same difference

∴ Average of these terms (i.e., 11^th term) = 726/3 = 242   ----(i)

Similarly, last three terms must be 19^th, 20^th and 21^st

 ∴ Average of last three terms (i.e., 20^th term) = 1320/3 = 440   ----(ii)

Now, a₁₁ = a + 10 d and a₂₀ = a + 19d

⇒ 242 = a + 10d [From (i)]   ----(iii)

⇒ 440 = a + 19d [From (ii)]   ----(iv)

By using elimination method of linear equation in two variables, solve equations (iii) and (iv)

Then, we get d = 22 and put the value of d in equation (iii) then we get a = 22

Now, we have the value of a and d then we can easily find the 17^th term

⇒ a₁₇ = a + 16d

⇒ a₁₇ = 22 + 16 × 22

⇒ a₁₇ = 374

∴ 17^th term is 374

Given: The sum of the first ten terms of an AP is 180 and that of the last ten is 960. The AP has 21 terms.

Formula used: In an AP, The middle term is equal to the average of the first and the last term.

Middle term = (a + l)/2

Sum of AP = S_n = n/2[2a + (n – 1)d] OR S_n = [(a + l)/2]n . where ‘l’ is the last term

Calculation:

Let the first term is ‘a’ and common difference is ‘d’ and the middle term is ‘x’.

According to question

Sum of first 10 terms + x + sum of last 10 terms = sum of 21 terms

⇒ 180 + x + 960 = {(a + l )/2} × 21

⇒ 1140 + x = 21x

⇒ 20x = 1140

x = 57

∴ required answer is 57

Given:

The series is:

1, 2, 4, 7, *, 16, 22?

Calculation:

The series follows the pattern:

1 + 1 = 2

2 + 2 = 4

4 + 3 = 7

7 + 4 = 11

11 + 5 = 16

16 + 6 = 22