# Tips And Tricks And Shortcut For Arithmetic Progressions

Here, are quick ans easy tips and tricks for you on Arithmetic Progression questions quickly, easily, and efficiently in competitive exams.

There are 4 types of questions asked in exams

## Type 1: Find nth term of series

### Question 1.

Find 10th term in the series 1, 3, 5, 7, …

Options:

1. 20
2. 19
3. 15
4. 21

#### Solution:

We know that,

tn = a + (n – 1)d
where tn = nth term, a= the first term , d= common difference, n = number of terms in the sequence

In the given series,

a (first term) = 1
d (common difference) = 2 …. (3 – 1, 5 – 3…..)

Therefore, 10th term = t10 = a + (n-1) d

t10 = 1 + (10 – 1) 2

t10 = 1 + 18

t10 = 19

#### Correct option: B ## Type 1: Find nth term of series

### Question 1.

Find 10th term in the series 1, 3, 5, 7, …

Options:

1. 20
2. 19
3. 15
4. 21

#### Solution:

We know that,

tn = a + (n – 1)d
where tn = nth term, a= the first term , d= common difference, n = number of terms in the sequence

In the given series,

a (first term) = 1
d (common difference) = 2 …. (3 – 1, 5 – 3…..)

Therefore, 10th term = t10 = a + (n-1) d

t10 = 1 + (10 – 1) 2

t10 = 1 + 18

t10 = 19

## Type 2: Find number of terms in the series

### Question 1.

Find the number of terms in the series 7, 11, 15, . . .71 Options:
1. 18
2. 19
3. 13
4. 17

#### Solution:

We know that, n=  [(l-a)/d]+ 1 where n = number of terms, a= the first term, l = last term, d= common difference In the given series, a (first term) = 7 l (last term) = 71 d (common difference) = 11 – 7 = 4 n=  [(71-7)/4]+ 1 n =  [64/4]+ 1 n = 16 + 1 n = 17

## Type 3: Find sum of first ‘n’ terms of the series

### Question 1.

Find the sum of the series 1, 3, 5, 7…. 201.

Options:

1. 18
2. 19
3. 13
4. 17

#### Solution:

We know that,

Sn = n/2[2a + (n − 1) d] OR n/2   (a+l)
where,

a = the first term,
d= common difference,
l = tn = nth term = a + (n-1)d

In the given series,

a = 1, d = 2, and l = 201

Since we know that, l = a + (n – 1) d

201 = 1 + (n – 1) 2

201 = 1 + 2n -2

202 = 2n = n = 101

n = 101

Sn =   n/2(a+l)

Sn =   101/2 (1+201)

Sn = 50.5 (1 + 201)

Sn = 50.5 * 202

Sn = 10201

## Type 4: Find the arithmetic mean of the series.

### Question 1.

Find the arithmetic mean of first five prime numbers.

Options:

1. 4.5
2. 6.5
3. 5.6
4. 6.4

#### Solution:

We know that
b =  1/2 (a + c)

Here, five prime numbers are 2, 3, 5, 7 and 11

Therefore, there arithmetic mean (AM) = (2+3+5+7+11)/5 = 5.6