# Tips and Tricks and Shortcuts for AP and GP And HP

## Tips and Tricks for AP and GP and HP

Here, are quick and easy Tips and Tricks for AP and GP and HP for you to help in AP, GP, and HP questions quickly, easily, and efficiently in competitive exams.

There are many Applications of AP and GP and HP as well as they are Very Important Topics from Examination Point of View.

## Tips and Tricks for AP and GP an HP

Here, are quick and easy Tips and Tricks for AP and GP and HP for you to help in AP, GP, and HP questions quickly, easily, and efficiently in competitive exams.

There are many Applications of AP and GP and HP as well as they are Very Important Topics from Examination Point of View.

### Tips and Tricks for AP

1. nth term of an AP = tn= a + (n – 1)d
2. Number of terms of an AP $(n) = [ \frac{(l-a)}{d} ]+ 1$
3. Sum of first n terms in an AP = Sn = $\frac{n}{2} [2a + (n − 1) d]$ OR  $\frac{n}{2} (a+l)$
4. While Solving three unknown Term in an A.P whose sum or product is given should be assumed as a-d, a, a+d.
5. While Solving Four terms in an A.P. whose sum or product is given should be assumed as a-3d, a-d, a+d, a+3d.

### Tips and Tricks for GP

1. nth term of an GP =  an= arn-1
2. Sum of first n terms in an GP = Sn = $= a \times \frac{r^{n} -1}{r-1}$ if r>1
3. Sum of first n terms in an GP = Sn = $= a \times \frac{1-r^{n}}{1-r}$ if r<1
4. Infinite term GP $\frac{a}{1-r}$

5. While Solving three unknown Term in an G.P whose sum or product is given should be assumed as $(\frac{a}{r}), a, ar$

### Tips and Tricks for HP

1. nth term of an HP$a_{n} = \mathbf{\frac{1}{(a + (n – 1)d)}}$
2. Harmonic Mean of two numbers a and b is $\mathbf{\frac{2ab}{a+b}}$
3. While Solving HP Questions we convert given Problem in an AP.

## Type 1: AP questions

### Question 1.

Find 10th term in the series 2, 5, 8, 11, 14……

Options:

1. 26
2. 29
3. 32
4. 27

#### Solution:

We know that,

tn = a + (n – 1)d
In the given series,

a = 2
d = 3 …. (5 – 2, 8 – 5…..)

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

t10 = 2 + (10 – 1)  3

t10 = 2 + 9 x 3

t10 = 29

Correct option: 2

### Question 2.

The sum of 3 numbers in arithmetic progression is 36 and product of their extreme is 80. Find the numbers.

Options:

1. 20 , 8 , 16
2. 4 , 12, 20
3. 12 , 20 , 28
4. None of these

Solution:

We know that,

tn = a + (n – 1)d

Assume the numbers as:

a-d, a, a+d

or a-d+a+a+d=36

or 3a=36

a=12

now (a-d) (a+d)=80

a²-d²=80

144-d²=80

d²=64

d=8

thus the numbers will be: 4, 12 and 20

Correct option: 2

## Type 2: GP question

### Question 1.

Find the number of terms in the series 5, 10, 20, . . ..320?

Options:

1. 5
2. 4
3. 6
4. 7

#### Solution:

We know that,

an= arn-1

a = 5
r = 2

an= 320
320 = 5 x 2n-1

64 = 2n-1

$2^{6}$ = 2n-1

n-1 = 6

n = 7

Correct Option. 4

Question 2.

The sum of three numbers is 14 and their product is 64. All the three numbers are in GP. Find all the three numbers when value of r is a whole number?

Options:

1.  3, 6, 4

2.  2, 4, 8

3.  2, 4, 6

4.  4, 4, 4

Solution: The three numbers can be written as  $(\frac{a}{r}), a, ar$

Sum of the three numbers =   $(\frac{a}{r}) + a + ar = 14$

Product of the three numbers = $(\frac{a}{r}) \times a \times ar = 64$

i.e.  a3 =64

Therefore, a = 4

Put the value of a in $\frac{a}{r} + a + ar = 14$

$a (\frac{ 1}{r} + 1 + r) = 14$$4(\frac{ 1}{r} + 1 + r) = 14$

2(1 + r + r2) = 7r

2r2 – 5r + 2 = 0

r = 2  or $\frac{1}{2}$

If r = 2, then numbers are 2, 4, 8.
If r = $\frac{1}{2}$ then numbers are 8, 4, 2

Correct Option. 2

## Type 3: HP questions

### Question 1.

Find the 15th term in the series $\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}$…….

Options:

1. 45
2. $\frac{1}{70}$
3. $\frac{1}{45}$
4. 70

#### Solution:

We know that,

$a_{n} = \frac{1}{a+(n-1)d}$

Convert the HP series in AP

We get 3, 6, 9, 12……
In the given series,

a = 3

d = 3…..(6 – 3)

Therefore, 15th term = $a_{15}$=  a + (n-1) d

$a_{15}$= 3 + (15– 1) 3

$a_{15}$ = 3 + 14 x 3

$a_{15}$ = 3 + 42

$a_{15}$ = 45

HP $a_{15}= \frac{1}{45}$

Correct option: 3

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