## ARITHMETIC PROGRESSION AND GEOMETRIC PROGRESSION FORMULAS

Arithmetic progression and geometric progression formulas :

On the webpage, we can find the formulas used in the topic arithmetic and geometric progression.

General form of arithmetic progression :

a , (a+d), (a+2d), (a+3d), ................

nth term or general term of the arithmetic sequence :

an = a+(n-1)d

here "n" stands for the required term.

Number of terms in the arithmetic sequence :

n = [(l- a)/d]+1

Common difference :

d = a₂ - a

In the above formula,

• "a" stands for the first term
• "d" stands for the common difference
• "l" stands for the last term
• and "n" stands for the total number of terms or required term.
• a₁ and a₂ are first and second term respectively.

Note:

Suppose if we want to find the 15th term of the given sequence, we need to apply n = 15 in the general term formula.

## Formula for geometric progression

General form of geometric progression :

a , ar, ar², .............

Common ratio :

r = a₂ / a

nth term or general term of the arithmetic sequence :

an = ar^(n-1)

In the above formula,

• "a" stands for the first term
• "r" stands for the common ratio
•  "n" stands for the required term
• a and a₂ are first and second term respectively.

## Arithmetic progression and geometric progression formulas - Examples

Question 1 :

Find the common difference and 15th term of an A.P 125 , 120 ,115 , 110 , ……….….

Solution :

First term (a) = 125

Common difference (d) = a2 – a1 ==> 120 – 125 ==>  -25

General term of an A.P (an) =  a + (n - 1) d

= 125 + (15 - 1) (-25) ==> 125 + 14 (-25) ==> 125 – 350

a₁₅   = -225

Therefore 15th term of A.P is -225

Question 2 :

Which term of the arithmetic sequence is 24  , 23 ¼ ,22 ½ , 21 ¾ , ………. Is 3?

Solution :

First term (a) = 24

Common difference = a2 – a1 ==> 23 ¼ – 24 ==> (93/4) – 24

d = -3/4

an =  a + (n - 1) d

Let us consider 3 as nth term

an = 3

3 = 24 + (n-1) (-3/4)

3 – 24 = (n-1)  (-3/4)

(-21 x 4)/(-3) = n -1 ==> 84/3 = n -1 ==> 28 = n – 1 ==> n=29

Hence,3 is the 29th term of A.P.

Question 3 :

The 10th and 18th terms of an A.P are 41 and 73 respectively. Find the 27th term and ap

Solution :

10th term = 41 ==>  a + 9 d = 41   ------- (1)

18th term = 73 ==> a + 17 d = 73   ------- (2)

Subtracting the second equation from first equation

aaaaaaaaaaaaaaaaa a + 17d = 73 aaaaaaaaaaaaaaaaaaaaaaaa

aaaaaaaaaa(-)aaaa aa + 9 d = 41 aaaaaaaaaaaaaaaaaaaaaaaa

aaaaaaaaaaaaaaaaa(-)a(-)aaa(-)aaaaaaaaaaaaaaaaaaaaaaaaa

aaaaaaaaaaaaaaaaa------------ aaaaaaaaaaaaaaaaaaaaaaaaa

aaaaaaaaaaaaaaaaa a8d = 32 aaaaaaaaaaaaaaaaaaaaaaaa

Substitute d = 4 in the first equation

a + 9 (4) = 41 ==> a + 36 = 41 ==> a = 5

Now, we have to find 27th term

an =  a + (n - 1) d

here n = 27

=  5 + (27-1) 4 ==> 5 + 26 (4) ==> 5 + 104 ==> 109

Hence, 27th term of the sequence is 109.

General form of ap:

a, (a+d), (a+2d),...............

5, (5+4), (5+8), ...............

5, 9, 13,....................

Let us see the next example on "Arithmetic progression and geometric progression formulas". Question 4 :

Find n so that the nth terms of the following two A.P’s are the same

1 , 7 ,13 ,19, ………………. and 100 , 95 , 90 ,………..

Solution :

an = a + (n - 1) d

nth term of the first sequence

a = 1   d = t₂-t₁ ==> 7-1 ==> d = 6

an = 1 + (n-1) 6 ==> 1 + 6 n – 6 ==> 6 n – 5   -----(1)

nth term of the second sequence

a = 100   d = a₂-a₁ ==> 95 - 100 ==> -5

an = 100 + (n-1) (-5) ==> 100 - 5 n + 5 ==> 105 - 5 n -----(2)

(1) = (2)

6 n – 5 = 105 – 5 n

6 n + 5 n = 105 + 5

11 n = 110 ==> 110/11 ==> 11

Hence, 11th terms of the given sequence are equal

Let us see the next example on "Arithmetic progression and geometric progression formulas".

Question 5 :

7, 13, 9,............................205

Solution :

First term a = 7,

common difference d = t2 - t1 = 13 - 7 = 6

l = 205

n = [(l-a)/d] + 1

n = [(205 - 7)/6] + 1

n = [198/6] + 1 ==> 33 + 1 ==> 34

Hence, total number of terms in the above sequence is 34

Question 6 :

Find the  10th term and the common ratio of the geometric sequence 1/4,-1/2,1,-2,............

Solution :

To find the 10th terms of the G.P we have to use the formula for general term that is

tn = a r^(n-1)

here a = 1/4   r = (-1/2)/(1/4) ==>(-1/2) x (4/1) = -2

n = 10

t₁₀ = (1/4) (-2)^(10-1)

=  (1/4) (-2)^9

= (1/4) (-512)

= -512/4

= -128

Question 7 :

If the 4th and 7th terms of a G.P are 54 and 1458 respectively, find the G.P

Solution :

4th term = 54

7th term = 1458

t₄ = 54

a r³ = 54   ----- (1)

t₇ = 1458

a r⁶ = 1458  ----- (2)

(2)/(1) = (a r⁶)/(a r³) = 1458/54

r³ = 27

r³ = 3³

r = 3

Substitute r = 3 in the first equation we get

a (3)³ = 54

a(27) = 54

a = 54/27

a = 2

The general form of G.P is   a, a r , a r ²,.........

= 2 ,2(3),2(3)²,..............

= 2,6,18,............

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