**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 - a₁**** **

**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.

**General form of geometric progression :**

**a , ar, ar**², .............

**Common ratio :**

**r = a₂ / a₁**** **

**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
- a₁ and a₂ are first and second term respectively.

**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

aaaaaaaaaaaaaaaaaaaaad = 4aaaaaaaaaaaaaaaaaaaaaaaaa

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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