Common difference of arithmetic progression or sequence is defined as the difference between the two consecutive terms in the arithmetic progression or sequence or series. It is constant across the whole series

if $a_1$,$a_2$ ,$a_3$.$a_4$ ….. are in arithmetic progression or sequence

then

$d=a_2 -a_1 = a_3 -a_2 =a_4 -a_3$….

If the common difference is not same across the series ,then series is not in Arithmetic progression. This is important condition for the series to be in A.P.

So common difference of the Arithmetic series can be found given the two consecutive terms in the A.P

**Examples**

Find the common difference of the below Arithmetic progression

a. $2 + \sqrt {3}$ , $2+ \sqrt {12}$ , $2+ \sqrt {27}$, ,$2 + \sqrt {48}$ ….

b. -10 , -7 , -4 ,-1 ,2…….

c. 2, 3.5 , 5, 6.5 , 8……..

d. 1 , 3/2 , 2 , 5/2 ……

**Solutions**

a.

Here $d= 2+ \sqrt {12} – 2 + \sqrt {3} = \sqrt {12} – \sqrt {3} = \sqrt {4 \times 3} – \sqrt {3} = 2 \sqrt {3} – \sqrt {3}= \sqrt {3}$

$d= 2+ \sqrt {27} – 2+ \sqrt {12} = 3 \sqrt {3} – 2 \sqrt {3}= \sqrt {3}$

So common difference is $\sqrt {3}$

b. d= -7 – (-10) = 3

d= -4 – (-7) = 3

…..

So common difference is 3

c. d= 3.5 -2 = 1.5

d= 5 -3.5 = 1.5

……

So common difference is 1.5

d. $d = \frac {3}{2} -1 =\frac {1}{2}$

$d= 2 – \frac {3}{2} = \frac {1}{2}$

……

So common difference is 1/2

**How to find the common difference if the nth term of the Arithmetic Progression is given**

Steps would be

a. First find the first term by putting n=1

b. Find the second term by putting n=2

Now common difference = Second term – First term

**Examples**

a. Nth term is given as

$T_n = 2n -1$

b. Nth term is given as

$T_n = 4n -3$

**Solutions**

a.

$T_n = 2n -1$

$T_1 = 2 \times 1 – 1 = 1$

$T_2 = 2 \times 2 -1 = 3$

Common difference = 3 -1 =2

b.

$T_n = 4n -3$

$T_1 = 4 \times 1 – 3 = 1$

$T_2 = 4 \times 2 – 3 = 5$

Common difference = 5 -1 = 4

**How to find the common difference if sum of the nth term of the Arithmetic Progression is given**

Steps would be

a. First sum up-to to first term by putting n=1 ($S_1$)

b. Find sum up-to to second term by putting n=2 ( $S_2$)

Now second Term= $S_2 – S_1$

Common difference = Second term – First term = $S_2 – S_1 – S_1 = S_2 – 2S_1$

**Examples**

$S_n = 3 n^2 + 5n $

**Solution**

$S_1 = 3 (1)^2 + 5 \times 1 = 8$

$S_2 = 3 (2)^2 + 5 \times 2 =22$

Now second term = $S_2 – S_1 = 22 -8=14$

Common difference = Second term – First term = $S_2 – S_1 – S_1$ = $22 – 2 \times 8=6$

**How to find the common difference if two terms are given in the Arithmetic Series**

Suppose we have 10th and 16th terms given as

$T_{10}= 20$

$T_{16}=32$

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

then using nth term formula

$a+ (10-1)d= 20$ or $a+9d =20$

$a+ (16-1)d =32$ or $a + 15 d=32$

Solving these we get a=2 ,d=2

I hope you find this article on common difference of arithmetic progression or sequence useful

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