Question

The nth term of an A.P. is 6n + 2. Find the common difference.

Answer 1

Answer:

The Common Difference is 6.

Step-by-step explanation:

Given :  

nth term of an A.P , an = 6n + 2……….(1)

On Putting n = 1,2 ,3 & 4 in eq 1,

n = 1

a1 = 6 × 1 + 2

a1 = 8

 

n = 2

a2 = 6 × 2 + 2  

a2 = 14

 

n = 3

a3 = 6 × 3 + 2

a3  = 20

 

n = 4  

a4 = 6 × 4 + 2

a4 = 24 + 2  

a4 = 26

A.P  8, 14, 20 , 24 ...........

Common Difference , d = a2 - a1

d = 14 - 8

d = 6

Hence, the Common Difference is 6.

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Answer 2

${\mathfrak{\red{\underline{\underline{Answer:-}}}}}$

$\sf{Common\;difference = 6}$

${\mathfrak{\red{\underline{\underline{Explanation:-}}}}}$

Given:-

$\sf{a_{n}=6n+2}$

$\sf{Therefore,}$

$\sf{a_{1}=6\times 1+2 = 8}$

$\sf{a_{2}=6\times 2+2 = 14}$

$\sf{a_{3}=6\times 3+2 = 20}$

$\sf{So,\;here\;a.p.\;is:8,\;14,\;20,.....}$

$\sf{Common\;difference(d)=a_{2}-a_{1}}$

$\sf{=14-8}$

$\sf{=6}$

${\boxed{\boxed{\bf{\red{So,\;the\;common\;difference\;is\;6}}}}}$

${\mathfrak{\red{\underline{\underline{Some\;important\;formulas\;related\;to\;chapter:-}}}}}$

$\sf{1).\;a_{n}=a+(n-1)d}$

$\sf{2).\;s_{n}=\dfrac{n}{2}[2a+(n-1)d]}$

$\sf{3).\;s_{n}=\dfrac{n}{2}[a+l]}$

$\sf{4).\;d=a_{2}-a_{1}}$

$\sf{Where,}$

$\sf{l=last\;term}$

$\sf{d=common\;difference}$

$\sf{a\;or\;a_{1}=first\;term}$

$\sf{s=sum\;of\;n\;terms}$

$\sf{a_{n}=number\;of\;terms}$