Question

the first term of an arithmetic sequence is 3 and common difference 2 how many times common difference should be added to its first term to its 10th term​

Answer 1

Step-by-step explanation:

given: a=3

d=2

an=a+(n-1)d

a10=3+(10-1)2

a10=3+(9)2

a10=3+18

a10=21.......

------>>>>I hope it will help.......u...

Answer 2

Given :

First term of an arithmetic sequence is 3 .

Common difference is 2 .

To Find :

10th term

Solution :

$\longmapsto\tt{First\:term\:(a)=3}$

$\longmapsto\tt{Common\:difference\:(d)=2}$

$\longmapsto\tt{No\:of\:terms\:(n)=10}$

Using Formula :

$\longmapsto\tt\boxed{{a}_{n}=a+(n-1)\times{d}}$

Putting Values :

$\longmapsto\tt{{a}_{10}=3+(10-1)\times{2}}$

$\longmapsto\tt{{a}_{10}=3+(9)\times{2}}$

$\longmapsto\tt{{a}_{10}=3+18}$

$\longmapsto\tt\bf{{a}_{10}=21}$

So , The 10th term of the sequence is 21 .

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$\tt{{a}_{n}=a+(n-1)\times{d}}$

$\tt{{s}_{n}=\dfrac{n}{2}[2a+(n-1)\times{d}}$

$\tt{{s}_{n}=\dfrac{n}{2}\:[a+l]}$

Here :

a = first term

d = common difference

n = number of terms

l = last term

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