Question

The first term of an arithmetic sequence is 3 and common difference 2 a) What is the 101th term of this sequence​

Answer 1

Step-by-step explanation:

$\mathtt{\huge{ \colorbox{gray}{Given :-}}}$

$\: \: \: : \implies \: first \: term ,a= 3$

$\: \: \: : \implies \: common \: difference,d= 2$

$\mathtt{\huge{ \colorbox{pink}{To find :-}}}$

${10}^{th} \: term$

$\color{red} \underline{ \color{red} \bold{ \mathfrak{We \: know \: that,}}}$

$a_n = a + (n - 1)d$

$a_{10} = 3 + (10 - 1)2$

$a_{10} = 3 + 18$

$\boxed{a_{10} = 21}$

$\mathtt{Hence \: the \: {10}^{th} \: term \: is \: 21}$

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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