Question

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

Answer 1

First term of an arithmetic sequence is 10 and its common difference 3. Write the first three terms of the sequence. Verify whether 100 is a term of this sequence.

Answer · 16 votes

First term of the sequence is 10 and common difference is 3. a1 = 10 and d = 3Next term = a2 = a1 + d = 10 + 3 = 13 a3 = a2 + d = 13 + 3 = 16 Thus, first three terms of the sequence are 10, 13 and 16.Let 100 be the nth term of the sequence. an = a1 + (n - 1)d 100 = 10 + (n - 1)3 90 = (n - 1)3 n - 1 = 30 n = 31 , which is a whole number.Therefore, 100 is the 31^st term of the sequence.

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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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