Question

find the 23rd term of AP whose 1st term is 6 and common difference is 2​

Answer 1

Step-by-step explanation:

ANSWER

We know that is a is first term and d is the common difference, then the arithmetic progression is

a,a+d,a+2d,a+3d,.....

Here, a=10 and d=3.

So, the arithmetic progression is 10,(10+3),(10+6),(10+9),(10+12)

i.e. 10,13,16,19,22,...

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Answer 2

AnswEr:

GivEn:

$\quad\bullet\normalsize\sf\ 1st \: term(a) \: = \: 6$

$\quad\bullet\normalsize\sf\ common \: difference(d) \: = \: 2$

To Find:

$\quad\bullet\normalsize\sf\ 23rd \: term \: of \: A.P.(a_{23})$

Solution:

$\underline{\bigstar\:\sf{According \: to \: given \: in \: Question :}}$

Using the $\normalsize\bf\ n^{th} \: formula \: of \: A.P.$;

$\normalsize\ : \implies\sf\ a_{n} = a + (n-1)d$

Here;

⋆ a = First term

⋆ n = No. of total terms

⋆ d = Common difference

Now, put n = 23 as we have to find 23rd term, Also put the other known values;

$\normalsize\ : \implies\sf\ a_{23} = 6 + (23-1)2$

$\normalsize\ : \implies\sf\ a_{23} = 6 + 22 \times\ 2$

$\normalsize\ : \implies\sf\ a_{23} = 6 + 44$

$\normalsize\ : \implies\sf\ a_{23} = 50$

$\normalsize\ : \implies{\underline{\boxed{\mathsf \pink{ a_{23} = 50}}}}$