Question

how many term are there in the arithmetic progression 201,208,215,.....369 ?

Answer 1

Answer:

25

Step-by-step explanation:

Given : 201,208,215,.....369

To Find: How many term are there in the arithmetic progression?

Solution:

A.P.: 201,208,215,.....369

First term = 201

Common difference = d = 208-201=215-208=7

Last term =$a_n=369$

Formula of nth term = $a_n=a+(n-1)d$

where a is the first term

d is the common difference

Substitute the value in the formula.

$369=201+(n-1)(7)$

$369=201+7n-7$

$369=194+7n$

$369-194=7n$

$175=7n$

$\frac{175}{7}=n$

$25=n$

Thus there are 25 terms in a given A.P.

Answer 2

Answer:

$25 \:terms \: are \:there \:in \:given \: A.P$

Step-by-step explanation:

Given A.P:

201,208,215,....,369

$First\:term (a)=201,\\common\: difference (d)=a_{2}-a_{1}\\=208-201\\=7\\Let \: n^{th}\:term (a_{n})=369$

$a+(n-1)d=369$

$\implies 201+(n-1)7=369$

$\implies (n-1)7=369-201$

$\implies (n-1)7=168$

$\implies n-1=\frac{168}{7}$

$\implies n-1=24$

$\implies n = 24+1$

$\implies n = 25$

Therefore,

$25 \:terms \: are \:there \:in \:given \: A.P$

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