Question

Which term of the AP: 12 , 16 ,20 . . . . . is 248

Answer 1

Answer:

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Answer 2

$Given \:A.P : 12,16,20,\ldots$

$First \:term (a = a_{1} ) = 12$

$Common \: difference (d) = a_{2} - a_{1}$

$= 16 - 12$

$= 4$

$\boxed{ \pink{ n^{th} \:term (a_{n}) = a+(n-1)d }}$

$Here, a + (n-1)d = 248 \:( given )$

$\implies 12 + (n-1) \times 4 = 248$

/* Dividing each term by 4 , we get */

$\implies 3 + n - 1 = 62$

$\implies 2 + n = 62$

$\implies n = 62 - 2$

$\implies n = 60$

Therefore.,

$\green{ 60^{th}\:term \: in \: given \:A.P \:is \:248 }$

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