Question

Which term of the A.P. 3, 7, 11, 15,.… is 63?​

Answer 1

Answer:

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Answer 2

$Given \: 3,7,11,15,\ldots .,63\\is\: in \: A.P$

$First \:term (a= a_{1}) = 3$

$Common \: difference (d) \\= a_{2} - a_{1} \\= 7 - 3 \\= 4$

/* We know that , */

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

$Here , a_{n} = 63 \: (given)$

$\implies a + (n-1)d = 63$

$\implies 3 + (n-1)\times 4 = 63$

$\implies (n-1)\times 4 = 63 - 3$

$\implies (n-1)\times 4 = 60$

$\implies n-1= \frac{60}{4}$

$\implies n = 15 + 1$

$\implies n = 16$

Therefore.,

$\green{ 16^{th} \: term \: of \: given }$

$\green { A.P \: is \: 63 }$

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