Question

The nth term of 3,7,11,15,....is

Answer 1

Answer:

hope it helps .........

Answer 2

Question:

$The \: {n}^{th}\: term \: of \: 3,7,11,15,....is$

To find:

$To \: find \: the \: {n}^{th}\: term$

Answer:

${n}^{th} \: term \: of \: the \: given \:A.P. \: is \: \underline{ \: \underline{ \: \sf \pink{ 4n - 1} \: } \: }$

Step-by-step explanation:

$3,7,11,15,.....$

The given consecutive terms is in A.P.

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

$Common \: difference \: \underline{ (d) = t _{2} - t _{1}}$

$t _{1} = 3$

$t _{2} = 7$

$d = 7 - 3$

$\underline{ \: d = 4 \: }$

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

$= 3 + (n - 1)4$

$= 3 + 4n - 4$

$\boxed{t _{n} = 4n - 1}$

$\therefore {n}^{th} \: term \: of \: the \: given \:A.P. \: is \: \underline{ \: \bold{ 4n - 1} \: }$

Related formula:

$\bigstar \:no.of \: terms \: \underline{ n = \frac{l - a}{d }+1}$