Question

t18 of an AP 6,13,20,27

Answer 1

$d = t_{2} - t_{1} \\ d = 13 - 6 \\ d = 7$
Here
First term(a)=6
Common difference(d)=7

${n}^{th} \: \: term \: \: of \: \: an \: \: ap$
$t_{n} = a + (n - 1)d \\ t_{18} = 6 + (18 - 1) \times 7 \\ t_{18} = 6 + 119 \\ t_{18} = 125$

Answer 2

$\boxed{\boxed{\large\mathsf{125}}}$
___________________________________

$\underline{\Large\mathbf{ Given :}}$

$\mathsf{In \: the \: A.P \: 6, 13, 20, 27}$

$\mathsf{First \: term(a) = 6}$

$\mathsf{Common \: Difference (d)} \\\mathsf{ => t_2 - t_1 = 13 - 6 = 7}$

$\underline{\Large\mathbf{To \: Find :}}$

$\mathsf{18^{th} term \: of \: t_{18}}$

$\underline{\underline{\huge\mathfrak{Solution :}}}$

$\underline{\textsf{Using the Formula,}}$

$\boxed{\Large\mathsf{a_n = a + (n-1)d}}$

$\mathsf{=> t_{18} = a + (t-1)d}$

$\underline\text{Putting the Values in the Formula, We get,}$

$\mathsf{=> t_{18} = 6 + (18-1)7 }$

$\mathsf{=> t_{18} = 6 + 17(7)}$

$\mathsf{=> t_{18} = 6 + 119}$

$\boxed{\mathbf{=> t_{18} = 125}}$

Hence, the $\mathsf{18^{th}}$ term of A.P is $\mathsf{125.}$
____________________________________