Question

Decide whether the following sequence is an A.P. or not. if so, find the 20th term of the progression :-
-12, -5, 2, 9, 16, 23, 30, ...​

Answer 1

$\fbox{\sf \huge \underline{solution}} \huge\rightarrow$

$\sf{here,a = t_1 = - 12,t_2 = - 5,t_3 = 2,t_4 = 9,t_5 = 16, \: ...}$

$\sf \implies{t_2 - t_1 = - 5 - ( - 12) = - 5 + 12 = 7</p><p></p><p>,}$

$\sf \implies{t_3 - t_2 = - 2 - ( - 5) = 2 + 5 = 7</p><p></p><p>, \: }$

$\sf \implies{t_4 - t_3 = 9 - 2 = 7</p><p></p><p>, \: }$

$\sf \implies{t_5 - t_4 = 16 - 9 = 7}$

$\sf{the \: common \: difference = d = 7} \\ \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(a \: constant \: number)}$

$\sf \therefore{the \: given \: sequence \: is \: an \: a.p.}$

$\sf \implies{t_n = a + (n - 1)d \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(formula)}$

$\sf{t_20 -12 + (20 - 1) \times 7 } \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(substituting \: the \: values)}$

$\sf \implies{ - 12 + 19 \times 7}$

$\sf \implies{ - 12 + 133}$

$\fbox{ \sf \implies\underline{t_20 = 121}}$

$\sf{ans \mapsto \: the \: given \: sequence \: is \: an \: a.p.} \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: the \: 20th \: term \: of \: the \: a.p.is \: 121.}$

Answer 2

Answer:

The 20th term of the progression is 121.