Question

find the nth of 13 ,8 ,3 -2, -7,....

Answer 1

Question :–

▪︎ Find the nth term of give series –

13 , 8 , 3 , -2 , -7, ......

ANSWER :–

$\\ \to { \green { \boxed { \bold { \: \: nth \: \: term(T_{n}) = 18 - 5n }}}}$

EXPLANATION :–

GIVEN :–

• A series =》 13 , 8 , 3 , -2 , -7 , .......

TO FIND :–

• nth term of given series .

SOLUTION :—

☞ Given series is an Arithmetic progression (A.P.) series , because it's Common difference(d) is constant .

• We know that nth term term of A.P. –

$\\ \implies{ \red{ \boxed { \bold { T_{n} = a + (n - 1) d}}}}$

• Here –

$\\ \: \: \: \: \: \: \: \: \: . \: \: \: \: { \bold { a = first \: \: term = 13 }}$

$\\ \: \: \: \: \: \: \: \: \: . \: \: \: \: { \bold { d = common \: \: difference = -5}}$

$\\ \: \: \: \: \: \: \: \: \: . \: \: \: \: { \bold { n = total \: \: number \: \: of \: \: terms = n}}$

$\\ \: \: \: \: \: \: \: \: \: . \: \: \: \: { \bold { T_{n} = n \th \: \: term}}$

• Now put the values –

$\\ \implies { \bold { T_{n} = 13 + (n - 1) ( - 5)}}$

$\\ \implies { \bold { T_{n} = 13 - 5n + 5 }}$

$\\ \implies { \red{ \boxed{ \bold { T_{n} = 18 - 5n }}}}$

Hence , nth term of A.P. ${ \underline { \bold { (T_{n}) = 18 - 5n }}}$

Answer 2

Given

First term (a) = 13

Common difference (d) = -5

We know that , the general formula or first n terms of an AP is given by

$\sf \star \: \: a_{n} = a + (n - 1)d$

Thus ,

$\sf \Rightarrow a_{n} = 13 + (n - 1)( - 5) \\ \\\sf \Rightarrow a_{n} =13 - 5n + 5 \\ \\\sf \Rightarrow a_{n} =18 - 5n$

$\therefore \sf \bold{ \underline{The \: nth \: term \: of \: given \: AP \: is \: 18 - 5n}}$