Check whether - 150 is a term of the ap 17 12 7 2

Is -150 a term of AP 17,12,7,2

Posted by Alok Raj 3 years, 1 month ago

 The given term of arithmatic progression is17, 12, 7, 2, ...... Here, a = 17, d = 12 - 17 = - 5 where a is first term and d is common difference

Suppose an = -150

a + (n - 1)d = -150 {tex}\Rightarrow{/tex} 17 + (n - 1)(-5) = -150
{tex}\Rightarrow{/tex} (n - 1)(-5) = -150 - 17 {tex}\Rightarrow{/tex} (n - 1)(-5) = -167
{tex}\Rightarrow{/tex} n - 1 = {tex}\frac { 167 } { 5 }{/tex} {tex}\Rightarrow{/tex} n = {tex}\frac { 167 } { 5 }{/tex} + 1
{tex}\Rightarrow{/tex} n = {tex}\frac { 167 + 5 } { 5 } = \frac { 172 } { 5 } = 34 \frac { 2 } { 5 }{/tex}where n is not a whole number.

{tex}\therefore{/tex} -150 is not a term of the A.P.

-150 is not a term of the given A.P.Given,

First term \(= a = 17\)

Common difference \(= d = -5\)As we know,

For the \(n\)th term,

\({T_n} = a + (n - 1)d\)
\( - 150 = 17 + (n - 1)( - 5)\)
\((n - 1) = \frac{{ - 167}}{{ - 5}} = 33.4\)
\(n = 33.4 + 1 = 34.4\)
Hence \(n\) = 34.4 which is in decimal so -150 is not a term of A.P.

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