Question

find the 29th term of 150,141,132,123,...

Answer 1

a=150,d=141-150=-9 and n=29
So,
an=a+(n-1)d
a29=150+(29-1)(-9)
=150+28×(-9)
=150-252
a29=-102

Hence the 29th term is -102.

Answer 2

Hey there !

Thanks for the question !

Here's the answer !

Given Sequence is an AP.

First Term ( a ) = 150

Common Difference = -9

We know that,

$General \ Form \ of \ any \ term \ in \ an \ AP \ is : \\\\ a_n = a + ( n - 1 ) d$

$=> a_{29} = a + ( 29 - 1 ) d \\\\ Substituting \ a = 150 \ and \ d = -9 \ we \ get, \\\\ => a_{29} = 150 + ( 28 ) ( -9 ) \\\\ a_{29} = 150 - 252 \\\\ a_{29} = -102$

$\boxed{ Hence \ the \ 29^{th} \ term \ is \ -102}$

Hope my answer helped !