Question

Question 16
Find the ${12}^{th}$ term from the end of the AP –2, –4, –6….. –100.

Answer 1

Given AP, - 2 , - 4 , - 6 ,……, - 100
Here, first term (a) = - 2, common difference (d) = - 4 – ( - 2 ) = - 2 and the last term (l) = - 100.
We know that, the nth term of an AP from the end is $a_n$ = l – ( n – 1)d, where l is the last term and d is the common difference.
$\therefore {12}^{th}$ term from the end,
$a_{12}$ = - 100 - (12 - 1)(-2)
= -100 + (11)(2) = -100 + 22 = -78
Hence, the ${12}^{th}$ term from the end is –78.