Find the $12 t h$ from the end of the A.P. $- 2, - 4, - 6, . . . . . - 100$ .

Open in App

Solution

Given $A P$ is $- 2, - 4, - 6, . . . . . ., - 100$
$a = - 2 d = - 2$ and $l = - 100$
$t_{n} = a + (n - 1) d - 100 = - 2 + (n - 1) (- 2) \Rightarrow - 98 = (n - 1) (- 2) \Rightarrow 49 = n - 1 \Rightarrow n = 49 + 1 = 50$
$12 t h$ term from end $= (50 - 12 + 1) t h$ term from beginning $39 t h$ from beginning .
$t_{39} = a + (39 - 1) d = - 2 + 38 \times - 2 = - 2 + (- 76) = - 78$

Suggest Corrections

Similar questions

12^{t h}

A . P . - 2, - 4, - 6, . . . ., - 100

12^{t h}

2, 4, 6, . . ., 100.

12^{t h}

Find the $12^{t h}$ term from the end of the A.P $- 2, - 4, - 6, . . . - 100.$

12^{t h}

2, 6, 10 . . . . . . . . . 58

Join BYJU'S Learning Program