An AP consists of 50 terms of which 3rd term is 12 and the last term is 106.

Question.

An AP consists of 50 terms of which $3 \mathrm{rd}$ term is 12 and the last term is 106 . Find the 29 th term.


Solution:

$\mathrm{t}_{3}=12, \mathrm{t}_{50}$ (last term) $=106$

$\Rightarrow a+2 d=12$ ..(1)

and $a+49 d=106$ ...(2)

Subtracting (1) from (2), we get

$47 \mathrm{~d}=106-12=94 \Rightarrow \mathrm{d}=2$

From (1), a + 2 × 2 =12 $\quad \Rightarrow a=8$

$\mathrm{t}_{29}=\mathrm{a}+28 \mathrm{~d}=8+28 \times 2=64$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now