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.
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 $\quad \Rightarrow a=8$
$\mathrm{t}_{29}=\mathrm{a}+28 \mathrm{~d}=8+28 \times 2=64$
$\mathrm{t}_{3}=12, \mathrm{t}_{50}$ (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 $\quad \Rightarrow a=8$
$\mathrm{t}_{29}=\mathrm{a}+28 \mathrm{~d}=8+28 \times 2=64$