Question

Find the ${12}^{\text{th}}$ term from the end of the following arithmetic progressions:
(i)$3,5,7,9,.......201$
(ii)$3,8,13,.........253$
(iii)$1,4,7,10,......88$

Answer 1

(i) $a=201$ $d=-3+5=+2$

$a_n=a+(n-1)d$

$201=3+(n-1)2$

$99=n-1$

$n=100$

$12th$ sum from end $\Rightarrow a+88d=3+88\times 2=3+176=179$

(ii) $3,8,\mathrm{18.................253}$

$a=3$

$d=5$

$253=3+(n-1)5$

$\frac{250}{5}$= $n-1\Rightarrow n=51$

$12th$ term from the end $\Rightarrow$ $a+39d$

= $3+39\times 5=3+195=198$

(iii) $1,4,7,\mathrm{10............88}$

$a=1$, $d=3$

$88=a+(n-1)d=1+(n-1)3$

$n=\frac{87}{3}=29$

$12th$ sum from the end $=a+17d$

$=1+17\times 3=52$