Question

In an AP:
(i) given $a=5,d=3,a_n=50$, find $nand{S}_n$.
(ii) given $a=7,a_{13}=35$, find $dand{S}_{13}$.
(iii) given $a_{12}=37,d=3$, find $aand{S}_{12}$.
(iv) given $a_{13}=15,S_{10}=125$, find $dand{a}_{10}$.

Answer 1

$\left(i\right)$

given that,

$a=5,d=3.a_n=50$

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

$=>50=5+(n-1)\times 3$

$=>n=\frac{48}{3}=16$

Now,

$s_n=\frac{n}{2}(2a+(n-1\left)d\right)$

$=>\frac{16}{2}(2\times 5+(16-1)\times 3)$

$=>8(10+45)$

$=>440$

$\left(ii\right)$

Given that,

$a=7,a_{(}13)=35$

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

$=>35=7+(13-1)\times d$

$=>d=\frac{7}{3}$

now

$s_n=\frac{n}{2}(a+l)$

$=>\frac{13}{2}(7+35)$

$=>\frac{13}{2}\times 42$

$=>273$

$\left(iii\right)$

given that

$a_{(}12)=37,d=3$

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

$=>37=a+(12-1)\times 3$

$=>a=4$

now

$s_n=\frac{n}{2}(a+l)$

$=>\frac{12}{2}(4+37)$

$=>6\times 41$

$=>246$

$\left(iv\right)$

Given that

$a_{(}13)=15,s_{(}10)=125$

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

$=>a_n=a_{(}13)=15$ and $n=3$

$=>15=a+2d$

$=>a=15-2d--------------\left(i\right)$

Now

$=>s_{(}10)=125$

$s_n=\frac{n}{2}(a+(n-1\left)d\right)$

$s_n=s_{(}10)=125,n=10$

$=>s_{(}10)=\frac{10}{2}(2a+(10-1)d)$

$=>125=5(2a+9d)$.

$=>\frac{125}{5}=2a+9d$

$=>25=2a+9d$

$=>a=\frac{25-9d}{2}$

from $\left(i\right)$ and $\left(ii\right)$ we get

$=>15-2d=\frac{25-9d}{2}$

$=>30-4d=25-9d$

$=>d=-1$

Now put the value of $d=-1$ in equation $\left(i\right)$ we get

$=>a=15-2d$

$=>a=15-2\times (-1)$

$a=17$

now we find $a_{(}10)$

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

$=>a_{(}10)=17+(10-1)\times (-1)$

$=>17+9\times (-1)$

$=>17-9$

$=>a_(10)=8$

Hence $d=-1$ and $a_{(}10)=8$