Question

In an AP

(i) Given $a=5,d=3,a_n=50$, find $n$ and $S_n$.

(ii) Given $a=7,a_{13}=35$, find $d$ and $S_{13}$.

(iii) Given $a_{12}=37,d=3$, find $a$ and $S_{12}$.

(iv) Given $a_3=15,S_{10}=125$, find $d$ and $a_{10}$.

(v) Given $d=5,S_9=75$, find $a$ and $a_9$.

(vi) Given $a=2,d=8,S_n=90$, find $n$ and $a_n$.

(vii) Given $a=8,a_n=62,S_n=210$, find $n$ and $d$.

(viii) Given $a_n=4,d=2,S_n=-14$, find $n$ and $a$.

(ix) Given $a=3,n=8,S_8=192$, find $d$.

(x) Given $l=28,S_n=144$ and there are total $9$ terms. Find $a$.

Answer 1

General Formula to be used:

1) $a_n=a+(n-1)d$ 2) $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ 3) $S_n=\frac{n}{2}[a+l]$

Now,

(i) $a=5,d=3,a_n=50$

$\Rightarrow a_n=a+(n-1)d=50$ [From formula (1)]

$\Rightarrow a_n=5+(n-1)\left(3\right)=50$

$\Rightarrow 5+3n-3=50$

$\Rightarrow n=\frac{50-2}{3}=16$

$\therefore S_{16}=\frac{16}{2}\left[2\right(5)+(16-1\left)\right(3\left)\right]$ [From formula (2)]

$\Rightarrow S_{16}=8[10+45]=440$

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

$\Rightarrow a_{13}=7+(13-1)d=35$ [From formula (1)]

$\Rightarrow 7+12d=35$

$\Rightarrow d=\frac{35-7}{12}=\frac{7}{3}$

$\therefore S_{13}=\frac{13}{2}\left[2\right(7)+(13-1\left)\frac{7}{3}\right]$ [From formula (2)]

$=\frac{13}{2}\times 2[7+6\times \frac{7}{3}]$

$\Rightarrow S_{13}=13[7+14]=273$

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

$\Rightarrow a_{12}=a+(12-1)3=37$ where $n=12$ [From formula (1)]

$\Rightarrow a+33=37$

$\Rightarrow a=4$

$\therefore S_{12}=\frac{12}{2}\left[2\right(4)+(12-1\left)3\right]$ [From formula (2)]

$\Rightarrow S_{12}=6[8+33]=246$

(iv) $a_3=15,S_{10}=125$

$\Rightarrow a_3=a+(3-1)d=15$ [From formula (1)]

$\Rightarrow a+2d=15$

$\Rightarrow a=15-2d$...(a)

Also,

$\Rightarrow S_{10}=\frac{10}{2}[2a+(10-1\left)d\right]=125$ [From formula (2)]

$\Rightarrow 5\left[2\right(15-2d)+9d]=125$ [From (a)]

$\Rightarrow 30-4d+9d=\frac{125}{5}$

$\Rightarrow 5d=25-30$

$\Rightarrow d=-1$

$\therefore a=15-2(-1)=17$

Hence,

$a_{10}=17+(10-1)(-1)=8$

(v) $d=5,S_9=75$

$\Rightarrow S_9=\frac{9}{2}[2a+(9-1\left)5\right]=75$ [From formula (2)]

$\Rightarrow \frac{9}{2}\times 2[a+20]=75$

$\Rightarrow a+20=\frac{75}{9}$

$\Rightarrow a=\frac{25}{3}-20=\frac{-35}{3}$

$\therefore a_9=\frac{-35}{3}+(9-1)5=\frac{-35}{3}+40$ [From formula (1)]

$\Rightarrow a_9=\frac{85}{3}$

(vi) $a=2,d=8,S_n=90$

$\Rightarrow S_n=\frac{n}{2}\left[2\right(2)+(n-1\left)\right(8\left)\right]=90$ [From formula (2)]

$\Rightarrow n[2+(n-1\left)\right(4\left)\right]=90$

$\Rightarrow 2n+4n^2-4n=90$

$\Rightarrow 2n^2-n-45=0$

$\Rightarrow 2n^2-10n+9n-45=0$

$\Rightarrow 2n(n-5)+9(n-5)=0$

$\Rightarrow (2n+9)(n-5)=0$

$\Rightarrow n=5$ as $n$ cannot be a fraction.

Hence, $a_5=2+(5-1)8=34$ [From formula (1)]

(vii) $a=8,a_n=62,S_n=210$

$\Rightarrow a_n=8+(n-1)d=62$ [From formula (1)]

$\Rightarrow (n-1)d=54$ ...(b)

Now,

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]=210$ [From formula (2)]

$\Rightarrow \frac{n}{2}\left[2\right(8)+54]=210$ [From (b)]

$\Rightarrow n[8+27]=210$

$\Rightarrow n=\frac{210}{35}=6$

$\therefore (6-1)d=54$

$\Rightarrow d=\frac{54}{5}$ or $10\frac{4}{5}$

(viii) $a_n=4,d=2,S_n=-14$

$\Rightarrow a_n=a+(n-1)2=4$ [From formula (1)]

$\Rightarrow a+2n=6$

$\Rightarrow a=6-2n$...(c)

Also, $S_n=\frac{n}{2}[2a+(n-1\left)2\right]=-14$ [From formula (2)]

$\Rightarrow \frac{n}{2}\left[2\right(6-2n)+(n-1\left)2\right]=-14$ [From (c)]

$\Rightarrow n\left[\right(6-2n)+(n-1\left)\right]=-14$

$\Rightarrow n[5-n]=-14$

Comparing with $7[5-7]=7\times (-2)=-14$, we get,

$\Rightarrow n=7$

$\therefore a=6-2\left(7\right)=-8$

(ix) $a=3,n=8,S_8=192$

$\Rightarrow S_8=\frac{8}{2}\left[2\right(3)+(8-1\left)d\right]=192$ [From formula (2)]

$\Rightarrow 4[6+6d]=192$

$\Rightarrow 1+d=\frac{192}{24}$

$\Rightarrow d=8-1=7$

(x) $l=28,S_n=144$ where $n=9$

$\Rightarrow S_n=\frac{n}{2}[a+l]=144$ [From formula (3)]

$\Rightarrow S_9=\frac{9}{2}[a+28]=144$

$\Rightarrow a+28=144\times \frac{2}{9}$

$\Rightarrow a+28=32$

$\Rightarrow a=4$