In an AP
(i) given a=5,d=3, $a_{n}$ =50, find n and $S_{n}$ .
(ii) given $a_{12}$ =37,d=3, find a and $S_{12}$ .
(iii) given d=5, $S_{9}$ =75, find a and $a_{9}$ .
(iv) given a=8, $a_{n}$ =62, $S_{n}$ =210, find n and d.
(v) given l=28,S=144, and there are total of 9 terms. Find a.

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 $a$ .

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Given a12=37,d=3, find a and S12

The correct option is D 4,24637=a+3(12−1)=a+33⇒a=37−33=4S12=122(4+37)=246

Given $a_{12} = 37, d = 3$ , find a and $S_{12}$

The correct option is D $4, 246$
$37 = a + 3 (12 - 1) = a + 33$
$\Rightarrow a = 37 - 33 = 4$
$S_{12} = \frac{12}{2} (4 + 37) = 246$