Question:

### (i)
Given: $a=5$, $d=3$, $a_n=50$.

1. $a_n = a+(n-1)d \Rightarrow 50 = 5 + (n-1)\cdot3$
2. $ (n-1)\cdot3 =45 \Rightarrow n-1 =15 \Rightarrow n =16$
3. $S_n = \frac{n}{2}(a+a_n) = \frac{16}{2}(5+50)=8\times55=440$

Answer: $n=16$, $S_{16}=440$.

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### (ii)
Given: $a=7$, $a_{13}=35$.

1. $a_{13}=a+12d \Rightarrow 35 = 7 +12d$
2. $12d =28 \Rightarrow d = \frac{28}{12}=\frac{7}{3}$
3. $S_{13}=\frac{13}{2}(a+a_{13}) = \frac{13}{2}(7+35)=\frac{13}{2}\times42 =273$

Answer: $d=\frac{7}{3}$, $S_{13}=273$.

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### (iii)
Given: $a_{12}=37$, $d=3$.

1. $a_{12}=a+11d \Rightarrow 37 = a +11\times3$
2. $a = 37-33 =4$
3. $S_{12}=\frac{12}{2}(a+a_{12}) =6\times(4+37)=6\times41=246$

Answer: $a=4$, $S_{12}=246$.

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### (iv)
Given: $a_3=15$, $S_{10}=125$.

1. $a_3 = a+2d \Rightarrow a = 15-2d$
2. $S_{10}=\frac{10}{2}(2a+9d)=5(2a+9d)=125$
3. $2a+9d =25$
4. Substitute $a$: $2(15-2d)+9d =25 \Rightarrow 30-4d+9d =25 \Rightarrow 5d = -5 \Rightarrow d = -1$
5. $a = 15-2(-1)=17$
6. $a_{10}=a+9d = 17+9(-1)=8$

Answer: $d=-1$, $a_{10}=8$.

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### (v)
Given: $d=5$, $S_9=75$.

1. $S_9 = \frac{9}{2}(2a+8d) = \frac{9}{2}(2a+40)=75$
2. Multiply by 2: $9(2a+40)=150 \Rightarrow 2a+40 = \frac{150}{9}=\frac{50}{3}$
3. $2a = \frac{50}{3}-40 = \frac{50-120}{3}= -\frac{70}{3}$
4. $a = -\frac{35}{3}$
5. $a_9 = a+8d = -\frac{35}{3}+40 = \frac{85}{3}$

Answer: $a = -\frac{35}{3}$, $a_9 = \frac{85}{3}$.

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### (vi)
Given: $a=2$, $d=8$, $S_n=90$.

1. $S_n = \frac{n}{2}\bigl(2a+(n-1)d\bigr) = \frac{n}{2}\bigl(4+8(n-1)\bigr) = \frac{n}{2}(8n-4) = 4n^2-2n$
2. Set equal to 90: $4n^2-2n =90 \Rightarrow 2n^2 - n -45 =0$
3. Solve quadratic: $\Delta = 1+360 =361$, $\sqrt{\Delta}=19$
4. $n = \frac{1\pm19}{4}$ → $n =5$ (positive integer)
5. $a_n = a+(n-1)d = 2+4\times8 = 34$

Answer: $n=5$, $a_5 =34$.

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### (vii)
Given: $a=8$, $a_n=62$, $S_n=210$.

1. $S_n = \frac{n}{2}(a+a_n) = \frac{n}{2}(8+62)=35n$
2. $35n =210 \Rightarrow n =6$
3. $a_n = a+(n-1)d \Rightarrow 62 = 8+5d \Rightarrow d = \frac{54}{5}$

Answer: $n=6$, $d=\frac{54}{5}$.

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### (viii)
Given: $a_n=4$, $d=2$, $S_n=-14$.

1. $a_n = a+(n-1)d \Rightarrow a = 4-2(n-1)=6-2n$
2. $S_n = \frac{n}{2}(a+a_n) = \frac{n}{2}\bigl((6-2n)+4\bigr)=\frac{n}{2}(10-2n)=5n-n^2$
3. Set $5n-n^2 = -14 \Rightarrow n^2-5n-14=0$
4. $\Delta =25+56=81$, $\sqrt{\Delta}=9$
5. $n = \frac{5\pm9}{2}$ → $n=7$ (positive integer)
6. $a = 6-2\times7 = -8$

Answer: $n=7$, $a=-8$.

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### (ix)
Given: $a=3$, $n=8$, $S=192$.

1. $S_n = \frac{n}{2}\bigl(2a+(n-1)d\bigr) = 4\bigl(6+7d\bigr)=24+28d$
2. $24+28d =192 \Rightarrow 28d =168 \Rightarrow d =6$

Answer: $d=6$.

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### (x)
Given: last term $l = a_9 =28$, total terms $n=9$, $S=144$.

1. $S_n = \frac{n}{2}(a + l) \Rightarrow 144 = \frac{9}{2}(a+28)$
2. Multiply by 2: $288 = 9(a+28)$
3. $a+28 = 32 \Rightarrow a =4$

Answer: First term $a =4$.