In an AP: (i) Given a=5, d=3, an=50, find n and Sn. (ii) Given a=7, a13=35, find d and S13. (iii) Given a12=37, d=3, find a and S12. (iv) Given a3=15, S10=125, find d and a10. (v) Given d=5, S9=75, find a and a9. (vi) Given a=2, d=8, Sn=90, find n and an. (vii) Given a=8, an=62, Sn=210, find n and d. (viii) Given an=4, d=2, Sn=-14, find n and a. (ix) Given a=3, n=8, S=192, find d. (x) Given l=28, S=144, and there are total 9 terms. Find a.

We know that in an AP
If first term is a1=a, common difference is d then the nth term is an=a+(n-1)d
Sum of first n terms of this AP is Sn=n/2(a1+an)=n/2(2a+(n-1)d)
Now,
(i) a1=a=5
d=3
an=50=a+(n-1)d →50=5+(n-1)3 →n=16
Sn=n/2(a1+an) →Sn=16/2(5+50) →Sn=8×55=440
(ii) a1=7
a13=35 →n=13
a13=a1+(n-1)d →35=7+(13-1)d →d=21/3=7
S13=13/2(a1+a13) →S13=13/2(7+35) →S13=13×21=273
(iii) a12=37
d=3 →n=12
a12=a1+(n-1)d=a+(12-1)3 →37=a+33 →a=4
S12=n/2(a1+a12) →s12=12/2(4+37)=6×41=246
(iv) a3=15 →a3=a+(3-1)d →15=a+2d..(1)
S10=125
10/2(2a+9d)=125 →2a+9d=25 (2)
From (1) and (2), we get d=-1, a=17
a10=17+(9)(-1)=8
(v) d=5
S9=75=9/2(2a+8d)=9/2(2a+40)=9(a+20) →a=-35/3
a9=a+8d=-35/3+8×5=85/3
(vi) a=2
d=8
Sn=90 →n/2(2a+(n-1)d)=90 →n/2(4+(n-1)8)=90 →n(4n-4)=180 →2n^2-n-90=0 →n=5 (n>0, we have neglected negative value of n=-9/2)
an=a5=a+(4)(d)=2+4×8=34
(vii) a=8
an=62
Sn=210=n/2(a+an) →420=n(8+62)=70n →n=6
an=a+(n-1)d
62=8+(5)d →d=54/5=10.8
(viii) an=4
d=2
an=a+(n-1)d
4=a+(n-1)2 →6=a+2n..(1)
Sn=-14
Sn=n/2(a+an)
-14=n/2(a+4)(2)
From (1) and (2), we get n=7 or n=-2, (n>0, we have to neglect negative value)
n=7, then a=-8
(ix) a=3
n=8
S=192
S=192=n/2(2a+(n-1)d) →192=8/2(6+7d) →48=6+7d →d=6
(x) l=an=28
Sn=144
n=9
Sn=n/2(a+an)
144=9/2(a+28) →a=4