(i) From the question,
First term a=5
Then common difference d=3
an​=50
We know that, an​=a+(n−1)d
50=5+(n−1)350=5+3n−350=2+3n3n=50−23n=48n=48/3​
n=16So,Sn​=====​(n/2)(2a+(n−1)d)(16/2)((2×5)+(16−1)×3)8(10+45)8(55)440​

(ii) given a=7,a13​=35, find d and S13​
From the question,
First term a=7
a13​=35
We know that, an​=a+(n−1)d
35=7+(13−1)d35=7+13d−d35=7+12d12d=35−712d=28d=28/12​
d=7/3
So,S13​​=(n/2)(2a+(n−1)d)=(13/2)((2×7)+((13−1)×(7/3)))=(13/2)((14+(12×7/3))=(13/2)(14+28)=(13/2)(42)=13×21=273​

(iii) given d=5,S9​=75, find a and a9​
From the question it is given that,
Common difference d=5
S9​=75
We know that, an​=a+(n−1)d
a9​=a+(9−1)5a9​=a+45−5a9​=a+40​
 Then, S9​=(n/2)(2a+(n−1)d)75=(9/2)(2a+(9−1)5)75=(9/2)(2a+(8)5)(75×2)/9=2a+40150/9=2a+402a=150/9−402a=50/3−402a=(50−120)/32a=−70/3a=−70/(3×2)a=−35/3​
Now, substitute the value of a in equation (i),
a9​​=a+40=−35/3+40=(−35+120)/3=85/3​

(iv) given a=8,an​=62,Sn​=210, find n and d
From the question, it is given that,
First-term a=8
an​=62 and Sn​=210
We know that, an​=a+(n−1)d
62=8+(n−1)d(n−1)d=62−8(n−1)d=54​ Then, Sn​=(n/2)(2a+(n−1)d)210=(n/2)((2×8)+54)210=(n/2)(16+54)420=n(70)n=420/70n=6​
Now, substitute the value of n in equation (i),
(n−1)d=54(6−1)d=545d=54d=54/5​
Therefore, d=54/5 and n = 6

(v) given a=3,n=8,S8​=192, find d
From the question, it is given that,
First term a=3
n=8S8​=192​
We know that, Sn​=(n/2)(2a+(n−1)d)
192=(8/2)((2×3)+(8−1)d)192=4(6+7d)192/4=6+7d48=6+7d48−6=7d42=7dd=42/7d=6​
Therefore, the common difference d is 6.