Given,
↪T7=12
↪a+(7-1)d=12
↪a+6d=12➡(i)
&,
↪T10=T2+8
↪a+(10-1)d=a+(2-1)d+8
↪a+9d=a+d+8
↪a-a+9d-d=8
↪8d=8
↪d=1➡ANS
▶Put value of d in (i),then,
↪a+6×1=12
↪a=12-6
↪a= 6➡ANS
❇To find the term which exceeds 400,
↪a+(n-1)d=400
↪6+(n-1)×1=400
↪(n-1)×1=394
↪n=394+1
↪n=395 for which the term becomes equal to 400. So n must be greater by 1 than current value of n so that the term becomes greater than 400.
❇Hence, n=396➡ANS
That means, 396th term exceeds the value 400.
❇NOTE:
▶Formula used:
↪Tn=a+(n-1)d
where,
a= 1st term
d=common difference
