sol1]

let the first series be a ,a+d,a+2d,.....

the second series be A,A+D, A+2D,.......

given s/S=(7n+1)4n+27)

TO FIND--------- t11/T11=? .............t11 denotes 11th term of first AP

and T11 denotes 11th term of Second AP

s=n/2*[2a+(n-1)d] ................................. sum of first AP upto n terms

similarly sum of second AP is

S=n/2*[2A+(n-1)D]

s/S=(7n+1)4n+27)

[n/2*{2a+(n-1)d}] / [n/2*{2A+(n-)D}]=(7n+1)4n+27)

n/2 gets cancelled we get

[2a+(n-1)d]/ [2A+(n-)D]=(7n+1)4n+27)

dividing nr.& dr. of LHS by2 we get

[a+(n-1)d/2] / [A+(n-1)D/2] = (7n+1)4n+27)

now put n=21 both sides we get

[a+10d]/[A+10D]=148/111

t11/T11=148/111