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i=j/m

PV
= FV (1+  i)^-n

i=j/m

FV=
PV(1+  i)^n

log105+log5x+1=log10x+5+1

log1055x+1=log1010x+5

55x+1=10x+5

5x+1=2x+10

3x=9

x=3

Let each installment be Rs.x. Then, 

(P.W. of Rs.x due 1 year hence) + (P.W of Rs.x due 2 years hence) + (P.W of Rs. X due 3 years hence) = 7620.

x1+503x100+x1 +503x1002+x1 +503x1003 = 7620

 6x7+36x49+216x343=7620

 294x + 252x + 216x = 7620 x 343

 => x = 3430

 Amount of each installment = Rs.3430

Let the sum be Rs.P.then
P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii)
On dividing,we get (1+R/100)^3=10025/6690=3/2.
Substituting this value in (i),we get:
P*(3/2)=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.

Interest for 1 year is the same whether it's simple interest or the compound interest.

Now interest of third year = 8575 - 7350 = 1225; means principal for this interest is 7350 if comound interest is takenIf 7350 is the principal interest = 1225

if 100 is the principal interest =(1225/7350) * 100 = 50/3 %

When a thing increases for two successive times the overall increase on initial amount = a + b + (a * b)/100

Therefore overall interest for two years = 50/3 + 50/3 + [(50/3)*(50/3)]/100 = 325/9 %

Therefore amount after 2 years = 100 + 325/9 = 1225/9

If 1225/9 is the amount principal =100

if 7350 is the amount principal =(900/1225)*7350 =5400

So sum = 5400; Rate = 50/3 %