TASK: to find out the number of the m th term of the original series.

Original Serial No.

* Here, all the serial no. divisible by 3 and their succeeding serial no. have the same no. in the series & if the original series is transformed into recast series, the numbers of recast series are in AP.

Recast Serial No.

If Original Serial no. is given, Corresponding Recast Serial no. can be found out by the following formula:--

* If last original serial no.(os1) is divisible by 3, then last recast serial no. (rs1) = os1-(os1/3) + 1 = (2os1+3)/3; otherwise rs1= os1 - [(immediately preceding os1 divisible by three)/3]

If Recast Serial no. is given(rs2), Corresponding Original Serial no.(os2) can be found out by the following formula:--

* If recast serial no. is odd, original serial no. are os2a & os2b and os2a==3(rs2-1)/2 and os2b=os2a+1=(3rs2-1)/2.

* If recast serial no. is even, original serial no. is os2a=(3rs2-2)/2 ;

* Suppose in the recast series, 1st no. is a, the series is in Arithmetic Progression (AP), common difference (cd) is d, then nth number is a+(n-1)d.

* Value under original serial no. is given by--

   If  original serial no. is divisible by 3, a1 +2*os1*d/3 where a1 is 1st term , d=common difference

   If original serial no. is not divisible by 3, a1+(2*os1-3 +os1%3)*(d/3);

* sum in AP of recast series=sum1=(rs1/2)*[2a1+(rs1-1)d];

* sum in AP of original series = (diff/2)[(2a1+4d) +(diff-1)2d]; where diff = os1-rs1;

Exa- Take original serial no. as 7, 1st term as 0 and cd=1.

Original  Serial No(os1) .- Given
First term of the Series (a1)
Common Difference (d)
Find out -->    
Recast  Serial No.(rs1)
Value  under the Given Original serial no.
Sum of recast series up to rs1 (sum1)
Sum of original series up to os1 (sum2)
Recast Serial no.- Given (rs2)
Original Serial No.(os2a,os2b) and