S1=a2+a4+a6+ upto 100 terms and S2=a1+a3+a5+ - Maths - Arithmetic Progressions

Question

S 1 = a 2 + a 4 + a 6 +   u p t o   100   t e r m s
  a n d   S 2 = a 1 + a 3 + a 5 +   u p t o  
100   t e r m s   o f   a   c e r t a i n
  A P   P r o v e   t h a t   t h e   c o
m m o n   d i f f e r e n c e   i s   S 1 - S 2 100

Lovina Kansal · Accepted Answer

Dear student
Note: an=a+(n-1)dSN=N22A+(N-1)DGiven:S1=a2+a4+
upto 100 terms=a+(2-1)d+a+(4-1)d+
upto 100 terms     =a+d+(a+3d)+
upto 100 terms=(a+a+ upto 100 terms)+d(1+3+5+
upto 100 terms)    =100a+d100221+(100-1)(2)      here A=1,N=100, D=3-1=2=100a+d502+198=100a+10000d       
(1)and S2=a1+a3+ upto 100 terms=a+(1-1)d+a+(3-1)d+
upto 100 terms     =a+0×d+(a+2d)+
upto 100 terms=(a+a+ upto 100 terms)+d(0+2+4+
upto 100 terms)    =100a+d100220+(100-1)(2)      here A=0,N=100, D=2-0=2=100a+d50198=100a+9900d       
(2)So, S1-S2=100a+10000d-100a-9900d=100dHence,d=S1-S2100
Regards

S 1 = a 2 + a 4 + a 6 + . . . . . . . . . . . . u p t o 100 t e r m s a n d S 2 = a 1 + a 3 + a 5 + . . . . . . . u p t o 100 t e r m s o f a c e r t a i n A P . P r o v e t h a t t h e c o m m o n d i f f e r e n c e i s S 1 - S 2 100

$S_{1} = a_{2} + a_{4} + a_{6} + . . . . . . . . . . . . u p t o 100 t e r m s a n d S_{2} = a_{1} + a_{3} + a_{5} + . . . . . . . u p t o 100 t e r m s o f a c e r t a i n A P . P r o v e t h a t t h e c o m m o n d i f f e r e n c e i s \frac{S_{1} - S_{2}}{100}$