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 Share with your friends Share 8 Lovina Kansal answered this 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 6 View Full Answer Najre Alam answered this bhai question samajh me nhi aa rha 1