this is new point? as I know CDCl3 give us sharp singlet at about 7.25 plus- minus 0.02 ppm. please give me an example.

I think you are talking about a 13C spectrum. Deuterium is a spin-1 nucleus and in this case the multiplicity is calculated by 2*n*I+1 with nuclear spin I and number of neighbors n. So you observe a triplett for chloroform due to one deuteron scalar coupled to the carbon. They have equal intensity because the spin-1 nuclei has the three states +1, 0 and -1.

13C NMR      2n+1      n=1

A common solvent for dissolving compounds for 1H and 13C NMR spectroscopy is deuteriochloroform, DCCl3. In 1H NMR spectra, the impurity of HCCl3 in DCCl3 gives a small signal at 7.2 ppm (see spectrum of methyl propanoate). In 13C spectroscopy 1.1% of the deuteriochloroform has a 13C isotope and it is bonded to a deuterium atom. The nucleus of the deuterium atom, the deuteron, has a more complicated nuclear spin than does the proton, and it has a gyromagnetic ratio ( ) 1/6 as large. This more complicated nuclear spin gives rise to three spin states instead of the two spin states for the proton, and the deuteron undergoes resonance at a different frequency than either the proton or 13C nucleus. These spin states are approximately all equally populated. Because the spin-spin coupling between the 13C and the deuterium is not eliminated during proton decoupling, the DCCl3 shows three equal peaks of low to moderate intensity at about 77 ppm. The separation is the carbon-deuterium coupling constant JCD. The intensity is low to moderate because the 13C receives no Nuclear Overhauser Enhancement from the proton decoupling. 

Thank you Prof Joshi, Jonathan Dickerhoff , Alaa J Mahrath & M. P. Filatova  for your appreciable answer, But still I have one doubt in my mind that I want to share with you. .

In 1H NMR, the nuclei which has half integer value is NMR active eg. H (1/2), C (1/2), F(1/2) ; due to this half value of spin angular moment quantum number the above nuclei is NMR active. .

While in case of dutterium the value is 1 which is NMR inactive so how can Duterrium couple with Hydrogen ?

Every nuclei with a nonzero spin is NMR active, so deuterium can couple with hydrogen. I added two links to Glenn Facey blog where he explains coupling in chloroform and to quadrupolar nuclides.

http://u-of-o-nmr-facility.blogspot.de/2015/01/multiplets-for-spin-i-12-nclides.html

http://u-of-o-nmr-facility.blogspot.ca/2007/09/isotope-shifts-for-chloroform.html

nuclei with non zero spin are nmr active.In CDCl3, triplet in 13C(Spin 1/2) NMR is due to coupling with deuterium nuclei having spin 1,multiplicity  rule 2nI+1,In proton NMR signal at 7.2 correspond to residual chloroform in CDCl3

Oky. Thank you very much to everyone to clear my doubts. 

It is simple: multiplicity rule 2nI+1=2x1x1+1=3

In 13C NMR cdcl3 gives a peak (not triplet because triplet will be in this ratio 1:2:1) in this ratio 1:1:1due to spin states of deuterium +1;0;-1

But in 13c NMR of DMSO its peak appeared as septate .So why the intensity varies.

CD3SOCD3 (DMSO)

n=3, I=1

2nI+1 = 7 (septet)