fie a=numar impar, n= nr termenilor
a, (a+2), (a+2·2), (a+3·2), .......,[a+(n-3)·2], [a+(n-2)·2], [a+(n-1)·2]
[a+(n-3)·2]+ [a+(n-2)·2]+[a+(n-1)·2]=309
si [a+(n-1)·2]-a=104
2n-2=104
n=106:2
n=53, e numarul termenilor
a+(53-3)·2+a+(53-2)·2+a+(53-1)·2=309
3a+100+102+104=309
3a+306=309
a=1, primul termen
ultimul termen este:
1+(53-1)·2=105
verificare: 101+103+105=309 suma ultimilor 3 termeni
105-1=104