Este o progresie aritmetica cu ratia =2, inlocuim in formula termenului general:
[tex] a_{n}= a_{1}+(n-1)r,obtinem:x+ \frac{121}{4}=x}+ \frac{1}{4}+(n-1)*2 [/tex]⇒ 30=2*(n-1)⇒ n=16. In formula sumei:[tex] S _{16}= \frac{ (a_{1}+ a_{16})*16 }{2}=340 [/tex], adica[tex](x+ \frac{1}{4}+x+ \frac{121}{4})*8=340 [/tex]⇒ 2x=12 ⇒x=6.
