The above series is an A.P. with a = 2, d = 5 - 2 = 3 and Sum = 155.
Let x be nth term so, an = a + (n - 1)d.
⇒ x = 2 + (n - 1)3
⇒ x = 2 + 3n - 3
⇒ x = 3n - 1.
Sum = 2n[a+l]
∴155=2n[2+x]⇒2n[2+3n−1]=155⇒2n[3n+1]=155⇒3n2+n=155×2⇒3n2+n−310=0⇒3n2−30n+31n−310=0⇒3n(n−10)+31(n−10)=0⇒(3n+31)(n−10)=0⇒3n+31=0 or n−10=0⇒n=−331 or n=10.
Since, number of terms cannot be in fraction so, n ≠ −331.
∴ x = 3n - 1 = 3 × 10 - 1 = 30 - 1 = 29.
Hence, the value of x is 29.