If the first n terms of an and Sn = (1 / 3) ^ Na + 1 / 6, a=

If the first n terms of an and Sn = (1 / 3) ^ Na + 1 / 6, a=

Sn=(1/3)^na+1/6,a=-1/6
Method 1
When the common ratio Q ≠ 1,
The sum of the first n terms of the equal ratio sequence Sn = A1 (1-Q & # 8319;) / (1-Q)
Let A1 / (1-Q) = t
Then Sn = t (1-Q & # 8319;) = - TQ & # 8319; + T
It can be seen that the coefficients of Q & #; and the constant terms are opposite to each other
If the constant term is 1 / 6, then the coefficient of (1 / 3) &; a = - 1 / 6
Method 2
a1=S1=a/3+1/6
a2=S2-S1=a/9-a/3=-2/9a
a3=S3-S2=a/27-a/9=-2a/27
∴a2/a1=a3/a2=1/3
∴-2a/9=1/3*(a/3+1/6)
∴a/3=-1/18
∴a=-1/6