Be the PG (the_{1}, a_{2}, a_{3}, a_{4},… , The_{no} ,…). For calculating the sum of the first n terms S_{no}, let's consider the following:**s _{no} = a_{1} + a_{2} + a_{3} + a_{4} +… + A_{n-1} + a_{no}**

Multiplying both members for the reason q we have:**s _{no}.q = a_{1} . q + a_{2} .q +… + a_{n-1} . q + a_{no} .q **

According to the definition of PG, we can rewrite the expression as:**s _{no} . q = a_{2} + a_{3} +… + A_{no} + a_{no} . what **

Note that the_{2} + a_{3} +… + An is equal to S_{no} - a_{1} . So instead, comes:**s _{no} . q = S_{no} - a_{1} + a_{no} . what **

Hence, simplifying conveniently, we will arrive at the following sum formula:

If we replace an = a_{1} . what^{n-1}, we will get a new presentation for the sum formula, namely:

Example:

Calculate the sum of the first 10 terms of PG (1,2,4,8,…)

We have:

Note that in this case the_{1} = 1.