The sequence of the number of girl attendees at each Barangay is nonzero, given by multiplying the number of attendees in the previous Barangay by 2
The required values are;
a. The completed table is presented as follows;
![\underline{\begin{array}{|l|c|c|c|c|c|c|c|}\mathbf{Barangay}&1&2&3&4&5&6&7\\\mathbf{Number \ of \ attandees}&4&8&16&32&64&128&256\end{array}}](https://tex.z-dn.net/?f=%5Cunderline%7B%5Cbegin%7Barray%7D%7B%7Cl%7Cc%7Cc%7Cc%7Cc%7Cc%7Cc%7Cc%7C%7D%5Cmathbf%7BBarangay%7D%261%262%263%264%265%266%267%5C%5C%5Cmathbf%7BNumber%20%5C%20of%20%5C%20attandees%7D%264%268%2616%2632%2664%26128%26256%5Cend%7Barray%7D%7D)
b. The sequence of the number of girls attendees aₙ = <u>4×2⁽ⁿ⁻¹⁾</u>
c. The sum of beneficiaries of the lecture from the twelve Barangay is <u>16,384 girls</u>
Reason:
Known parameters are;
Number of girls that attended the lecture at Barangay 1 = 4 girls
Number of girls that attended from Barangay 2 = 8 girls
Number that attended from Barangay 3 = 16 girls
a. The required table with the assumption that the number of girls that attended at each Barangay from Barangay 1 to Barangay 7 doubles, is presented as follows;
b. The sequence of the number of girls who attended the lecture from Barangay 1 to Barangay 7 is presented as follows;
The sequence is a geometric progression having the form, aₙ = a·r⁽ⁿ⁻¹⁾
The first term of the sequence, a = 4
The common ratio, r = 2 (each term is twice the previous term)
n = The number of terms
The sequence is therefore;
c. The total number of girls in the geometric progression, (G. P.) is given by the sum of a G. P. as follows
![S_n = \dfrac{a \cdot (r^n - 1 )}{r - 1}](https://tex.z-dn.net/?f=S_n%20%3D%20%5Cdfrac%7Ba%20%5Ccdot%20%28r%5En%20-%201%20%29%7D%7Br%20-%201%7D)
Where;
n = The number of terms = 7 + 5 = 12
Which gives;
The total number of girls who will benefit from the lectures, S₁₂ = <u>16,384 girls</u>
Learn more about geometric progression here:
brainly.com/question/14256037