The total number of books that we had in out bookcase when we were 11 years old was 2,187 books.
What is a geometric sequence and how to find its nth terms?
Suppose the initial term of a geometric sequence is [tex]a[/tex]
and the term by which we multiply the previous term to get the next term is [tex]r[/tex]
Then the sequence would look like
[tex]a, ar, ar^2, ar^3, \cdots[/tex]
(till the terms to which it is defined)
Thus, the nth term of such sequence would be
[tex]T_n = ar^{n-1}[/tex] (you can easily predict this formula, as for nth term, the multiple r would've multiplied with initial terms n-1 times).
What is the sum of terms of a geometric sequence?
Lets suppose its initial term is [tex]a[/tex] , multiplication factor is [tex]r[/tex]
and let it has total n terms, then, its sum is given as:
[tex]S_n = \dfrac{a(r^n-1)}{r-1}[/tex]
(sum till nth term)
For this case the initial number of books (2) is multiplied by 3 each year.
So we got:
a = 2,
r = 3
Since a = 2 was for age 5, and if we take age 5 at 1st position, then age 11 is at the 7th position.
The number of books we read at each age is a geometric sequence with initial term 2 and multiplication factor 3.
The sum of its term(which would make it geometric series) will give the total number of books we will have.
Since age 11 is at 7th position, so n = 7
Thus, the total number of books we had in the bookcase when we were 11 years old is:
[tex]S_n = \dfrac{a(r^n-1)}{r-1}\\\\S_7 = \dfrac{2((3)^7-1)}{3-1} = 2187[/tex]
We could've derived it without formula as:
- Age 5: 2 books
- Age 6: [tex]2 \times 3[/tex] books
- Age 7: [tex]2 \times 3^2[/tex] books
- Age 8: [tex]2 \times 3^3[/tex] books
- Age 9: [tex]2 \times 3^4[/tex] books
- Age 10: [tex]2 \times 3^5[/tex] books
- Age 11: [tex]2 \times 3^6[/tex] books
Their sum is 2187.
Thus, the number of books that we had in out bookcase when we were 11 years old was 2187 books.
Learn more about sum of terms of geometric sequence here:
brainly.com/question/1607203
