Geometric series word problem help please?

by the way your problem is not geometric but arithmetic and the formula for sum is:
S = (n/2)[(2a + (n-1)d]

S = total of the arithmetic progression
n = total count of numbers ; 200
a = 1st number ; 25
d = difference of every other number ; 27-25 or 29-27 that is 2

S = (200/2)[(2*25) + (200-1)*2]
S = 100[50+199*2]
S = 100[50+398]
S = 100(448)
S = 44,800

hope this helps

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Two ways of solving
1)
25 + 27 + 29 + ...
the terms of the sum are in arithmetic progression
(24 + 1) + (24 + 3) + (24 + 5) + ... (200 terms)

this is
24· 200 + (1 + 3 + 5 + ...)
the sum of the first 200 odd numbers is 200² [*]

the seating capacity of the stadium is
24· 200 + 200² = 44,800

[*]
the sum of the first n odd numbers is n²

proof
suppose n is even
1 + 3 + 5 + ... + (2n - 5) + (2n - 3) + (2n - 1)
if we add the first and last term we get 2n
the second and the last but one is 2n
and so on
As terms are n, pairs are n/2
the sum is (n/2)(2n) = n²
²
if n is odd, n - 1 is even and previous formula applies
the sum of the first (n - 1) odd numbers is (n - 1)²
now add the last one (2n - 1) and we get
(n - 1)² + (2n - 1) = n² - 2n + 1 + 2n - 1 = n²

QED

2)
in general if you have a sum of n terms in arithmetic progression
a + b·i , i=1,2,...,n
ₙ
Σ(a+b·i) = n (2a + b(n + 1))/2 [**]
ⁱ⁼¹

₂₀₀
Σ(23+2·i) = 200(2·23 + 2(200 + 1))/2 = 44,800
ⁱ⁼¹

[**]
ₙ
Σ(a+b·i) =
ⁱ⁼¹
= na + bn(n+1)/2 = n (2a + b(n + 1))/2

n(n + 1)/2 is the sum of the integers from 1 to n
1 + 2 + ... + (n - 1) + n
(the proof is similar to the previous one)

if they are
b + 2b + ... + b(n - 1) + bn
collect b
b(1 + 2 + ... + n) = bn(n + 1)/2

This is the question on arithmetic progression .
series is 25 ,27 ,29,.............
sum of this series = n/2(2a + (n-1)d)
where n = no of terms = 200
a = first term = 25
d = common difference = 27 -25 =2
~ S = 200/2( 2*25 + (200-1)2)
~ = 100(50 + (199*2))
~ =100(50 + 398)
~ = 100*448
~ = 44800
Hence the seating capacity of stadium is 44800

25 = 25 + 2(1 - 1)
27 = 25 + 2(2 - 1)
29 = 25 + 2(3 - 1)

Thus, the number of seats is
25 + 27 + 29 + ... + (25 + 2 * 199)
= 25 + (25 + 2) + (25 + 4) + ... + (25 + 2(200 - 1))
= (25 x 200) + (2 + 4 + ... + 398)
= 5000 + 2(1 + 2 + ... + 199)
= 5000 + 2(1 + 199)(199)/2
= 5000 + 39800
= 44800.