# the sum of 50 consecutive positive odd integers is 3800. What is the largest of these integers?

### 19 Answers

- formengLv 65 months ago
Well, let’s first identify the terms.

So, let’s let a be the first number, L be the last number d be the difference, and n be the number of numbers.

We don’t know the first or last, but we know the difference, so we can write

First =a and last = a + (n-1)d

= a+ (50-1)2

=a +98

Now is we played around with sequences, which we won’t, we could come up with a formula for the sum of the numbers.

S = [n(a +L)]/2

3800 =[50(a +a+98]/2

3800 =25(2a +98)

50a =1350

a=27

Now we can go back to the formula for L.

L=a+(n-1)d

=27 +49*2

=125

- lenpol7Lv 75 months ago
S(n) = (n/2)(2a + ( n-1)d)

Where

n = 50

S(n) = 3800

a = NOt known

d = 2

Hence

3800 = (50/2)(2(a) + ( 50 -1)2))

3800 = (25)(2a + ( 49(2))

3800 = 25(2a + 98))

3800/25 = 2a + 98)

152 = 2a + 98

2a = 54

a = 27

Hnce largest is 54 + 98 = 152

- VamanLv 75 months ago
Let x be the first number. x+2, x+2n etc wil be the next number. 5o number require

49 n values. The sum will 2*49*50/2=2450 remove this from 3800 which gives 1350. This will 50 x values sum. x=1350/50=25. The largest wiil be 25+2*49=123 I hope, I am correct.

- King LeoLv 75 months ago
Arithmetic series:

First term a₁, common difference d = 2, number of terms n = 50 and sum S = 3800.

S = ½n [ 2a₁ + d( n - 1 ) ]

3800 = ½ * 50 [ 2a₁ + 2( 50 - 1 ) ]

a₁ = 27

the nth term:

a𝓃 = a₁ + d( n - 1 )

a𝓃 = 27 + 2( n - 1 )

a𝓃 = 25 + 2n

last term ( largest term ), i.e n = 50

a𝓃 = 25 + 2n

a𝓃 = 25 + 2*50

a₅₀ = 125

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- Anonymous5 months ago
just make the first integer the largest.

x1+x2+x3...+x50=3800 on a decreasing scalevthen its simple. on a decreasing scale what odds add to 3800. they can all be equal too btw.

- roderick_youngLv 75 months ago
My strategy would be to pair the biggest number with the smallest number, then the second biggest with second smallest, and so forth. So I would have 25 pairs of odd numbers.

3800/25 = 152, which will be the total for each pair.

Half of 152 is 76, so if I go up and down from that by 1, (75, 77) is one pair of odd numbers. The next pair out from that would be (73, 79). Notice that to generate one more pair, I subtracted 2 from the first number, and added 2 to the second number. Besides the original pair of (75,77), I need 24 more pairs to get 50 numbers total. 24*2 = 48, so I must offset the first and second number by 48 to arrive at the last pair. 77 + 48 = 125, which is the largest of the integers.

Source(s): 75 - 48 = 27 is the smallest, by the way - KrishnamurthyLv 75 months ago
The sum of 50 consecutive positive odd integers is 3800.

What is the largest of these integers?

The largest of these integers is 125.

27, 29, 31, 33, 35, ...

a_n = 2 n + 25

27 + 29 + 31 + .... + 125 = 3800

- llafferLv 75 months ago
The arithmetic sequence of a list of odd numbers will have a common difference of 2. But we don't know where this starts. So we can set up this general expression:

a(n) = a + b(n - 1)

a(n) = a + 2(n - 1)

a(n) = a + 2n - 2

The sum of the first "n" terms of an arithmetic sequence is:

S(n) = [ a + a(n)] n / 2

We don't know "a" so will leave that as an unknown and have an expression for a(n). Substitute and simplify:

S(n) = (a + a + 2n - 2) n / 2

S(n) = (2a + 2n - 2) n / 2

S(n) = 2(a + n - 1) n / 2

S(n) = (a + n - 1)n

S(n) = an + n² - n

S(n) = n² + an - n

We are told the sum of the first 50 terms if 3800, so we can substitute what we know and solve for "a":

S(50) = 50² + a(50) - 50

3800 = 2500 + 50a - 50

3800 = 2450 + 50a

1350 = 50a

27 = a

The first term is now known to be "a". Put this back into our starting function, simplify, then solve for a(50):

a(n) = a + 2n - 2

a(n) = 27 + 2n - 2

a(n) = 2n + 25

a(50) = 2(50) + 25

a(50) = 100 + 25

a(50) = 125

The last term in your sequence is 125.

- RobertLv 75 months ago
Call the first integer 2n + 1, so the numbers are

2n + 1

2n + 3

2n + 5

2n = 7

. . . .

2n + 99

99 is the 50th odd number; a well-known (?)

formula says the sum of the first k odd numbers

is k^2

So the above sum is

100n + 50^2 = 3800

100n = 1300, so

n = 13, and so the largest of the numbers is

2*13 + 99 = 125