The sum of certain consecutive odd numbers is 57^2 – 13^2. Find the numbers.

**chakravarthiponmudi** · April 29th 2009, 06:52 PM

Problem.
The sum of certain consecutive odd numbers is $57^2$ – $13^2.$ Find the numbers.

solution
$\sum 1+3+5+....+(2n-1) = n^2 [1]$

$57^2=n^2$

$n=57$

also from equ 1

$n=\frac{l+1}{2}$

therefore
$\frac{l_1+1}{2} = 57$

$l_1=113$

similarly
$13^2=n^2$

$l_2=25$

$57^2 - 13 ^2 = 27^2+29^2+....113^2$

a = 27 , d = 2
l= 113
a+nd = 113
27 + 2n = 113
n = 43

this is where I am facing the problem. To satisfy the condition I should have n = 44 and not 43.
because 57 - 13 = 44. other way to find the n.

Please correct my understanding...

**Opalg** · April 30th 2009, 03:12 AM

Originally Posted by chakravarthiponmudi

a+nd = 113

The formula for the n'th term of an arithmetic progression is a + (n–1)d, not a + nd.

**Soroban** · April 30th 2009, 06:28 AM

Hello, chakravarthiponmudi!

You already have all the information you need.

The sum of certain consecutive odd numbers is $57^2 - 13^2$ . . Find the numbers.

Solution

$\sum^n_{k=1}(2k-1) \:=\:1+3+5+ \hdots+(2n-1) \:=\: n^2$

That formula says:
. . the sum of the first $n$ odd numbers is $n^2.$

So $57^2$ is the sum of the first 57 odd numbers.
. . and $13^2$ is the sum of the first 13 odd numbers.

Hence: . $57^2 - 13^2 \:=\:\text{(first 57 odd numbers)} - \text{(first 13 odd numbers})$

. . . . . . . . . . . . $= \;(1 + 3 + 5 + \hdots + 113) - (1 + 3 + 5 + \hdots + 25)$

. . . . . . . . . . . . $= \;27 + 29 + 31 + \hdots + 113$

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