# arithmetic sequence

• August 27th 2007, 09:30 AM
arithmetic sequence
hello
'find the arithmetic progression of nine continuous numbers where the sum total is 135'
kindly get me detailed solution
sincerely
anil
• August 27th 2007, 10:25 AM
ThePerfectHacker
Quote:

hello
'find the arithmetic progression of nine continuous numbers where the sum total is 135'
kindly get me detailed solution
sincerely
anil

The first number is $a$ the next is $a+k$ the next is $a+2k$ and so on ... the ninth is $a+8k$.

In sum we have,
$\underbrace{(a+a+...+a)}_9 + k(1+2+...+8) = 9a+36k$.

So we want,
$9a+36k=135$
Divide by 9,
$a+4k=15$.
So the possible such sequences are:
$a=11,k=1$
$a=7,k=2$
$a=3,k=3$
• August 28th 2007, 05:09 PM
Soroban
Hello, Anil!

Quote:

Find the arithmetic progression of nine continuous numbers where the sum total is 135
Did you mean "consecutive" numbers?
. . Then the first term is $a$, and the common difference is $d = 1$.

The sum of the first n terms of an A.P. is: . $S_n\:=\:\frac{n}{2}\left[2a + (n-1)d\right]$
. . Then we have: . $S_9 \:=\:\frac{9}{2}\left[2a + 8(1)\right] \:=\:135\quad\Rightarrow\quad a \:=\:11$

Therefore, the progression is: . $11,\,12,\,13,\,14,\,15,\,16,\,17,\,18,\,19$