hello

'find the arithmetic progression of nine continuous numbers where the sum total is 135'

kindly get me detailed solution

sincerely

anil

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- Aug 27th 2007, 08:30 AManilpadavilarithmetic sequence
hello

'find the arithmetic progression of nine continuous numbers where the sum total is 135'

kindly get me detailed solution

sincerely

anil - Aug 27th 2007, 09:25 AMThePerfectHacker
The first number is $\displaystyle a$ the next is $\displaystyle a+k$ the next is $\displaystyle a+2k$ and so on ... the ninth is $\displaystyle a+8k$.

In sum we have,

$\displaystyle \underbrace{(a+a+...+a)}_9 + k(1+2+...+8) = 9a+36k$.

So we want,

$\displaystyle 9a+36k=135$

Divide by 9,

$\displaystyle a+4k=15$.

So the possible such sequences are:

$\displaystyle a=11,k=1$

$\displaystyle a=7,k=2$

$\displaystyle a=3,k=3$ - Aug 28th 2007, 04:09 PMSoroban
Hello, Anil!

Quote:

Find the arithmetic progression of nine*continuous*numbers where the sum total is 135

*"consecutive"*numbers?

. . Then the first term is $\displaystyle a$, and the common difference is $\displaystyle d = 1$.

The sum of the first n terms of an A.P. is: .$\displaystyle S_n\:=\:\frac{n}{2}\left[2a + (n-1)d\right]$

. . Then we have: .$\displaystyle S_9 \:=\:\frac{9}{2}\left[2a + 8(1)\right] \:=\:135\quad\Rightarrow\quad a \:=\:11$

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