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 $\displaystyle a$ 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:
$\displaystyle a=11,k=1$
$\displaystyle a=7,k=2$
$\displaystyle a=3,k=3$
Hello, Anil!
Did you mean "consecutive" numbers?Find the arithmetic progression of nine continuous numbers where the sum total is 135
. . 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: .$\displaystyle 11,\,12,\,13,\,14,\,15,\,16,\,17,\,18,\,19$