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how to solve the equation
- Thread starter manisha
- Start date
D
Deleted member 4993
Guest
hello there
I want to know each and every step in detail to solve this equation.
please reply quickly
n
/2[2a + (n - 1)d] = 1800
thanks,
Solve for what? n, a or d?
Please read the post titled "Read before Posting".
We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")
Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.
ok then,
actually i was solving
The sum of how may tersm of the series 6 + 12 + 18 + 24 + ... is 1800 ? this question which is an example of arthmetic progression
where
a = 6, d = 6 and Sn = 1800 then i came across this equation which i dident understand how to solve
so please explain me in detail....
thanks and regards
actually i was solving
The sum of how may tersm of the series 6 + 12 + 18 + 24 + ... is 1800 ? this question which is an example of arthmetic progression
where
a = 6, d = 6 and Sn = 1800 then i came across this equation which i dident understand how to solve
| Then, | n | [2a + (n - 1)d] = 1800 |
| 2 |
thanks and regards
I am presuming that what is in the box above is:ok then,
actually i was solving
The sum of how may tersm of the series 6 + 12 + 18 + 24 + ... is 1800 ? this question which is an example of arthmetic progression
where
a = 6, d = 6 and Sn = 1800 then i came across this equation which i dident understand how to solve
so please explain me in detail....
Then,
n
[2a + (n - 1)d] = 1800
2
thanks and regards
\(\displaystyle \{n[2a + (n - 1)d]\}/2 = 1800?\) It helps to use grouping symbols and / for division
If I substitute in a = 6 = d and Sn = 1800
\(\displaystyle 1800 = \dfrac{n\{2*6 + 6(n-1)\}}{2} = \dfrac{6n(2 + n - 1)}{2} = 3n(n + 1) \implies n(n + 1) = 600.\)
What is so hard about that? Of course, that is going all around Robin Hood's barn.
\(\displaystyle 6 + 12 + 18 +\ ...\ = 1800 \implies\)
\(\displaystyle 6(1 + 2 + 3 +\ ...) = 6(300) \implies\)
\(\displaystyle 1 + 2 + 3 +\ ...\ = 300 \implies \dfrac{n(n + 1)}{2} = 300 \implies n(n + 1) = 600?\)
Or are you having trouble solving \(\displaystyle n^2 + n - 600 = 0?\)
I am presuming that what is in the box above is:
\(\displaystyle \{n[2a + (n - 1)d]\}/2 = 1800?\) It helps to use grouping symbols and / for division
If I substitute in a = 6 = d and Sn = 1800
\(\displaystyle 1800 = \dfrac{n\{2*6 + 6(n-1)\}}{2} \implies\)
\(\displaystyle 1800 = \dfrac{6n(2 + n - 1)}{2} \implies\)
\(\displaystyle 1800 = 3n(n + 1) \implies \)
\(\displaystyle n(n + 1) = 600.\)
\(\displaystyle 6 + 12 + 18 +\ ...\ = 1800 \implies\)
\(\displaystyle 6(1 + 2 + 3 +\ ...) = 6(300) \implies\)
\(\displaystyle 1 + 2 + 3 +\ ...\ = 300 \implies \)
\(\displaystyle \dfrac{n(n + 1)}{2} = 300 \implies \)
\(\displaystyle n(n + 1) = 600\)
It's from an equation to an equation using implications.