Find number of terms in arithmetic progression
$begingroup$
In a arithmetic progression sum of first four terms sum :
$$a_1+a_2+a_3+a_4=124$$
and sum of last four terms :
$$a_n+a_{n-1}+a_{n-2}+a_{n-3}=156$$
and sum of arithmetic progression is :
$$S_n=350$$
$$n=?$$
How to find $n$? I tried using arithmetic progression sum formulas but getting negative or fractional numbers.
sequences-and-series summation arithmetic-progressions
edited Dec 14 '18 at 16:19
Lord Shark the Unknown
105k1160133
asked Dec 14 '18 at 16:15
Serif YaohimSerif Yaohim
32
$endgroup$
add a comment |
$begingroup$
In a arithmetic progression sum of first four terms sum :
$$a_1+a_2+a_3+a_4=124$$
and sum of last four terms :
$$a_n+a_{n-1}+a_{n-2}+a_{n-3}=156$$
and sum of arithmetic progression is :
$$S_n=350$$
$$n=?$$
How to find $n$? I tried using arithmetic progression sum formulas but getting negative or fractional numbers.
sequences-and-series summation arithmetic-progressions
edited Dec 14 '18 at 16:19
Lord Shark the Unknown
105k1160133
asked Dec 14 '18 at 16:15
Serif YaohimSerif Yaohim
32
$endgroup$
add a comment |
$begingroup$
In a arithmetic progression sum of first four terms sum :
$$a_1+a_2+a_3+a_4=124$$
and sum of last four terms :
$$a_n+a_{n-1}+a_{n-2}+a_{n-3}=156$$
and sum of arithmetic progression is :
$$S_n=350$$
$$n=?$$
How to find $n$? I tried using arithmetic progression sum formulas but getting negative or fractional numbers.
sequences-and-series summation arithmetic-progressions
edited Dec 14 '18 at 16:19
Lord Shark the Unknown
105k1160133
asked Dec 14 '18 at 16:15
Serif YaohimSerif Yaohim
32
$endgroup$
In a arithmetic progression sum of first four terms sum :
$$a_1+a_2+a_3+a_4=124$$
and sum of last four terms :
$$a_n+a_{n-1}+a_{n-2}+a_{n-3}=156$$
and sum of arithmetic progression is :
$$S_n=350$$
$$n=?$$
How to find $n$? I tried using arithmetic progression sum formulas but getting negative or fractional numbers.
sequences-and-series summation arithmetic-progressions
sequences-and-series summation arithmetic-progressions
edited Dec 14 '18 at 16:19
Lord Shark the Unknown
105k1160133
asked Dec 14 '18 at 16:15
Serif YaohimSerif Yaohim
32
edited Dec 14 '18 at 16:19
Lord Shark the Unknown
105k1160133
asked Dec 14 '18 at 16:15
Serif YaohimSerif Yaohim
32
edited Dec 14 '18 at 16:19
Lord Shark the Unknown
105k1160133
edited Dec 14 '18 at 16:19
Lord Shark the Unknown
105k1160133
edited Dec 14 '18 at 16:19
Lord Shark the Unknown
105k1160133
105k1160133
asked Dec 14 '18 at 16:15
Serif YaohimSerif Yaohim
32
asked Dec 14 '18 at 16:15
Serif YaohimSerif Yaohim
32
asked Dec 14 '18 at 16:15
Serif YaohimSerif Yaohim
32
32
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
One has $a_1+a_2+a_3+a_4+a_{n-3}+a_{n-2}+a_{n-1}+a_n=280$. The mean of those
eight terms is $35$, so the mean of $a_1,ldots,a_n$ is also $35$. The sum
of those $n$ terms is $n$ times their mean, and is $350$. So now you can read
off $n$.
answered Dec 14 '18 at 16:23
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
hohohoh man thank you a lot. God bless you
$endgroup$
– Serif Yaohim
Dec 14 '18 at 16:25
add a comment |
$begingroup$
Use the formula $$a_n=a_1+(n-1)d$$ where $$a_{n+1}-a_{n}=d$$
answered Dec 14 '18 at 16:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
add a comment |
$begingroup$
$$350=dfrac{n(a_1+a_n)}2$$
Now $a_1+a_n=a_2+a_{n-1}=cdots=dfrac{124+156}4$
answered Dec 14 '18 at 16:28
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Use standard formula of A.P.
which is
$a_n = a + (n-1)d$
and a simple difference formula
$a_n - a_{n-1} = d$
edited Dec 18 '18 at 19:46
Prakhar Nagpal
752318
$endgroup$
add a comment |
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4 Answers
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oldest
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4 Answers
4
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
One has $a_1+a_2+a_3+a_4+a_{n-3}+a_{n-2}+a_{n-1}+a_n=280$. The mean of those
eight terms is $35$, so the mean of $a_1,ldots,a_n$ is also $35$. The sum
of those $n$ terms is $n$ times their mean, and is $350$. So now you can read
off $n$.
answered Dec 14 '18 at 16:23
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
hohohoh man thank you a lot. God bless you
$endgroup$
– Serif Yaohim
Dec 14 '18 at 16:25
add a comment |
$begingroup$
One has $a_1+a_2+a_3+a_4+a_{n-3}+a_{n-2}+a_{n-1}+a_n=280$. The mean of those
eight terms is $35$, so the mean of $a_1,ldots,a_n$ is also $35$. The sum
of those $n$ terms is $n$ times their mean, and is $350$. So now you can read
off $n$.
answered Dec 14 '18 at 16:23
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
hohohoh man thank you a lot. God bless you
$endgroup$
– Serif Yaohim
Dec 14 '18 at 16:25
add a comment |
$begingroup$
One has $a_1+a_2+a_3+a_4+a_{n-3}+a_{n-2}+a_{n-1}+a_n=280$. The mean of those
eight terms is $35$, so the mean of $a_1,ldots,a_n$ is also $35$. The sum
of those $n$ terms is $n$ times their mean, and is $350$. So now you can read
off $n$.
answered Dec 14 '18 at 16:23
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
One has $a_1+a_2+a_3+a_4+a_{n-3}+a_{n-2}+a_{n-1}+a_n=280$. The mean of those
eight terms is $35$, so the mean of $a_1,ldots,a_n$ is also $35$. The sum
of those $n$ terms is $n$ times their mean, and is $350$. So now you can read
off $n$.
answered Dec 14 '18 at 16:23
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 14 '18 at 16:23
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 14 '18 at 16:23
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 14 '18 at 16:23
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
$begingroup$
hohohoh man thank you a lot. God bless you
$endgroup$
– Serif Yaohim
Dec 14 '18 at 16:25
add a comment |
$begingroup$
hohohoh man thank you a lot. God bless you
$endgroup$
– Serif Yaohim
Dec 14 '18 at 16:25
$begingroup$
hohohoh man thank you a lot. God bless you
$endgroup$
– Serif Yaohim
Dec 14 '18 at 16:25
$begingroup$
hohohoh man thank you a lot. God bless you
$endgroup$
– Serif Yaohim
Dec 14 '18 at 16:25
add a comment |
$begingroup$
Use the formula $$a_n=a_1+(n-1)d$$ where $$a_{n+1}-a_{n}=d$$
answered Dec 14 '18 at 16:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
add a comment |
$begingroup$
Use the formula $$a_n=a_1+(n-1)d$$ where $$a_{n+1}-a_{n}=d$$
answered Dec 14 '18 at 16:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
add a comment |
$begingroup$
Use the formula $$a_n=a_1+(n-1)d$$ where $$a_{n+1}-a_{n}=d$$
answered Dec 14 '18 at 16:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
Use the formula $$a_n=a_1+(n-1)d$$ where $$a_{n+1}-a_{n}=d$$
answered Dec 14 '18 at 16:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered Dec 14 '18 at 16:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered Dec 14 '18 at 16:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered Dec 14 '18 at 16:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
76.8k42866
add a comment |
add a comment |
$begingroup$
$$350=dfrac{n(a_1+a_n)}2$$
Now $a_1+a_n=a_2+a_{n-1}=cdots=dfrac{124+156}4$
answered Dec 14 '18 at 16:28
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
$$350=dfrac{n(a_1+a_n)}2$$
Now $a_1+a_n=a_2+a_{n-1}=cdots=dfrac{124+156}4$
answered Dec 14 '18 at 16:28
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
$$350=dfrac{n(a_1+a_n)}2$$
Now $a_1+a_n=a_2+a_{n-1}=cdots=dfrac{124+156}4$
answered Dec 14 '18 at 16:28
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$$350=dfrac{n(a_1+a_n)}2$$
Now $a_1+a_n=a_2+a_{n-1}=cdots=dfrac{124+156}4$
answered Dec 14 '18 at 16:28
lab bhattacharjeelab bhattacharjee
226k15157275
edited Dec 14 '18 at 16:54
answered Dec 14 '18 at 16:28
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 14 '18 at 16:28
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 14 '18 at 16:28
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
$begingroup$
Use standard formula of A.P.
which is
$a_n = a + (n-1)d$
and a simple difference formula
$a_n - a_{n-1} = d$
edited Dec 18 '18 at 19:46
Prakhar Nagpal
752318
$endgroup$
add a comment |
$begingroup$
Use standard formula of A.P.
which is
$a_n = a + (n-1)d$
and a simple difference formula
$a_n - a_{n-1} = d$
edited Dec 18 '18 at 19:46
Prakhar Nagpal
752318
$endgroup$
add a comment |
$begingroup$
Use standard formula of A.P.
which is
$a_n = a + (n-1)d$
and a simple difference formula
$a_n - a_{n-1} = d$
edited Dec 18 '18 at 19:46
Prakhar Nagpal
752318
$endgroup$
Use standard formula of A.P.
which is
$a_n = a + (n-1)d$
and a simple difference formula
$a_n - a_{n-1} = d$
edited Dec 18 '18 at 19:46
Prakhar Nagpal
752318
edited Dec 18 '18 at 19:46
Prakhar Nagpal
752318
edited Dec 18 '18 at 19:46
Prakhar Nagpal
752318
edited Dec 18 '18 at 19:46
Prakhar Nagpal
752318
752318
answered Dec 18 '18 at 19:32
user578581
add a comment |
add a comment |
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