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Given the sum of first 4 terms and last 4 terms of arithmetic progression
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Given the sum of the first 4 terms of the arithmetic progression,it equals to 124. Also given the sum of last 4 terms of the progression, it equals to 156. If the sum of arithmetic progression equal to 490, how many terms have arithmetic progression?
Please help me with solution of this problem.
arithmetic-progressions
asked Jun 5 '16 at 11:08
user345768
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Given the sum of the first 4 terms of the arithmetic progression,it equals to 124. Also given the sum of last 4 terms of the progression, it equals to 156. If the sum of arithmetic progression equal to 490, how many terms have arithmetic progression?
Please help me with solution of this problem.
arithmetic-progressions
asked Jun 5 '16 at 11:08
user345768
91
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Interesting question! (+1). See alternative approach and much shorter solution posted below.
– hypergeometric
Dec 30 '16 at 8:35
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Given the sum of the first 4 terms of the arithmetic progression,it equals to 124. Also given the sum of last 4 terms of the progression, it equals to 156. If the sum of arithmetic progression equal to 490, how many terms have arithmetic progression?
Please help me with solution of this problem.
arithmetic-progressions
asked Jun 5 '16 at 11:08
user345768
91
Given the sum of the first 4 terms of the arithmetic progression,it equals to 124. Also given the sum of last 4 terms of the progression, it equals to 156. If the sum of arithmetic progression equal to 490, how many terms have arithmetic progression?
Please help me with solution of this problem.
arithmetic-progressions
arithmetic-progressions
asked Jun 5 '16 at 11:08
user345768
91
asked Jun 5 '16 at 11:08
user345768
91
asked Jun 5 '16 at 11:08
user345768
91
asked Jun 5 '16 at 11:08
user345768
91
asked Jun 5 '16 at 11:08
user345768
91
91
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Interesting question! (+1). See alternative approach and much shorter solution posted below.
– hypergeometric
Dec 30 '16 at 8:35
add a comment |
Interesting question! (+1). See alternative approach and much shorter solution posted below.
– hypergeometric
Dec 30 '16 at 8:35
Interesting question! (+1). See alternative approach and much shorter solution posted below.
– hypergeometric
Dec 30 '16 at 8:35
Interesting question! (+1). See alternative approach and much shorter solution posted below.
– hypergeometric
Dec 30 '16 at 8:35
add a comment |
3 Answers
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With $;n,a_1,d;$ the number of elements, first element and common difference respectively, the following is what you're given
$$begin{align*}&frac42left(2a_1+3dright)=124implies 2a_1+3d=62\{}\&frac42left(a_{n-3}+3dright)=156implies a_{n-3}+3d=78\{}\&frac n2left(2a_1+(n-1)dright)=290end{align*}$$
You have three equations and three unknowns. Observe that $;a_{n-3}=a_1+d(n-4);$ ...
answered Jun 5 '16 at 11:14
DonAntonio
175k1491224
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If there are $n$ terms with the first term, common difference being $a,d$ respectively,
we have $$124=dfrac42{2a+(4-1)d}iff2a+3d=62 (1)$$
and $$156=dfrac42{a+(n-1)d+3(-d)}iff78=a+(n-4)d (2)$$
Adding we get $$62+78=2a+(n-1)d$$
Finally, $$490=dfrac n2{2a+(n-1)d} (3)$$
$$implies490=dfrac n2cdot(62+78)iff n=?$$
Generalization: Find $a,d$ in terms of $n$ from $(1),(2)$ and replace the values in $(3)$
answered Jun 5 '16 at 11:14
lab bhattacharjee
220k15154271
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Here's an alternative and much shorter approach.
Note that the sum of groups of $m$ terms of an AP form another AP.
For the question given, use $m=4$.
Let $N$ be the number of terms of the new AP, of which the first term is $124$ and the last term $156$. Using the formula for an AP sum, we have
$$290=frac N2 (124+156);;Rightarrow
N=frac 72$$
For the original AP,
$$n=Nm=frac 72cdot 4=color{red}{14}$$
i.e. there are $14$ terms in the original AP.
Check
As a check, by substitution of $n=14$ into standard AP formulas, one can easily show that $a=29.8$ and $d=0.8$. The AP is
$$
underbrace{underbrace{29.8,;;
30.6,;;
31.4,;;
32.2}_{Sigma=124},;;
33.0,;;
33.8,;;
34.6,;;
35.4,;;
36.2,;;
37.0,;;
underbrace{37.8,;;
38.6,;;
39.4,;;
40.2}_{Sigma=156}}_{Sigma=490}$$
answered Dec 30 '16 at 8:34
hypergeometric
17.3k1755
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
With $;n,a_1,d;$ the number of elements, first element and common difference respectively, the following is what you're given
$$begin{align*}&frac42left(2a_1+3dright)=124implies 2a_1+3d=62\{}\&frac42left(a_{n-3}+3dright)=156implies a_{n-3}+3d=78\{}\&frac n2left(2a_1+(n-1)dright)=290end{align*}$$
You have three equations and three unknowns. Observe that $;a_{n-3}=a_1+d(n-4);$ ...
answered Jun 5 '16 at 11:14
DonAntonio
175k1491224
add a comment |
up vote
0
down vote
With $;n,a_1,d;$ the number of elements, first element and common difference respectively, the following is what you're given
$$begin{align*}&frac42left(2a_1+3dright)=124implies 2a_1+3d=62\{}\&frac42left(a_{n-3}+3dright)=156implies a_{n-3}+3d=78\{}\&frac n2left(2a_1+(n-1)dright)=290end{align*}$$
You have three equations and three unknowns. Observe that $;a_{n-3}=a_1+d(n-4);$ ...
answered Jun 5 '16 at 11:14
DonAntonio
175k1491224
add a comment |
up vote
0
down vote
up vote
0
down vote
With $;n,a_1,d;$ the number of elements, first element and common difference respectively, the following is what you're given
$$begin{align*}&frac42left(2a_1+3dright)=124implies 2a_1+3d=62\{}\&frac42left(a_{n-3}+3dright)=156implies a_{n-3}+3d=78\{}\&frac n2left(2a_1+(n-1)dright)=290end{align*}$$
You have three equations and three unknowns. Observe that $;a_{n-3}=a_1+d(n-4);$ ...
answered Jun 5 '16 at 11:14
DonAntonio
175k1491224
With $;n,a_1,d;$ the number of elements, first element and common difference respectively, the following is what you're given
$$begin{align*}&frac42left(2a_1+3dright)=124implies 2a_1+3d=62\{}\&frac42left(a_{n-3}+3dright)=156implies a_{n-3}+3d=78\{}\&frac n2left(2a_1+(n-1)dright)=290end{align*}$$
You have three equations and three unknowns. Observe that $;a_{n-3}=a_1+d(n-4);$ ...
answered Jun 5 '16 at 11:14
DonAntonio
175k1491224
answered Jun 5 '16 at 11:14
DonAntonio
175k1491224
answered Jun 5 '16 at 11:14
DonAntonio
175k1491224
answered Jun 5 '16 at 11:14
DonAntonio
175k1491224
175k1491224
add a comment |
add a comment |
up vote
0
down vote
If there are $n$ terms with the first term, common difference being $a,d$ respectively,
we have $$124=dfrac42{2a+(4-1)d}iff2a+3d=62 (1)$$
and $$156=dfrac42{a+(n-1)d+3(-d)}iff78=a+(n-4)d (2)$$
Adding we get $$62+78=2a+(n-1)d$$
Finally, $$490=dfrac n2{2a+(n-1)d} (3)$$
$$implies490=dfrac n2cdot(62+78)iff n=?$$
Generalization: Find $a,d$ in terms of $n$ from $(1),(2)$ and replace the values in $(3)$
answered Jun 5 '16 at 11:14
lab bhattacharjee
220k15154271
add a comment |
up vote
0
down vote
If there are $n$ terms with the first term, common difference being $a,d$ respectively,
we have $$124=dfrac42{2a+(4-1)d}iff2a+3d=62 (1)$$
and $$156=dfrac42{a+(n-1)d+3(-d)}iff78=a+(n-4)d (2)$$
Adding we get $$62+78=2a+(n-1)d$$
Finally, $$490=dfrac n2{2a+(n-1)d} (3)$$
$$implies490=dfrac n2cdot(62+78)iff n=?$$
Generalization: Find $a,d$ in terms of $n$ from $(1),(2)$ and replace the values in $(3)$
answered Jun 5 '16 at 11:14
lab bhattacharjee
220k15154271
add a comment |
up vote
0
down vote
up vote
0
down vote
If there are $n$ terms with the first term, common difference being $a,d$ respectively,
we have $$124=dfrac42{2a+(4-1)d}iff2a+3d=62 (1)$$
and $$156=dfrac42{a+(n-1)d+3(-d)}iff78=a+(n-4)d (2)$$
Adding we get $$62+78=2a+(n-1)d$$
Finally, $$490=dfrac n2{2a+(n-1)d} (3)$$
$$implies490=dfrac n2cdot(62+78)iff n=?$$
Generalization: Find $a,d$ in terms of $n$ from $(1),(2)$ and replace the values in $(3)$
answered Jun 5 '16 at 11:14
lab bhattacharjee
220k15154271
If there are $n$ terms with the first term, common difference being $a,d$ respectively,
we have $$124=dfrac42{2a+(4-1)d}iff2a+3d=62 (1)$$
and $$156=dfrac42{a+(n-1)d+3(-d)}iff78=a+(n-4)d (2)$$
Adding we get $$62+78=2a+(n-1)d$$
Finally, $$490=dfrac n2{2a+(n-1)d} (3)$$
$$implies490=dfrac n2cdot(62+78)iff n=?$$
Generalization: Find $a,d$ in terms of $n$ from $(1),(2)$ and replace the values in $(3)$
answered Jun 5 '16 at 11:14
lab bhattacharjee
220k15154271
edited Jun 5 '16 at 11:23
answered Jun 5 '16 at 11:14
lab bhattacharjee
220k15154271
answered Jun 5 '16 at 11:14
lab bhattacharjee
220k15154271
answered Jun 5 '16 at 11:14
lab bhattacharjee
220k15154271
220k15154271
add a comment |
add a comment |
up vote
0
down vote
Here's an alternative and much shorter approach.
Note that the sum of groups of $m$ terms of an AP form another AP.
For the question given, use $m=4$.
Let $N$ be the number of terms of the new AP, of which the first term is $124$ and the last term $156$. Using the formula for an AP sum, we have
$$290=frac N2 (124+156);;Rightarrow
N=frac 72$$
For the original AP,
$$n=Nm=frac 72cdot 4=color{red}{14}$$
i.e. there are $14$ terms in the original AP.
Check
As a check, by substitution of $n=14$ into standard AP formulas, one can easily show that $a=29.8$ and $d=0.8$. The AP is
$$
underbrace{underbrace{29.8,;;
30.6,;;
31.4,;;
32.2}_{Sigma=124},;;
33.0,;;
33.8,;;
34.6,;;
35.4,;;
36.2,;;
37.0,;;
underbrace{37.8,;;
38.6,;;
39.4,;;
40.2}_{Sigma=156}}_{Sigma=490}$$
answered Dec 30 '16 at 8:34
hypergeometric
17.3k1755
add a comment |
up vote
0
down vote
Here's an alternative and much shorter approach.
Note that the sum of groups of $m$ terms of an AP form another AP.
For the question given, use $m=4$.
Let $N$ be the number of terms of the new AP, of which the first term is $124$ and the last term $156$. Using the formula for an AP sum, we have
$$290=frac N2 (124+156);;Rightarrow
N=frac 72$$
For the original AP,
$$n=Nm=frac 72cdot 4=color{red}{14}$$
i.e. there are $14$ terms in the original AP.
Check
As a check, by substitution of $n=14$ into standard AP formulas, one can easily show that $a=29.8$ and $d=0.8$. The AP is
$$
underbrace{underbrace{29.8,;;
30.6,;;
31.4,;;
32.2}_{Sigma=124},;;
33.0,;;
33.8,;;
34.6,;;
35.4,;;
36.2,;;
37.0,;;
underbrace{37.8,;;
38.6,;;
39.4,;;
40.2}_{Sigma=156}}_{Sigma=490}$$
answered Dec 30 '16 at 8:34
hypergeometric
17.3k1755
add a comment |
up vote
0
down vote
up vote
0
down vote
Here's an alternative and much shorter approach.
Note that the sum of groups of $m$ terms of an AP form another AP.
For the question given, use $m=4$.
Let $N$ be the number of terms of the new AP, of which the first term is $124$ and the last term $156$. Using the formula for an AP sum, we have
$$290=frac N2 (124+156);;Rightarrow
N=frac 72$$
For the original AP,
$$n=Nm=frac 72cdot 4=color{red}{14}$$
i.e. there are $14$ terms in the original AP.
Check
As a check, by substitution of $n=14$ into standard AP formulas, one can easily show that $a=29.8$ and $d=0.8$. The AP is
$$
underbrace{underbrace{29.8,;;
30.6,;;
31.4,;;
32.2}_{Sigma=124},;;
33.0,;;
33.8,;;
34.6,;;
35.4,;;
36.2,;;
37.0,;;
underbrace{37.8,;;
38.6,;;
39.4,;;
40.2}_{Sigma=156}}_{Sigma=490}$$
answered Dec 30 '16 at 8:34
hypergeometric
17.3k1755
Here's an alternative and much shorter approach.
Note that the sum of groups of $m$ terms of an AP form another AP.
For the question given, use $m=4$.
Let $N$ be the number of terms of the new AP, of which the first term is $124$ and the last term $156$. Using the formula for an AP sum, we have
$$290=frac N2 (124+156);;Rightarrow
N=frac 72$$
For the original AP,
$$n=Nm=frac 72cdot 4=color{red}{14}$$
i.e. there are $14$ terms in the original AP.
Check
As a check, by substitution of $n=14$ into standard AP formulas, one can easily show that $a=29.8$ and $d=0.8$. The AP is
$$
underbrace{underbrace{29.8,;;
30.6,;;
31.4,;;
32.2}_{Sigma=124},;;
33.0,;;
33.8,;;
34.6,;;
35.4,;;
36.2,;;
37.0,;;
underbrace{37.8,;;
38.6,;;
39.4,;;
40.2}_{Sigma=156}}_{Sigma=490}$$
answered Dec 30 '16 at 8:34
hypergeometric
17.3k1755
answered Dec 30 '16 at 8:34
hypergeometric
17.3k1755
answered Dec 30 '16 at 8:34
hypergeometric
17.3k1755
answered Dec 30 '16 at 8:34
hypergeometric
17.3k1755
17.3k1755
add a comment |
add a comment |
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Interesting question! (+1). See alternative approach and much shorter solution posted below.
– hypergeometric
Dec 30 '16 at 8:35