# Another arithmetic progression problem

• Government$In summary, the sum of the first 15 members of an increasing arithmetic progression is 1410, with the first term being -4 and the common difference being 14. This can be solved by setting up a system of equations, or by using the simplified method of setting the second term as "a" and solving from there. The textbook method uses the second term as "a" and yields a first term of -4 and a common difference of 14. However, the textbook incorrectly states the first term as 10. Government$

## Homework Statement

Sum of first three members of increasing arithmetic progression is 30 and sum of their squares is 692. What is the sum of the first 15 members?

## The Attempt at a Solution

So i have system of equations:
a1 + a2 + a3 = 30
(a1)^2 + (a2^2) + (a3^2) = 692
----------------------------------
3a1 + 3d = 30
3(a1)^2 +6(a2)d + d^2= 691
---------------------
d = 10 - a1

and if i plug that in 3(a1)^2 +6(a2)d + d^2= 691 i get

2(a1)^2 - 40a1 -192 = 0

now when i find roots of this equation i get 24 and -4 none of them are correct since it says that at the end of the book that a1 = 10 and d=14. Now i do get d=14 with a1=-4 d=-14 with a=-24 but since it is increasing progression i can rule out d=-14, but apparently my solution is not correct. And my solutions satisfy equation 3a1 + 3d = 30 while theirs doesnt.

P.S. This is how they solved this problem:

let a2=a then we get
3a=30 and 3a^2 + 2d^2=692 from that we get a1=10 and d=14 so S15 = 1410

thank you

Government$said: ## Homework Statement Sum of first three members of increasing arithmetic progression is 30 and sum of their squares is 692. What is the sum of the first 15 members? ## The Attempt at a Solution So i have system of equations: a1 + a2 + a3 = 30 (a1)^2 + (a2^2) + (a3^2) = 692 ---------------------------------- Neither of these equations uses the fact that this is an arithmetic progression 3a1 + 3d = 30 I think you are trying to use it here but are doing it incorrectly. Starting from a1 the numbers are a1, a1+ d and a1+ 2d which add to 3a1+ 2d, not "3d". 3a1+ 2d= 30 3(a1)^2 +6(a2)d + d^2= 691 And for the squares, you want a1^2+ (a1+ d)^2+ (a1+ 2d)^2= a1^2+ a1^2+ 4a1d+ d^2+ a1^2+ 4a1d+ 4d^2 which gives 3(a1)^2+ 8a1d+ 4d^2= 692. --------------------- d = 10 - a1 No, as above, d= 15- (3/2)a1. and if i plug that in 3(a1)^2 +6(a2)d + d^2= 691 i get 2(a1)^2 - 40a1 -192 = 0 now when i find roots of this equation i get 24 and -4 none of them are correct since it says that at the end of the book that a1 = 10 and d=14. Now i do get d=14 with a1=-4 d=-14 with a=-24 but since it is increasing progression i can rule out d=-14, but apparently my solution is not correct. And my solutions satisfy equation 3a1 + 3d = 30 while theirs doesnt. P.S. This is how they solved this problem: let a2=a then we get 3a=30 and 3a^2 + 2d^2=692 from that we get a1=10 and d=14 so S15 = 1410 thank you I don't understand, how come a1 + a2 + a3= a1 + a1 +d + a1 + 2d = 3a1 + 2d and not 3a1 + 3d. What happened to that one d Government$ said:
I don't understand, how come a1 + a2 + a3= a1 + a1 +d + a1 + 2d = 3a1 + 2d and not
3a1 + 3d. What happened to that one d

3a1 + 3d is right.

Can we drop the "a1"? I'm used to calling the first term of an AP simply "a".

I get a = -4, d = 14 to satisfy the condition that this is an increasing progression.

The sum of the first 15 terms is 1410.

I'm not sure why the textbook is saying the first term is 10, because that's wrong. But the sum IS 1410.

How did you calculate S15? What answer did you get?

EDIT: The textbook method MASSIVELY simplifies the algebra, so I like it. But remember that what they're calling a is the *second* term, and that's calculated to be 10. So the first term is still 10 - 14 = -4.

The AP goes: -4, 10, 24,...,192 up to the first 15 terms.

## 1. What is an arithmetic progression?

An arithmetic progression is a sequence of numbers in which the difference between consecutive terms is constant. For example, 2, 4, 6, 8, 10 is an arithmetic progression with a common difference of 2.

## 2. How do you find the next term in an arithmetic progression?

To find the next term in an arithmetic progression, you can use the formula an = a1 + (n-1)d, where a1 is the first term, n is the term number, and d is the common difference.

## 3. What is the formula for the sum of an arithmetic progression?

The formula for the sum of an arithmetic progression is Sn = n/2(a1 + an), where n is the number of terms, a1 is the first term, and an is the last term.

## 4. How is an arithmetic progression different from a geometric progression?

An arithmetic progression has a constant difference between consecutive terms, while a geometric progression has a constant ratio between consecutive terms. For example, 2, 4, 8, 16, 32 is a geometric progression with a common ratio of 2.

## 5. How can I use arithmetic progressions in real life?

Arithmetic progressions can be used in various real-life scenarios, such as calculating financial growth or calculating the growth of a population over time. They are also used in computer algorithms and in solving mathematical problems.

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