# Arithmetic Progression Problem

• z.js
In summary, to find the sum of n terms in an arithmetic progression, use the formula Sn = (n/2)[2a+(n-1)d], where a is the first term, d is the common difference, and n is the number of terms. In this problem, we are given that the sum of the first 7 terms is 28 and the sum of the first 15 terms is 90. By solving the equations, we find that the common difference is 7.75 and the first term is -18.5. Thus, the sum of n terms can be expressed as n/4(n + 9).
z.js

## Homework Statement

If the sum of the first 7 terms of an arithmetic progression is 28 and the sum of the first 15 terms is 90, find the sum of n terms.

## Homework Equations

Sn = 0.5n[2a+(n-1)d]
a is the first term and d is the common difference. n is the number of terms.
nth term = a + (n-1)d

## The Attempt at a Solution

(here goes )
(1)...a + 6d = 28
(2)...a + 14d = 90
(2)-(1)...8d = 62
...d = 7.75 (ok )
(1)...a + 6(7.75) = 28
.....a = -18.5 (yea! )
(drumrollllllllllll)
Sn = 0.5n(-37 + (n-1)7.75)
= -18.5n +n(n-1)3.875
= n[3.875(n-1) - 18.5]
... now what? the answer was n/4(n + 9)

z.js said:

## Homework Statement

If the sum of the first 7 terms of an arithmetic progression is 28 and the sum of the first 15 terms is 90, find the sum of n terms.

## Homework Equations

Sn = 0.5n[2a+(n-1)d]
a is the first term and d is the common difference. n is the number of terms.
nth term = a + (n-1)d

## The Attempt at a Solution

(here goes )
(1)...a + 6d = 28
(2)...a + 14d = 90
(2)-(1)...8d = 62
...d = 7.75 (ok )
(1)...a + 6(7.75) = 28
.....a = -18.5 (yea! )
(drumrollllllllllll)
Sn = 0.5n(-37 + (n-1)7.75)
= -18.5n +n(n-1)3.875
= n[3.875(n-1) - 18.5]
... now what? the answer was n/4(n + 9)

You have used formula of the nth term instead of using that of sum of n terms .28 is not the 7th term .Instead it is the sum of first 7 terms .Similarly with 90 being the sum of first 15 terms.

Use Sn = (n/2)[2a+(n-1)d]

Last edited:
ahh! I forgot! thanks

## 1. What is an arithmetic progression?

An arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is constant. For example, the sequence 3, 6, 9, 12, 15 is an arithmetic progression with a common difference of 3.

## 2. How do you find the common difference in an arithmetic progression?

The common difference in an arithmetic progression can be found by subtracting any two consecutive terms in the sequence. The result will be the constant difference between all terms in the progression.

## 3. What is the formula for finding the nth term of an arithmetic progression?

The formula for finding the nth term of an arithmetic progression is a + (n-1)d, where a is the first term and d is the common difference.

## 4. How can arithmetic progressions be used in real life?

Arithmetic progressions are commonly used in financial calculations, such as compound interest and loan repayments. They can also be used in solving problems involving time, distance, and speed.

## 5. Is there a limit to the number of terms in an arithmetic progression?

There is no limit to the number of terms in an arithmetic progression. As long as the common difference remains constant, the progression can continue indefinitely.

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