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How many terms of an ap 45,39,33,.must be taken so thier sum is 180?
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How many terms of an ap 45,39,33,.must be taken so thier sum is 180?
Okk

a = 45

D = -6

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

180 = n/2 ( 2*45 + (-6n +6)

360 = n ( 90 - 6n + 6)

360 = n ( 96 - 6n)

360 = 96 n - 6n^2

6n^2 - 96n + 360 = 0............1


Solve This equation U will Get Ur Answer
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How many terms of an ap 45,39,33,.must be taken so thier sum is 180?
Problem Statement

Find out how many terms of an AP 45, 39, 33, ... must be taken so that their sum is 180.


Explanation

Here, we need to find the number of terms of an AP whose sum is 180. To solve this problem, we can use the formula for the sum of n terms of an AP:

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

where Sn is the sum of n terms of the AP, a is the first term, d is the common difference between the terms, and n is the number of terms.

Solution

Let's apply the formula to the given AP:

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

Sn = n/2 [2(45) + (n - 1)(-6)]

Sn = n/2 [90 - 6n + 6]

Sn = n/2 [96 - 6n]

Simplifying further, we get:

Sn = 48n - 3n2

Now, we need to find out the value of n for which Sn = 180. So we can set up the equation:

48n - 3n2 = 180

3n2 - 48n + 180 = 0

Solving this quadratic equation, we get:

n = 6 or 10

So, we need to take either 6 or 10 terms of the given AP to get a sum of 180.


Conclusion

We can conclude that to get a sum of 180, we need to take either 6 or 10 terms of the given AP. This problem can be solved easily using the formula for the sum of n terms of an AP.
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How many terms of an ap 45,39,33,.must be taken so thier sum is 180?
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