CBSE BOARD X, asked by rs8208005633, 1 year ago

How many term of arithmetic progression 45, 39, 33 must be taken so that their sum is 180 explain the double answer

Answers

Answered by sankruno1
105

Hello there ...

here a = 45 , d = -6 , Sn = 180

To find : n

We know ,

sn \: = \frac{n}{2} \times 2a + (n - 1)d

Therefore,

180 = n / 2 × 2 ( 45) + (n - 1 ) -6

180 = n/2 × 90 - 6n + 6

180 × 2 = n ( 96 - 6n )

360 = 96n - 6n^2

6n^2 - 96n + 360 = 0

Dividing by 6

n^2 - 16n + 60 = 0

n^2 - 10n - 6n + 60 = 0 .......( factorizing )

n ( n - 10 ) - 6 ( n - 10 ) = 0 .......{ taking common out }

( n - 6 ) (n - 10 ) = 0

n - 6 = 0 or. n - 10 = 0

n = 6. or. n = 10

Therefore 6 or 10 terms must be taken so that their sum is 180

Answered by harendrachoubay
42

The number of arithmetic progression are "6 and 10".

Explanation:

Here, first term (a) = 45 , common difference (d) = 39 - 45 = - 6 and

S_{n} =180

Let the number of term = n

To find, the umber of term = ?

We Know that,

S_{n} =\dfrac{n}{2} (2a+(n-1)d)

\dfrac{n}{2} (2(45)+(n-1)(-6))=180

\dfrac{n}{2} (90-6n+6)=180

⇒  96n -6n^{2} =180\times 2=360

6n^{2}-96n +360=0

n^{2}-16n +60=0 [divided by 6]

n^{2}-10n-6n +60=0

(n-6)(n-10)=0

[(n-6)=0 or (n-10)=0

⇒ n = 6  or 10

n = 6  or 10

Hence, the number of arithmetic progression are 6 and 10.

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