Question

How many terms of the Arithmetic Progression 45, 39, 33, ... must be taken so that their sum is 180?
Explain the double answer.
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Answer 1

Answer:

Explanation:

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

S_{n} =180S

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)S

n

=

2

n

(2a+(n−1)d)

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

2

n

(2(45)+(n−1)(−6))=180

⇒ \dfrac{n}{2} (90-6n+6)=180

2

n

(90−6n+6)=180

⇒ 96n -6n^{2} =180\times 2=36096n−6n

2

=180×2=360

⇒ 6n^{2}-96n +360=06n

2

−96n+360=0

⇒n^{2}-16n +60=0n

2

−16n+60=0 [divided by 6]

⇒ n^{2}-10n-6n +60=0n

2

−10n−6n+60=0

⇒ (n-6)(n-10)=0(n−6)(n−10)=0

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

⇒ n = 6 or 10

∴ n = 6 or 10

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