Question

If an AP S9 = 45 then find a5
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Answer 1

The value of $a_{5}$ is equal to 5.

Step-by-step explanation:

Given,

$S_9$ = 45

To find, the value of $a_5$ = ?

Let the first term = a and common difference = d

We know that,

The nth term of an AP,

$a_{n}$ = a + (n -1)d

∴ The nth term of an AP,

$a_{5}$ = a + (5 -1)d

$a_{5}$ = a + 4d           ......... (1)

The sum of nth term of an AP,

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

The sum of 9th term of an AP,

$S_{9} =\dfrac{9}{2}[2a+(9-1)d]$

⇒ $\dfrac{9}{2}[2a+8d]=45$

⇒ $\dfrac{9}{2}\times 2(a+4d)=45$

⇒ $9\times (a+4d)=45$

⇒ a + 4d = 5            ....(2)

Putting the value of (2) in (1), we get

∴ $a_{5}$ = 5

Thus, the value of $a_{5}$ is equal to 5.

Answer 2

\= 5

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