The 9th term of an A.P. is equal to 6 times its second term.
Question:

The $9^{\text {th }}$ term of an A.P. is equal to 6 times its second term. If its $5^{\text {th }}$ term is 22 , find the A.P.

Solution:

Let a be the first term and d be the common difference.

We know that, nth term = an a + (n − 1)d

According to the question,

a9 = 6a2
⇒ a + (9 − 1)d =  6(a + (2 − 1)d)
⇒ a + 8d =  6a + 6d
⇒ 8d − 6=  6a − a
⇒ 2= 5a

$\Rightarrow a=\frac{2}{5} d$


Also, a5 = 22
⇒ a + (5 − 1)d = 22
⇒ a + 4d = 22   ….(2)

On substituting the values of (1) in (2), we get

$\frac{2}{5} d+4 d=22$

$\Rightarrow 2 d+20 d=22 \times 5$

 

$\Rightarrow 22 d=110$

⇒ = 5

$\Rightarrow a=\frac{2}{5} \times 5 \quad[$ From (1) $]$

⇒ a = 2

Thus, the A.P. is 2, 7, 12, 17, …. .

 

 

 

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