Question

The 9th term of an AP is equal to six times it’s second term . If it’s 15th term is 22 find the AP

Answer 1

Question :

The 9th term of an AP is equal to six times it’s second term . If it’s 15th term is 22 find the AP.

Theory :

Genral term ( nth term ) of an AP is given by:

${\purple{\boxed{\large{\bold{T_{n} = a + (n - 1)d}}}}}$

Solution :

Let the first term be a and common difference d.

Nineth term

$t _{9} = a + (9- 1)d$

= a+8 d ...(1)

Second term

$t _{2} = a + (2- 1)d$

= a+ d ...(2)

According to the question :

nineth term =6× second term

a+8d = 6( a+d)

a + 8d = 6a + 6d

2d = 5a

d = 5a/2 ......(3)

Given fifteen term = 22

a+14d = 22

put d = 5a/2

$36a = 22$

$a = \frac{22}{36} = \frac{11}{13}$

⇒

$d = \frac{5a}{2} = \frac{5}{2} \times \frac{11}{13} = \frac{55}{36}$

first term ,a = 11/13

second term ,a+d = 77/36

third term ,a+2d = 66/13

The required Ap series ;

$</p><p> \frac{11}{13} , \frac{77}{36} , \frac{66}{13} ..... ...$