Question

Find the A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.

Answer 1

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

It is given that a₄ = 9 and a₆ + a₁₃ = 40. 

a₄ = 9 

`rArr a+(4-1)d=9`             `a_n=a+(n-1)d`

`rArr a+3d=9`   

`a_6+a_13=9`

`rArr {a+(6-1)d}+{a+(13-1)d}=40`

`rArr {a+5d}+{a+12d}=40`

`rArr 2a+17d=40`

From (1): 

a = 9 − 3d

Substituting the value of a in (2): 

2 (9 − 3d) + 17d = 40

⇒ 18 + 11d = 40

⇒ 11d = 22

⇒ d = 2

∴ a = 9 − 3 × 2 = 3 

Thus, the given A.P. is a, a + d, a + 2d …, where a = 3 and d = 2. 

Thus, the A.P. is 3, 5, 7, 9 …