Two A.P.’s are given 9, 7, 5, . . . and 24, 21, 18, . . . . If nth term of both the progressions are equal then find the value of n and nth term.

Given A.P. is 9, 7, 5….

Whose first tern a = 9

Second term t₁ = 7

Third term t₃ = 5

Common difference d = t₃ – t₂ = 5 – 7 = – 2

Another A.P. is 24, 21, 18, . . .

Whose first tern a = 24

Second term t₁ = 21

Third term t₃ = 18

Common difference d = t₃ – t₂ = 18 – 21 = – 3

We have been given, n^th term of both the A.P. is same

thus, by using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Hence, by given condition we get,

⇒ 9 + (n – 1) × ( – 2) = 24 + (n – 1) × ( – 3)

⇒ 9 – 2n + 2 = 24 – 3n + 3

⇒ 11 – 2n = 27 – 3n

⇒ 3n – 2n = 27 – 11

⇒ n = 16

Thus, value of n^th term where a = 9, d = – 2, n = 16 is

⇒ t_n = 9 + (16 – 1) × ( – 2)

⇒ t_n = 9 – 15 × 2

⇒ t_n = 9 – 30 = – 21