Third term of an A.P. Is 16 and 7th term is 12 more than 5th term,

1.	Third term of an A.P. Is 16 and 7th term is 12 more than 5th term, then find AP.
Answer» Let first term of A.P. is a and common difference is d. Given a₃ = 16 a + (3 – 1)d = 16 ⇒ a + 2d = 16 …(i) According to question, a₇ = 12 – a₅ ⇒ a₇ – a₅ = 12 [a + (7 – 1)d] – [a+(5 – 1)d] = 12 ⇒ a₇ + 6d – a – 4d = 12 ⇒ a + 6d – a – 4d = 12 ⇒ 2d = 12 d = 12/2 = 6 Substituting value of d in equation (i) a + 2(6) = 16 ∴ a = 16 – 12 = 4 AP. a, a + d, a + 2d,… = 4, 4 + 6, 4 + 2 × 6,… = 4, 10, 16,… Hence, required A.P. is 4, 10, 16, 22,…

Third term of an A.P. Is 16 and 7th term is 12 more than 5th term, then find AP.

Answer»

Let first term of A.P. is a and common difference is d.

Given a₃ = 16

a + (3 – 1)d = 16

⇒ a + 2d = 16 …(i)

According to question, a₇ = 12 – a₅

⇒ a₇ – a₅ = 12

[a + (7 – 1)d] – [a+(5 – 1)d] = 12

⇒ a₇ + 6d – a – 4d = 12

⇒ a + 6d – a – 4d = 12

⇒ 2d = 12

d = 12/2 = 6

Substituting value of d in equation (i)

a + 2(6) = 16

∴ a = 16 – 12 = 4

AP. a, a + d, a + 2d,…

= 4, 4 + 6, 4 + 2 × 6,…

= 4, 10, 16,…

Hence, required A.P. is 4, 10, 16, 22,…