In an A.P. 5th term is 17 and 9th term exceeds 2nd term by 35. Find the A.P.

Formula Used.

a_n = a + (n–1)d

a₅ = a + (5–1)d

a₅ = a + 4d

If 5^th term of A.P is given as 17

Then,

a + 4d = 17

we get a = 17–4d ......eq 1

a₉ = a + (9–1)d

a₉ = a + 8d

If 9^th term of A.P is given as 35 + 2^nd term

Then,

a₂ = a + (1)d

= a + d

a + 8d = 35 + [a + d]

we get

a–a + 8d–d = 35

7d = 35

d = = 5 ......eq 2

Putting d in eq 1 we get ;

a = 17–(4×5)

= 17–20 = –3

As a = –3 and d = 5

Then;

a₁ = a + (n–1)d = –3 + (1–1)(5) = –3

a₂ = a + (n–1)d = –3 + (2–1)(5) = –3 + (5)×1 = –3 + 5 = 2

a₃ = a + (n–1)d = –3 + (3–1)(5) = –3 + (5)×2 = –3 + 10 = 7

a₄ = a + (n–1)d = –3 + (4–1)(5) = –3 + (5)×3 = –3 + 15 = 12

a_n = a + (n–1)d = –3 + (n–1)(5) = –3 + 5n–5 = 5n–8

∴ The A.P is –3, 2, 7, 12, ……, 5n–8