Question

The last three terms of an A.P are 297,303,309.Find the fourth and fifth term from the end ,respectively.

Answer 1

Answer:

291, 285

Step-by-step explanation:

if a, b, c, d, so on x, y, z are in an A. P, then z, y, x, so on c, b, a are in an A. P but with a negative common difference {Where a, b, c, so on x, y, z are terms of an A.P

so 309, 303, 297, so on are in an ap

for 4th term,                                                     for 5th term

a              = 309                                                it's a4 + d

d(C. D)     = 303 - 309 = - 6                                which is

n              =  4                                                     291 - 6

so                                                                         = 285

a4 = a + (n - 1)d

a4 = 309 + (4-1) (- 6)

    = 309 + 3(- 6)

    = 309 - 18  

    = 291