Question

plss answer...this maths question...#macha..refer attachment​

Answer 1

$</p><p>\huge{\fbox{\fbox{\bigstar{\mathfrak{\blue{The \: AP \: Is \: 3, \: 7, \: 11, \: ...... \: 147.}}}}}}$

Correction to the question: It would be ' last ' and not 'past '

The three middlemost terms of this AP are 18 th, 19th and 20 th terms.

The sum of these terms is 225.

Hence,

a + 17d + a + 18d + a + 19d = 225

Hence,

a + 18d = 75...............[|]

The sum of the last three terms is 429.

35 th + 36th + 37th = 429

a + 34d + a + 35d + a + 36d = 429

Hence,

a + 35d = 143...............[||]

Solving these two equations, we get:

d = 4, a = 3.

Hence, the required AP is 3, 7, 11, 15, 19........147