Question

An arithmetic progression consists of 37 terms . the sum of first three terms is 12 and the sum of last 3 terms is 318 find the first three numbers and no last three numbers​

Answer 1

Answer:

Let a,d and n be the first term, common difference and the no. of terms of given AP.

Here n=37

so middle most term is

2

n+1

⇒

2

37+1

⇒18

th

∴ three middle most terms are 18

th

,19

th

,20

th

a

18

+a

19

+a

20

=225⇒a+17d+a+18d+a+19d=225

⇒3a+54d=225 ...(1)

last three terms are 35

th

,36

th

,37

th

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

⇒3a+105d=429 ...(2)

subtracting (1) from (2), we get

51d=204

⇒d=4

putting d in (1), we get

3a+54×4=225

3a=225−216⇒a=3

Therefore the AP is a,a+d,a+2d,a+3d,....