Question

an arithmetic progression consists of 37 terms . the sum of the first 3 terms of it is 12 and the sum of its last 3 terms is 318, then find the first and last terms of the progression​

Answer 1

SOLUTION:- Let the first term be 'a' and common difference be 'd' respectively.

Since the A.P. contains 37 terms. So, the middle most term is (37+1)/2th term =19th term.

Thus, three middle most terms of A.P are 18th, 19th and 20th terms.

Given a+(18−1)d+a(19−1)d+a(20−1)d=225

⇒a+17d+a+18d+a+19d=225

⇒3a+54d=225

⇒3(a+18d)=225

⇒a+18d=75

⇒a=75−18d ……….(1)

According to given information

a+(35−1)d+a+(36−1)d+a+(37−1)d=429

⇒a+34d+a+35d+a+36d=429

⇒3a+105d=429

⇒3(a+35d)=429

⇒(75−18d+35d)=143

⇒17d=143−75=68

⇒17d=68

⇒d=4 ∴ Thus, the A.P. 3,7,11,15,....

Put d=4 in (1)

a=75−(8(4))

a=75−72

a=3.

Answer 2

Answer:

According to question :

a1 + a2 + a3 = 12

=> a + a+d + a+2d = 12

3a + 3d = 12

a + d = 4 -------------- (1)

a35 + a36 +a37 = 318

a+34d + a+35d + a+36d =318

3a + 105d = 318

a + 35d = 106 -----------------(2)

Now subtract (1) from (2)

34d = 102

d = 3

put this d in (1) we get

a = 1 (which is first

we know, An = a + (n-1)d

= 1 + ( 37-1)(3)

= 1 + ( 36)(3)

=> An = 109 ( which is last term of A.P.)