Question

An AP consists of 37 terms . The sum of the three middle most terms is 225 and the sum of the last term is 429. Find the AP

Answer 1

Let the first term and the common difference of the A.P are a and d respectively.

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

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

Given a 18 + a 19 + a 20 = 225

⇒ (a + 17d) + (a + 18d) + (a + 19d) = 225

⇒ 3(a + 18d) = 225

⇒ a + 18d = 75

⇒ a = 75 – 18d … (1)

According to given information

a 35 + a 36 + a 37 = 429

⇒ (a + 34d) + (a + 35d) + (a + 36d) = 429

⇒ 3(a + 35d) = 429

⇒ (75 – 18d) + 35d = 143

⇒ 17d = 143 – 75 = 68

⇒ d = 4

Substituting the value of d in equation (1), it is obtained

a = 75 – 18 × 4 = 3

Thus, the A.P. is 3, 7, 11, 15 …

Hope! This will help you.

Cheers!

Answer 2

so, n=37 (given)

middle term will be ((n-1)/2)+1 = 19th
so the thre middle most terms will be 18th, 19th & 20th

nth term in AP is written as: a+(n-1)d
where a is the first term and d is the common difference of the given arithmetic progression.

So, the sum of 18th 19th and 20th terms will be
(a+17d)+(a+18d)+(a+19d) = 225
3a+ 54d=225
a+18d=75..........(i)

by "sum of the last terms" I suppose you meant sum of the last three terms.

(a+34d)+(a+35d)+(a+36d) = 429
3a+105d= 429
a+35d=143 ...........(ii)

no Sab strecting equation (i) from equation (ii) we get.
17d=68
d=4

to find a let's substitute the value of d in equation (ii)
a=143-(35x4)
a=143-140
a=3

show the arithmetic progression will be
3 7 11 15 19 23 27 31 35 39 43 47 51 55.... and it goes up to 143.