Question

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

Answer 1

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Question: An AP consists of 37 terms. the sum of the three middle most tern is 225 and the sum of the last three terms is 429. Find the AP

Method of Solution ⏪

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


According to the Question Statement:→

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


→ Since the A.P contains 37 terms.

→ Therefore,the middle most term is (37+1)÷2
= 19th term.

Hence Required three middle most terms of this Arithmetic Progression are 18th, 19th and 20th terms.

Given T18+ T19 + T20 = 225

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

→ 3(a + 18d) = 225

→ a + 18d = 75


→ a = 75 – 18d. ------- (A)

 
→ Adding All Arithmetic Sequence or Progression!→


T 35 + T36 + T37 = 429

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

→ 3(a + 35d) = 429

→ (75 – 18d) + 35d = 143

→ 17d = 143 – 75 = 68

d = 4

 
→ Substitute the value of d in Equation! (A)→

a = 75 – 18d 
→ = 75 – (18 × 4) = 3

 

Hence, Required Arithmetic Progression is 3, 7, 11, 15 ..


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Answer 2

Let Number terms : [a+ 17d + a + 18d + a + 19d = 225]

= 3a + 54d = 225

= a + 18d = 75.......(1)

Sum of the last 3 terms = 429

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

3a + 105d = 429

a + 35d = 143.........(2)

subtracting (ii) from i ----(1)

=> 17d= 68

=> d = 4

Substituting in (1)

a= 3

hence the ap is 3, 7,11..............