24. An A.P. consists of 37 terms. The sum of the three middle most ter

1.	24. An A.P. consists of 37 terms. The sum of the three middle most terms is 225 and the sumof the last three terms is 429. Find the A.P
Answer» Let the first term and the common difference of the A.P areaanddrespectively. Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term = 19thterm. Thus, three middle most terms of this A.P.are 18th, 19thand 20thterms. Givena18+a19+a20= 225 ⇒ (a+ 17d) + (a+ 18d) + (a+ 19d) = 225 ⇒ 3(a+ 18d) = 225 ⇒a+ 18d= 75 ⇒a= 75 – 18d… (1) According to given information a35+a36+a37= 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 ofdin equation (1), it is obtained a= 75 – 18 × 4 = 3 Thus, the A.P. is 3, 7, 11, 15 …

24. An A.P. consists of 37 terms. The sum of the three middle most terms is 225 and the sumof the last three terms is 429. Find the A.P

Answer»

Let the first term and the common difference of the A.P areaanddrespectively.

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

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

Givena18+a19+a20= 225

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

⇒ 3(a+ 18d) = 225

⇒a+ 18d= 75

⇒a= 75 – 18d… (1)

According to given information

a35+a36+a37= 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 ofdin equation (1), it is obtained

a= 75 – 18 × 4 = 3

Thus, the A.P. is 3, 7, 11, 15 …