n A P consists of 37 terms. The sum of the three middle terms is 225 a

1.	n A P consists of 37 terms. The sum of the three middle terms is 225 and thesum of last three terms is 429. Find the AP23. A
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 …

n A P consists of 37 terms. The sum of the three middle terms is 225 and thesum of last three terms is 429. Find the AP23. A

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 …