Seems to be confusing.. How do I get it?
Attachments:
Answers
Answered by
2
Heyaa!!!
Consider a and d as the first term and the common difference of an A.P. respectively.
n th term of an A.P., an = a + ( n – 1)d
Sum of n terms of an A.P., S n = n/ 2 [2a + (n – 1)d]
Given that the sum of the first 10 terms is 210.
⇒ 10 / 2 [2a + 9d ] = 210
⇒ 5[ 2a + 9 d ] = 210
⇒2a + 9d = 42 ----------- (1)
15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning
⇒ a36 = a + 35d
Sum of the last 15 terms = 15/2 [2a36 + ( 15 – 1)d ] = 2565
⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565
⇒ 15 [ a + 35d + 7d ] = 2565
⇒a + 42d = 171 ----------(2)
From (1) and (2), we have d = 4 and a = 3.
Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199.
Consider a and d as the first term and the common difference of an A.P. respectively.
n th term of an A.P., an = a + ( n – 1)d
Sum of n terms of an A.P., S n = n/ 2 [2a + (n – 1)d]
Given that the sum of the first 10 terms is 210.
⇒ 10 / 2 [2a + 9d ] = 210
⇒ 5[ 2a + 9 d ] = 210
⇒2a + 9d = 42 ----------- (1)
15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning
⇒ a36 = a + 35d
Sum of the last 15 terms = 15/2 [2a36 + ( 15 – 1)d ] = 2565
⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565
⇒ 15 [ a + 35d + 7d ] = 2565
⇒a + 42d = 171 ----------(2)
From (1) and (2), we have d = 4 and a = 3.
Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199.
Similar questions