The sum of four consecutive numbers in an A P is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7 : 15 . Find the numbers.

Let the four consecutive number be a , a + d , a + 2 d , a + 3 d Given, that sum of four consecutive numbers in A . P is 32 . Therefore, a + a + d + a + 2 d + a + 3 d = 32 4 a + 6 d = 32 2 a + 3 d = 16 a = 16 - 3 d 2 . . . . . . . . . ( 1 ) The ratio of the product of the first and the last term to the product of two middle term is 7 : 15 . Therefore, a a + 3 d a + d a + 2 d = 7 15 ⇒ 15 a a + 3 d = 7 a + d a + 2 d ⇒ 15 a 2 + 45 a d = 7 a 2 + 21 a d + 14 d 2 ⇒ 8 a 2 + 24 a d - 14 d 2 = 0 ⇒ 4 a 2 + 12 a d - 7 d 2 = 0 ⇒ 4 a 2 + 14 a d - 2 a d - 7 d 2 = 0 ⇒ 2 a 2 a + 7 d - d 2 a + 7 d = 0 ⇒ 2 a + 7 d 2 a - d = 0 ∴ a = - 7 d 2 & a = d 2 . . . . . . . . . . . ( 2 ) Substitute the value of a in equation 2 So, a = - 7 d 2 ⇒ 16 - 3 d 2 = - 7 d 2 ⇒ - 4 d = 16 ⇒ d = - 4 and a = d 2 ⇒ 16 - 3 d 2 = d 2 ⇒ 4 d = 16 ⇒ d = 4 When d = 4 Then, a = d 2 = 4 2 = 2 Therefore, four consecutive number in an A P are 2 , 2 + 4 , 2 + 2 × 4 , 2 + 3 × 4 = 2 , 6 , 10 , 14 If we take d = - 4 then also we will get same numbers.

Solve: 1 + 4 + 7 + 10 + . . . + x = 287

1 + 4 + 7 + 10 + … + x = 287 Here a = 1 ; d = 3 & T n = x ∵ T n = x ⇒ a + n - 1 d = x ⇒ 1 + n - 1 × 3 = x ⇒ 3 n - 2 = x ⇒ n = x + 2 3 … ( i ) S n = n 2 a + l ⇒ 287 = n 2 1 + x ⇒ 574 = x + 2 3 ( 1 + x ) from – ( i ) ⇒ 1722 = x 2 + 3 x + 2 ⇒ x 2 + 3 x - 1720 = 0 ⇒ x + 43 x - 40 = 0 ⇒ x = - 43 , & x = 40 ∵ Sequence is increasing ∴ x = - 43 (reject) ∴ x = 40

Show that a - b 2 , a 2 + b 2 and a + b 2 are in A . P .

Assume that a - b 2 , a 2 + b 2 and a + b 2 are in A . P . So, difference between two consecutive terms will be same. a 2 + b 2 - a - b 2 = a + b 2 - a 2 + b 2 a 2 + b 2 - a 2 + b 2 - 2 a b = a 2 + b 2 + 2 a b - a 2 + b 2 a 2 + b 2 - a 2 - b 2 + 2 a b = a 2 + b 2 + 2 a b - a 2 - b 2 2 a b = 2 a b which is true. Hence, given terms are in A . P .

If the sum of first four terms of an AP is 40 and that of first 14 terms is 280 . Find the sum of its first n terms.

Let a be the first term and d be the common difference in AP , sum of n terms of the AP is denoted by S n . We know that S n = n 2 2 a + n - 1 d Given, S 4 = 40 ⇒ 4 2 2 a + 4 - 1 d = 40 ⇒ 2 2 a + 3 d = 40 ⇒ 2 a + 3 d = 20 . . . . ( i ) S 14 = 280 ⇒ 14 2 2 a + 14 - 1 d = 280 ⇒ 7 2 a + 13 d = 280 ⇒ 2 a + 13 d = 40 . . . . ( i i ) Subtracting ( i ) from ( i i ) , we get 10 d = 20 ⇒ d = 2 Substituting d value in i 2 a + 3 d = 20 ⇒ 2 a + 3 2 = 20 ⇒ 2 a + 6 = 20 ⇒ 2 a = 14 ⇒ a = 14 2 ⇒ a = 7 Now sum of first n terms, S n = n 2 14 + ( n - 1 ) 2 S n = n 2 14 + 2 n - 2 = n 2 ( 2 n + 12 ) = n ( n + 6 ) Hence, the sum of first n terms is n ( n + 6 ) .