Find the 15 t h term of the series 3 + 1 3 + 1 3 3 + … .

Given series is 3 + 1 3 + 1 3 3 + ⋯ ⋯ This is a GP with a 1 = 3 and common ratio r = a 2 a 1 = 1 3 × 1 3 = 1 3 n t h term in a GP is a n = a 1 r n - 1 , where a 1 is first term and r is common ratio. ∴ a 15 = a r 14 = 3 1 3 14 ∴ 15 t h term of given GP is 3 1 3 14 .

Find the 6 t h and the n t h terms of the sequence 3 2 , 3 4 , 3 8 , … .

Given sequence is 3 2 , 3 4 , 3 8 , ⋯ This is a GP with a 1 = 3 2 and common ratio r = a 2 a 1 = 3 4 × 2 3 = 1 2 n t h term in a GP is a n = a 1 r n - 1 , where a 1 is first term and r is common ratio. ∴ a 6 = a r 5 = 3 2 1 2 5 = 3 2 6 ∴ 6 t h term of given GP is 3 64 . a n = a r n - 1 ⇒ a n = 3 2 1 2 n - 1 ∴ n t h term of a given GP is 3 2 n .

The first term of an A.P is 5 , the last term is 45 and the sum is 400 . Find the number of terms and the common difference.

Given, a = 5 . Last Term of an A.P l = 45 . Sum of n terms, S n = 400 . We know that: S n = n 2 a + l ⇒ 400 = n 2 5 + 45 n = 800 50 = 16 Hence, there are 16 terms in the given A.P. We know that general terms of an arithmetic progression as: a n = a + n - 1 d a 16 = a + 15 d 45 = 5 + 15 d d = 45 15 = 8 3 Hence, the common difference of an A.P is 8 3 .

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