Find the sum of the series 1 - a + a 2 - … to n terms ( a ≠ - 1 ) .

Given the series : 1 - a + a 2 - … to n terms ( a ≠ - 1 ) . Since , sum of a G . P . series is represented by the formula, S n = a 1 r n - 1 r - 1 , when r ≠ 1 . S n represents the sum of the G . P . series upto n t h terms, a 1 represents the first term, r represents the common ratio and n represents the number of terms. Here, a 1 = 1 and r = - a 1 = - a ∴ S n = 1 × ( - a ) n - 1 - a - 1 ⇒ S n = 1 - ( - a ) n 1 + a

Find the sum of the sequence - 5 4 , 5 16 , - 5 64 , …

Given the sequence - 5 4 , 5 16 , - 5 64 , … We have to find out S = - 5 4 + 5 16 - 5 64 + . . . . . , which is an infinite G . P . with first term a = - 5 4 and common ratio r = 5 / 16 - 5 / 4 = - 1 4 Since , S = a 1 - r = - 5 4 1 - - 1 4 = - 5 4 × 4 5 = - 1 .

Find the 18 t h and 25 t h terms of the sequence defined by T n = n n + 2 , if n ˙ is even natural number 4 n n 2 + 1 , if n is odd natural number .

Given that, T n = n n + 2 , if n ˙ is even natural number 4 n n 2 + 1 , if n is odd natural number . Now T 18 = 18 18 + 2 [ ∵ 18 is even ] ⇒ T 18 = 360 T 25 = 4 ( 25 ) 25 2 + 1 [ ∵ 25 is odd ] ⇒ T 25 = 100 626 ⇒ T 25 = 50 313

Find a 10 for the sequence given by a 1 = 1 , a 2 = 6 and a n = a n - 1 a n - 2 , n ≥ 3

The sequence is given by a 1 = 1 , a 2 = 6 and a n = a n - 1 a n - 2 , n ≥ 3 . Put n = 3 , we get a 3 = a 2 a 1 = 6 1 = 6 and similarly a 4 = a 3 a 2 = 6 6 = 1 and a 5 = a 4 a 3 = 1 6 and a 6 = a 5 a 4 = 1 6 1 = 1 6 and a 7 = a 6 a 5 = 1 and a 8 = a 7 a 6 = 1 1 / 6 = 6 and a 9 = a 8 a 7 = 6 1 = 6 and a 10 = a 9 a 8 = 6 6 = 1 .