A’ is the nth term of the given series and ‘B’ is the (n + 1)th term of the given

series. Choose the correct statement from the following statements (i), (ii) & (iii).

Series: 1402, 698, 346, 170, 82, 38, 16, 5

(i) A = Two times of B + 6

(ii) 6B = Three times of A + √64 3

(iii) Difference between 1st and 2nd term from left side is two times of the difference

between 2nd and 3rd term and so on.

Solution

In the given series – If we take A = 1402 So, B = 698 Similarly, A = 698 So, 346 = B and so on -----   For (i) 1402 = 2 ×698+6                      1402 = 1402 Similarly, 698 = 346 ×2+6   698 = 698 So, if we take any term of given series, then statement (i) following.   For (ii) 6 × 698 = 1402 ×3+4             4188 ≠ 4210 So, statement (ii) did not follow if we take any term of the given series as ‘A’ & ‘B’   For (iii) (1402 – 698) = 2 ×(698−346)            704 = 2 ×352             704 = 704 So, if we take any term of given series as A and B, then statement (iii) following