In the following sequence 11, 88, 16, 80, 21, 72, _ , _ , _ , _ the blanks are two digit numbers. No number in the blank ends with
(i) 1
(ii) 4
(iii) 6
(iv) 7
Given:
Given sequence is 11, 88, 16, 80, 21, 72, _ , _ , _ , _
To do:
We have to find the numbers in the blanks do not end with which among the given options.
Solution:
The given sequence is formed by two series together.
The first series is 11, 16, 21......
The second series is 88, 80, 72.....
In the first series,
$a=11, d=16-11=5$
Therefore, the next two terms in the series are $21+5=26, (21+5)+5=31$.
 Similarly,
In the second series,
$a=88, d=80-88=-8$
Therefore, the next two terms in the series are $72-8=64, (72-8)-8=56$.
The sequence so formed is 11, 88, 16, 80, 21, 72, 26, 64, 31, 56....
As we can see the numbers in the blanks do not end with 7 among the given options.
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