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accountability conceptual digit numbers which leave a remainder 1 on division by 5 write down the sequence please help me​

Answer»

Step-by-step explanation:

The SMALLEST and the largest numbers of two digits, which when divided by 5 LEAVE remainder 1 are 11 and 96 respectively. 

So, the sequence of two digit numbers which when  divide by 5 leave remainder 1 are 11,16,21,...,96. 

Clearly, it is an AP with first term a=11 and COMMON DIFFERENCE d=5.

Let there be n terms in this sequence. 

Then,

an=96⇒a+(n−1)d=96⇒11+(n−1)×5=96⇒n=18

Now, Required sum=2n[2a+(n−1)d]

                                 =218[2×11+(18−1)×5]

                                 =9×107=963.


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