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6.  Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

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Numbers divisible by 4, yield remainder as 1 from 10 to 100 are 13,17,..................97

This sequence is an A.P.

Here , first term =a =13

common difference = 4.

We know , a_n = a+(n-1)d

              97 = 13+(n-1)4

      \Rightarrow \, \, 84 = (n-1)4

    \Rightarrow \, \, 21 = (n-1)

 \Rightarrow \, \, n=21+1=22

 

S_n = \frac{n}{2}[2a+(n-1)d]

       = \frac{22}{2}[2(13)+(22-1)4]

     = \frac{22}{2}[2(13)+21(4)]

      = 11\times 110

     =1210

The  sum of numbers divisible by 4 yield 1 as remainder  from 10 to 100 is 1210.

Posted by

seema garhwal

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