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Sum of all two digit numbers which when divided by 4 yield unity as remainder is. |
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Answer» The two-digit numbers, which when divided by `4`, yield `1` as remainder, are, `13, 17, … 97`. This series forms an A.P. with first term, `a = 13` and common difference, `d =4`. Let `n` be the number of terms of the A.P. It is known that the nth term of an A.P. is given by,` a_n = a + (n -1) d`. `:. 97 = 13 + (n -1) (4)` `=> 4 (n –1) = 84` `=> n – 1 = 21` `=> n = 22` Now, `S_22 = n/2(2a+(n-1)d) = 22/2(2(13)+(22-1)4) = 11**110 = 1210` So, the required sum will be `1210`. |
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