Find the sum of all two-digit numbers which,w | vedclass

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Question

(A) The two-digit numbers,which when divided by $4$ yield $1$ as a remainder,are $13, 17, \ldots, 97$.This series forms an $A.P.$ with the first term

Vedclass · Accepted Answer

$1210$

Vedclass · Answer

(A) The two-digit numbers,which when divided by $4$ yield $1$ as a remainder,are $13, 17, \ldots, 97$.This series forms an $A.P.$ with the first term $a = 13$ and common difference $d = 4$.Let $n$ be the number of terms in this $A.P.$The $n^{th}$ term formula is $a_{n} = a + (n - 1)d$.Substituting the values,we get $97 = 13 + (n - 1)(4)$.$84 = (n - 1)(4) \implies n - 1 = 21 \implies n = 22$.The sum of $n$ terms of an $A.P.$ is $S_{n} = \frac{n}{2}[a + a_{n}]$.$S_{22} = \frac{22}{2}[13 + 97] = 11 	imes 110 = 1210$.Thus,the required sum is $1210$.

Find the sum of all two-digit numbers which,when divided by $4$,yield $1$ as a remainder.

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