Question

Find the sum of all 3-digit numbers which leave the remainder 1, when divided by 4.
[Hint: First term =101, last term =997]

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Answer 1

$\small \bf \red{hey \: mate \: here \: is \: ur \: ans} \\ \\ \small \bf \red{ \huge \: solution} \\ \\ \small \bf \red{given \: that} \\ \\ \small \bf \red{first \: term \: (a1) = 101} \\ \\ \small \bf \red{2nd \: term (a2)=104 } \\ \\ \small \bf \red{ \therefore \: common \: difference \: (d) = (a2 - a1)} \\ \\ \small \bf \red{d = 105 - 101 = 4} \\ \\ \small \bf \red{last \: term \: (an) =997 } \\ \\ \small \bf \red{ \boxed{ \bf \: using \: this \: formula}} \\ \\ \small \bf \red{an = a + (n - 1) \times d} \\ \\ \small \bf \red{997 = 101 + (n - 1) \times 4} \\ \\ \small \bf \red{997 - 101 = 4n - 4} \\ \\ \small \bf \red{896 + 4= 4n} \\ \\ \small \bf \red{900 = 4n} \\ \\ \small \bf \red{n = \frac{900}{4} } \\ \\ \small \bf \red{n = 225} \\ \\ \small \bf \red{ \boxed{again \: using \: to \: this \: formula}} \\ \\ \small \bf \red{sn = \frac{n}{2}(2a + (n - 1) \times d) } \\ \\ \small \bf \red{sn = \frac{225}{2}(2 \times 101 + (225 - 1) \times 4) } \\ \\ \small \bf \red{sn = \frac{225}{2} (202 + 224 \times 4)} \\ \\ \small \bf \red{sn = \frac{225}{2} ( 202 + 896)} \\ \\ \small \bf \red{sn = \frac{225}{2} \times 1098 } \\ \\ \small \bf \red{sn = 225 \times 549} \\ \\ \small \bf \red{sn = 123525 \: \: ans}$