Question

Find the sum of all two digits no. Greater than 50. Which when divided by 7 leave remainder 4.

Answer 1

The First and last no. between 50 and 99 which on divided by 7 leave remainder 4 are 53 and 95 respectively.
Since, all the no. between 50 and 99 which are divisible by 7 and leaves remainder 4 are in AP, where a = 53, l = 95, d = 7
Now, l=a+(n-1)×d
=>95=53+(n-1)7
=>95-53=(n-1)7
=>42=(n-1)7
=>42/7=n-1
=>6=n-1
Therefore, n=7
Hence, there are 7 no.s between 50 and 99 which on divided by 7 leave remainder 4 are 53 and 95 respectively.
Now, sum of all 7 no.s is find as:
=>S7=7/2[2×53+(7-1)×7]
=>S7=7/2×148
=>7×74=518

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Answer 2

so here is the answer.
a=53 as it gives a reminder 4 when divided by 7(to get last number start from 99)
d=7
last term=95
a+(n-1)d=95
53+(n-1)7=95
7n=(95+7)-53
7n=49
n=7
n/2(2a+(n-1)d)
7/2(106+6x7)
7/2(148)
7x74=518.
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