Question

Find the sum of all two digit number greater than 50 which when divided by 7 leaves remainder 4. Please it is an urgent .....

Answer 1

53 +60+67+74+81+88+95=518

Answer 2

Answer:

518

Step-by-step explanation:

Two digit numbers greater than 50, divisible by 7 leaving remainder 4 are:

53(49 + 4), 60(56 + 4), 67(63 + 4), 74(70 + 4).... 95(91 + 4).

So the series starts with 53 and ends with 95.

So, the AP will be 53, 60, 67, 74... 95.

Here,

First term = a = 54

Common difference = 7

Last term = an = 95.

(i) nth term of an AP :

an = a + (n - 1) * d

95 = 53 + (n - 1) * 7

95 = 53 + 7n - 7

95 - 46 = 7n

7n = 49

n = 7.

(ii) Sum of n terms of an AP:

Sn = (n/2)[a + l]

    = (7/2)[53 + 95]

    = (7/2)[148]

    = 518.

Therefore, Sum of all two digit numbers = 518.

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