Question

find the sum of all two digit number greater than 50 which when divided by 7 leaves remainder 4

Answer 1

Answer:  Sum of all two digit number greater than 50 is 518.

Step-by-step explanation:

Since we have given that

Arithmetic series greater than 50 which when divided by 7 leaves remainder 4 :

53,60,67,...........95

Here, first term = a = 53

common difference = d is given by

$a_2-a_1=60-53=7$

First, we need to find the number of terms:

$95=a+(n-1)d\\\\95=53+(n-1)7\\\\95=53+7n-7\\\\95=46+7n\\\\95-46=7n\\\\49=7n\\\\n=\frac{49}{7}\\\\n=7$

Now, we will find the sum of all two digit numbers:

$S_7=\frac{7}{2}(2\times 53+(7-1)7)\\\\S_7=\frac{7}{2}(106+6\times 7)\\\\S_7=\frac{7}{2}(106+42)\\\\S_7=\frac{7}{2}\times 148\\\\S_7=518$

Hence, Sum of all two digit number greater than 50 is 518.

Answer 2

Answer:

53,60..95

a=53

d=7

an=a+(n-1)d

95=53+(n-1)7

42/7=n-1

n=7

S7=7/2+(2*53+6d)

=7/2×148

=518

hope it helps...