Question

please solve question number 8
answer is 518

Answer 1

Let a be the first term and d be the common difference.

Here First term a = 53(53 - 7 * 7 = 53 - 49 = 4)

Common difference d = 7.

Last term l = 95.


Now,

We know that Tn = a + (n - 1) * d

                       95 = 53 + (n - 1) * 7

                      95 = 53 + 7n - 7

                       95 - 53 + 7 = 7n

                       49 = 7n

                         n = 7.



Now,

Sum of n terms of an AP sn = n/2[2a + (n - 1) * d)

                                               = 7/2[2(53) + (7 - 1) * 7]

                                               = 7/2[106 + 42]

                                               = 7/2[148]

                                               = 7 * 74

                                               = 518.




Therefore, Sum of all two digit numbers which when divided by 7 yields remainder 4 is 518.



Hope this helps!

Answer 2

First use I= a+(n-1)×d
               =53+(n-1)×7
               (n-1) = 6
               n= 7

now use Sn= n/2[2a+(n-1)×d]
               S7=7/2[2×53+(7-1)×7]
               S7=7/2[106+42]
               S7=7/2[148]
               S7=7×74
               S7=518