Question

find the sum of integers between 100 and 700 which on dividing by 11 leaves a remainder 7. please show the solution step by step

Answer 1

110 is completely devides by 11
then 110+7 = 117 is number which devides by 11 and leaving remainder 7

117 + 11 = 128

we know 693 is completely devided by 11 so
693+7 = 700 so 700 is Last term
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117+128+..............+700

it is in A.P so

first term = a = 117

common difference = d = 128-11 = 11

number of terms = n = ?

last term = Tn = 700

Sum of the terms = Sn = ?

Tn = a+(n-1)d

700 = 117+(n-1)11

700-117 = 11n -11

583 = 11n - 11

583 + 11 = 11n

594 = 11n

594/11 = n

54 = n
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Sn = n/2[a+last term]

S54 = n/2[a+T54]

S54 = 54/2[117+700]

S54 = 27[817]

S54 = 22059
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sum of integers between 100 and 700 which on dividing by 11 leaves a remainder 7 is 22059

Answer 2

Answer: 110 is completely divisible by 11 so 11 7 is the first number and 693 is completely Davis with by 7

So 700 is the last term.

Step-by-step explanation:

a =117

d=11

last no.=700

n=?

sn=?

proof::::::

an=a+(n-1)d

700=117 +11n-11

11n=594

so. n=54

now ,

sn= n/2 ( a+ an )

s54 = 27(117+700)

=27 × 817

=22059