Find the sum of all integers between 40 and 250 which are multiples of 7
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Answered by
3
All integers between 50 and 500, which are divisible by 7 are
56,63,70,...…,497
which forms an A.P
first term of this A.P is a
1
=56
second term of this A.P is a
2
=63
last term of this A.P is a
n
=497
common difference
d=a
2
−a
1
⟹d=63−56=7 .
nth term of this A.P is given by
a
n
=a
1
+(n−1)d
put a
n
=497;a
1
=56 and d=7 in above equation we get,
⟹497=56+(n−1)7
⟹7n−7+56=497
⟹7n=497−49=448
n=
7
448
=64 number of terms in this A.P
now, sum of these n=64 terms is given by
S
n
=
2
n
(a
1
+a
n
)
put values of n=64;a
1
=56;a
n
=497 we get
S
64
=
2
64
(56+497)
⟹S
64
=
2
64
×553
⟹S
64
=32×553
⟹S
64
=17696
hence the sum of all integers between 50 and 500 which are divisible by 7 , is S
64
=17696
Answered by
2
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