find the sum of all natural number between 50 to 300 which are divible by 7
Answers 2
Answer:
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
Step-by-step explanation:
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Author:
dimash41o
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Answer:
Step-by-step explanation:
Given,
50 to 300 divisible by
To Find,
sum of all natural numbers between 50 to 300 divisible by 7
Solution,
The series of integers divisible by 7 between 50 and 300 are 56, 63, 70, …, 294.
Let the number of terms be ‘n’ So, a = 56, d = 63-56 = 7, an = 294
an = a + (n-1)d
294 = 56 + (n-1)7
294= 56 + 7n – 7
7n = 294– 56 + 7
7n = 245
n= 245/7 = 35
By using the formula, Sum of n terms,
S = n/2 [a + l]
= 35/2 [56 + 295]
= 35/2 [350]
=6125 ns
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Author:
max664
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Rate an answer:
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