middle term
of a
sequence
found by
all
3
digit number
which
live a
reminder 5
when divided by 7
Answers
Answered by
1
Step-by-step explanation:
The three digit number when divided by 7 divided reminder 5 are.
103, 110, 117,----- 999.
This forms an A.P with common difference = 7 and first term = 103.
Now a
n
=a+(n−1)d
999=103+(n−1)7
7
896
=n−1 ⇒n=128+1=129.
The middle term of sequence is 65
th
term.
and it is
a
65
=103+64×7=103+448
[a
65
=551].
Now sum of term before 65
th
term
s
n
=
2
n
[2a+(n−1)d] =
2
64
[2×103+63×7]
s
n
=32[206+441]=20,704.
sum of term after 65
th
term
S
n
=
2
64
[2×558+63×7]=49,824.
∴[S before middle term = 20,704] & [S after middle term = 49824].
and middle term is [a
65
=551]
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