3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 are consecutive terms in an arithmetic sequence
Number of terms = ......................
Middle term = ......................
If we add the terms which are at same distances from the two ends , then
x1 + x9 = 3 + 27 = 30
x2 + x8 = ..6.... + ...24.... = 30
x3 + x7 = ....9.. + .....21.. = 30
x4 + x6 = ..12.... + ....18... = 30
a) Can any relation between sums of each pair?
b) How the sum is related to the middle term?
please say the answers of option a and b
Answers
Answered by
2
Answer:
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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Answered by
1
Answer:
3'6'9'12'15'18'21'24'27'30
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