Question

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​

Answer 1

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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Answer 2

Answer:

3'6'9'12'15'18'21'24'27'30