Question

Find the middle term of the sequence formed by all three-digit numbers

which leave a remainder 2, when divided by 7. ​

Answer 1

Answer:

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A.P 103,110,117,……..,999

a=103

d=110−103=7

n be the number of terms

a

n

​

=999

103+(n−1)7=999

(n−1).7=999−103

⇒(n−1).7=896

⇒n−1=896/7

⇒n−1=128

⇒n=128+1

⇒n=129

Middle term =

2

n+1

​

th

term =

2

129+1

​

=65

th

term.

=a+6d=103+64×7=551

Sn=

2

n

​

[2a+(n−1)d]

S

64

​

=

2

64

​

[2.103+63.7]

⇒S

64

​

=20704

S

129

​

=

2

129

​

[2.103+128.7]

⇒S

129

​

=71079

S

129

​

−(S

64

​

+551)=71079−(20704+551)

=49824.

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