Question

find the middle term of the sequence formed by all the numbers between 9 and 95 which leaves a remainder 1 when divided by 5 also find the sum of the numbers on both sides of the middle term separately

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Answer 1

Answer:

ANSWER

Here, the required numbers are,

⇒10,13,16,......,94

Here,

a=10

d=3

l=94

Then,

l=a+(n−1)d

94=10+(n−1)3

3n−3=84

n=

3

87

n=29

Here, the number of terms in the AP is odd. So, it has only one middle term and it will be 15

th

term.

So,

a

15

=10+(15−1)3

a

15

=52

Now, sum of first 14 terms is,

S

14

=

2

14

[2×10+(14−1)3]

S

14

=

2

14

[59]

S

14

=413

Again, sum of last 14 terms is,

⇒S

29

−[S

14

+a

15

]

⇒

2

29

[10+84]−(413+52)

⇒1508−465

⇒1043

Hence, the required middle term is 52, the sum of first half terms is 413 and the sum of last half terms is 1043.