Question

the first term of an arithemetic sequence is 1 & the sum of 1st 4 term is 100. write the sequence.​

Answer 1

Step-by-step explanation:

First term of the sequence is 10 and common difference is 3.

a

1

= 10 and d = 3

Next term = a

2

=a

1

+d=10+3=13

a

3

=a

2

+d=13+3=16

Thus, first three terms of the sequence are 10, 13 and 16.

Let 100 be the nth term of the sequence.

a

n

=a

1

+(n−1)d

100=10+(n−1)3

90=(n−1)3

n−1=30

n=31, which is a whole number.

Therefore, 100 is the 31

st

term of the sequence.

Answer 2

Step-by-step explanation:

First number be n

Try with different A.P with different difference

1)With difference d=2

n+n+2+n+4+n+6=100

4n+12=100

n=88/4

n=22

Therefore first A.P 22,24,26,28….

2)With difference d=4

n+n+4+n+8+n+12=100

4n=100–24

4n=76

n=19

Second A.P 19,23,27,31…

3)With difference d=6

n+n+6+n+12+n+18=100

4n=100–36

4n=64

n=16

Third A.P 16,22,28,34…

4)With difference d=8

n+n+8+n+16+n+24=100

4n=100–48

4n=52

n=13

Fourth A.P 13,21,29,37