Question

FoRM CLAsS CHAPTER "arithmetic progression "

How many term are three in the A.P's given below 1)5 ,9, 13....................81? 2)3, 6, 9................. 111?​

Answer 1

Answer:

The given sequence is an A.P. with first term a=3 and common difference d=3. Let there be n terms in the given sequence. Then,

nth term=111

a+(n−1)d=111

3+(n−1)×3=111

n=37

Thus, the given sequence contains 37 terms.

Answer 2

Step-by-step explanation:

1)last term is 81and first term is 5

using the formula

x=a+(n-1)d

81-5/4=n-1

19=n-1

n=20

2)similar as the first answer putting the value

111-3/3=n-1

36+1=n

n=37

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