Question

insert 6 numbers between 3 and 34 that the resolving sequence is AP​

Answer 1

Answer:

for 6 numbers

Let A1 A2 A3 ...A6 be the required numbers

3 A1 A2 etc A6 34

so totally 8 terms

with a=3

a8=a+(8-1)d=34

3+7d=34

34-3/7=31/7

so A1=3+31/7

A2=3+2(31/7)

.

.

.etc

A6=3+5(31/7)