Question

inser A.M.s between 7 and 71 in such a way that the 5th A.M. is 27. find the number of A.M.s.

Answer 1

$\huge{Hey Mate!!!}$

☆☞ Here is ur answer ☜☆

☆☞ Here , a=7 , c=71    [ Where 'a' 1st term of A.P. & 'c' last term]

5th A.M. is = 27

         a+5d=27         [ d= common difference]

         7+5d=27

              5d=20

                 d=4

 Now we have, d=(c-a)/(n+1)  [ n=Number of A.M]

                           4=(71-7)/(n+1)

                   4(n+1)=64

                           4n =60

                             n=15

Therefor number of A.M. is 15 


HOPE IT HELPS!!!

@BinDaSSboY

Answer 2

hey, here is your answer......

let the progression be ;

7, a, a+d, a+2d, a+3d,........, a+(n-1)d, 71 .....

given, 5th A.M is 27..... i.e., a+4d = 27....

a-7 = d.... ., a+4(a-7) = 27......

a+4a-28 = 27.....

5a-28 = 27......

5a = 27+28...….

5a = 55....

a = 11....

given, a+4d = 27....

11+4d = 27...

4d = 27-11.....

4d = 16....

d = 4.....

71- [a+(n-1)d] = d....

71-[11+(n-1)4] = 4.....

71 - [11+4n-4] = 4.....

71 - [4n+7] = 4......

71-4n-7 = 4.....

64-4n = 4.....

64-4 = 4n.....

60 = 4n....

n = 15.......

therefore, 15 A.Ms are inserted between 7, 71...

hope this helps you.....
thank you.....

plzz mark me as brainliest....