1.

) Consider an arithmetic sequence whose 7th term is 34 and 15th term is 66a) Find the common differenceb) Find the 20th term.

Answer»

the correct value of common differences in this question is 4 and 20th term is 86

7th term =34, 15th tern =66; a+6d=34; a+14d=66/8d=32; d=32/4=8, a+6d=34 :; a+14d=66/8d=32; d=32/8=4; a+6d=34=a+6(4)=35; a=35-24=10; 25th term; a+( n-1)d=10+( 25-1)4=10+24(4)=10+96=106

20 th term ; a=10, d=4; 10+(19)(4)=10+76=86

c d is 4 and 20th terms is 86

10,14,18,22,26,30,34,38,42,46,50,54,58,62,66 is the arithmetic sequence

common difference---4

82 is the 20th term



Discussion

No Comment Found