Question

the sum of first 6 terms of an ap is 49. the ratio of its 10th term to its 30th term is 1:3 calculate the first and 13th term of this A.P​

Answer 1

Answer:

a=49/21

and T13=637/21

Step-by-step explanation:

S6=6/2(2a+5d)=49

6a+15d=49.............(1)

T10/T30=1/3

a+9d/(a+29d)=1/3

a+29d=3(a+9d)

2a=2d

a=d

from(1)

6a+15a=49

21a=49

a=d=49/21

now T13=a+12d

=49/21+12*49/21

=49/21( 1+12)=13*49/21=637/21