Question

The sum of the 1st 6 termsof an ap is 42.The ratio of the 10th term to its 30th term is 1:3. Find the 1st term and the thirteenth term os the ap

Answer 1

Answer:

s6=42

n/2[2a+(n-1)d]=42

6/2[2a+(6-1)d]=42

3[2a+5d]=42

2a+5d=14..........…eq(1)

a10/a30=1/3

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

by cross multiplying we get

3a+27d=a+29d

3a-a=29d-27d

2a=2d

a=d.............eq(2)

substitute eq(2)in (1)

2a+5a=14

7a=14

a=2

from eq(2),we get ,d=2

the first term of the AP is 2

a13=a+12d

a13=26