Question

The sum of first 6 terms of AP is 42 the ratio of its 10th term to 30th terms is 1:3 calculate the first and 13th term of the AP​

Answer 1

given

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

n=6

S6=6/2[2a+5d]

given S6=42

42=3[2a+5d]

42/3=2a+5d

14=2a+5d. ---->mark as equation 1

t10/t30=1/3

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

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

3a+27d=a+29d

3a-a=29d-27d

2a=2d

a=d. ------->mark as equation 2

substitute a=d in equation 1

14=2a+5d

14=2d+5d

14=7d

d=2 -----> mark as equation 3

from equation 2

we get a=2----->equation 4

we need to find t13

t13=a+12d

using equation 3 and 4

d=2,a=2

t13=2+12*2

t13=2+24

t13=26

answer is t13=26

thanks for asking