Question

The sum of first six terms of an arithmetic progression is 42. The ratio of its 10th term to its 30th term is 1:3. Calculate the first and the 13th term of the AP

Answer 1

a10/a30 =1/3
a+9d/a+29d =1/3

by cross multiplying
3a+27d = a+29d
2a=2d
a=d

now;

S6 =42
n/2 (2a+(n-1)d) =42
6/2 (2a+(6-1)d) =42
3 (2a+5d) =42
2a+5d = 14
now put the value of a as d in this term
2a+5d =14
2d +5d =14
7d =14
d =2
also,
now put the value of d as 2 in this

14=2a+5d
14=2a+5 (2)
14=2a+10
4=2a
2 =a
now we have to find the value of 13 term
so
a13 = a+12d
= 2+12(2)
= 2+24
= 26
plz mark it as brainliest

Answer 2

let first term is a and differencing is d
given
S(6)=42
6/2[2a+(6-1)d]=42
3[2a+5d]=42
2a+5d= 14-----------------(1)

a(10):a(30)=1:3
a+9d: a+29d=1:3
3a+27d=a+29d
2a=2d
a=d

from (1)
2a=14 - 5a
7a=14
a=2
d=2

a(13)=a+12d
. =2+12x2
. =26