Question

the sum of first six term of an A.P is 42. the ratio of it's 10 term to it's 30 term is 1:3 calculate the first and the 13 term of A.P​

Answer 1

Step-by-step explanation:

S6 = 42

a + 9d. 1

---------- = ----

a + 29d. 3

cross multiply we get

3a + 27d = a + 29d

2a - 2d = 0 -----------> eq 1

its given that

sum of first six terms of an AP is 42

S6 = n/2 ( 2a + ( n - 1 )d)

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

42 = 3 ( 2a + 5d )

14 = 2a + 5d ------------> eq 2

solve eq 1 and eq 2

2a - 2d = 0

2a + 5d = 14

we get

d = 2

a = 2

13th term of AP

= a + ( n - 1 ) d

= 2 + ( 13 - 1 ) 2

= 2 + 24

= 26

Hope its correct

plz mark me as brainlist and follow me