Question

The sum of first six terms of an A.P. is 42. The ratio of its 10th term to its 30th

term is 1 : 3 calalate the first and thirteenth term of the A.P.

Answer 1

S6 =42

a + 9d                    1

------------     =       -------

a + 29d                  3

cross multiply we  get

3a + 27d = a +29 d

2a - 2d = 0 ------------------ (1)

 

its given that  

sum of first six terms of an AP is 42

therefore  

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

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

42 = 3 (2a + 5d )

14 = 2a +5d

2a +5d = 14 ------------------- (2)

solve eq 1 and 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

-----------------------------------------

Answer 2

GIVEN:-
S6=42
10thterm:30thterm=1:3

TO FIND:-
First term and 13th term

Solution:-

S6 =42

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

cross multiply we  get

3a + 27d = a +29 d

2a - 2d = 0 ------------------ (1)
 
its given that 
sum of first six terms of an AP is 42

therefore 

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

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

42 = 3 (2a + 5d )

14 = 2a +5d

2a +5d = 14 ------------------- (2)

solve eq 1 and 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 it helps you
✌✌✌