The sum of first 6 terms of an arithmetic progression is 42 the ratio of its 10th term to the 30th term is 1 is to 3 calculate the 1st and 13th term of the AP

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

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