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
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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
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