11. The sum of the first six terms of an A.P. is 42. The ratio of the 10th term to the 30th terms
of the A.P. is 1/3. Calculate the first and the 13th term.
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Answer:
a = 2
a13 = 26
Step-by-step explanation:
Given:
S6 = 42
a10 : a30 = 1:3
To find:
a1 and a13
Solution:
a10 : a30 = 1:3
a10/a30 = 1/3
a + 9d / a + 29d = 1/3
3(a+9d) = a + 29d
3a + 27d = a + 29d
3a - a = 29d - 27d
2a = 2d
a = d
S6 = 42
n/2[2a + (n-1)d ] = 42
6/2[2a + 5d] = 42
3(2a + 5a) = 42 .... since a = d..substitute...
7a = 42/3
7a = 14
a = 14/7
a = 2
First term = 2
d = 2 since a = d
a13 = a + 12d
2 + 12(2) = 2 + 24 = 26
13th term = 26
Hope it helps you:)
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