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Question
The sum of first 6 terms of an AP is 42. The ratio of its 10th term to its 30th term is 1:3 . Calculate the first and the thirteenth term of the AP.
The sum of first 6 terms of an AP is 42. The ratio of its 10th term to its 30th term is 1:3 . Calculate the first and the thirteenth term of the AP.
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Solution
As usual, Let the first term be a and the common difference be d
thus, A to Q,
S6 = 42
=> 6/2 (2a + 5d) = 42
=> 3 (2a + 5d) = 42
=> 2a + 5d = 14 ----- (i)
Again, T10/T30 = 1 : 3
=> (a + 9d)/(a + 29d) = 1/3
=>a + 29d = 3a + 27d
=> 2a - 2d = 0
=> a - d = 0 --------- (ii)
On solving (i) and (ii) we get,
a = 2 and d = 2
Thus, first term = 2 and
Thirteenth term = 2 + 12(2) = 2 + 24 = 26
As usual, Let the first term be a and the common difference be d
thus, A to Q,
S6 = 42
=> 6/2 (2a + 5d) = 42
=> 3 (2a + 5d) = 42
=> 2a + 5d = 14 ----- (i)
Again, T10/T30 = 1 : 3
=> (a + 9d)/(a + 29d) = 1/3
=>a + 29d = 3a + 27d
=> 2a - 2d = 0
=> a - d = 0 --------- (ii)
On solving (i) and (ii) we get,
a = 2 and d = 2
Thus, first term = 2 and
Thirteenth term = 2 + 12(2) = 2 + 24 = 26
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