The sum of first six terms of ap is 42 . The ratio of its 10th term to its 30th term is 1:3. Calculate the first and the thirteenth term of an ap
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Heya!
Given that sum of 6 terms is 42. That means S₆= 42. Also, n (number of terms)= 6.
We know that,
Sₐ= n/2 [2a + (n-1)d ]
⇒ 42= 6/2 [ 2a + (6-1) d]
⇒ 42= 3 [ 2a + 5d ]
⇒ 2a + 5d= 14 ................. (1)
Now, Ratio of 10th and 30th term is 1:3.
We know that 10th term= a + 9d. Also, 30th term= a+ 29d
ATQ, (a+ 9d) / (a+29d) = 1/3
So, 3a + 27d = a + 29 d (cross multiplication)
2a - 2d= 0
⇒ a - d= 0 ............ (2)
Solvind eq (1) and (2) we get,
a= 2 and d= 2
Thus first term will be 2.
And thirteenth term will be a+ 12d = 2+ 12* 2= 2+ 24 = 26.
Hope it helps :)
Given that sum of 6 terms is 42. That means S₆= 42. Also, n (number of terms)= 6.
We know that,
Sₐ= n/2 [2a + (n-1)d ]
⇒ 42= 6/2 [ 2a + (6-1) d]
⇒ 42= 3 [ 2a + 5d ]
⇒ 2a + 5d= 14 ................. (1)
Now, Ratio of 10th and 30th term is 1:3.
We know that 10th term= a + 9d. Also, 30th term= a+ 29d
ATQ, (a+ 9d) / (a+29d) = 1/3
So, 3a + 27d = a + 29 d (cross multiplication)
2a - 2d= 0
⇒ a - d= 0 ............ (2)
Solvind eq (1) and (2) we get,
a= 2 and d= 2
Thus first term will be 2.
And thirteenth term will be a+ 12d = 2+ 12* 2= 2+ 24 = 26.
Hope it helps :)
Anonymous:
nice ans :)
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