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Question
Determine the A.P. Whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
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Solution
For given A.P
`t_3 = a + 2d = 16` ....(i)
Now
`t_7 - t_5 = 12`
`=> (a + 6d) - (a + 4d) = 12`
⇒ 2d = 12
⇒ d = 6
Substituting the value of din (i) we get
`a + 2 xx 6 = 16`
⇒ a + 12 = 16
`=> a = 4`
Thus the required A.P = a, a + d, a + 2d, a + 3d,..
= 4, 10, 16, 22,...
RELATED QUESTIONS
Find the sum of the first 15 terms of each of the following sequences having the nth term as
`a_n = 3 + 4n`
The sum of the first n terms of an AP in `((5n^2)/2 + (3n)/2)`.Find its nth term and the 20th term of this AP.
If Sn denote the sum of n terms of an A.P. with first term a and common difference dsuch that \[\frac{Sx}{Skx}\] is independent of x, then
Q.3
In an A.P. (with usual notations) : given a = 8, an = 62, Sn = 210, find n and d
If an = 3 – 4n, show that a1, a2, a3,... form an AP. Also find S20.
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
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The 5th term and the 9th term of an Arithmetic Progression are 4 and – 12 respectively.
Find:
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- common difference
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