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Question
In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.
Solution
Here, we are given that 24th term is twice the 10th term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d
We have to prove that `a_72 = 2a_34`
So, let us first find the two terms.
As we know
`a_n = a + (n - 1)d`
For 10th term (n = 10)
`a_10 = a + (10 - 1)d`
= a + 9d
For 24 th term (n = 24)
`a_24 = a + (24 - 1)d`
= a + 23d
Now we given that `a_24 = 2a_10`
So we get
a + 23d = 2(a + 9d)
a + 23d = 2a + 18d
23d - 18d = 2a - a
5d = a .....(1)
Further, we need to prove that the 72nd term is twice of 34th term. So let now find these two terms,
For 34th term (n = 34),
`a_34 = a + (34 - 1)d`
= 5d + 33d (Using 1)
= 38d
For 72nd term (n = 72)
`a_72 = a + (72 - 1)d`
= 5d + 71d (using 1)
= 76d
= 2(38d)
Therefore `a_72 = 2a_34`
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