Question

in a certain A.P. the 24th term is twice the 10th term.prove that the 72nd term is twice the 34th term

Answer 1

Using fomula of n th term = a + (n - 1)d

given a + (24 - 1)d = 2(a + (9 - 1)d)
solving the equation  we get a = 5d
again taking
a + (24 - 1)d = 2(a + (9 - 1)d)
⇒3a + (72 - 3)d = 6a +54d
⇒(a + (72 - 1)d) + 2 (a - d) = 6a + 54d
⇒(a + (72 - 1)d) = 2a + 10d + 56d
⇒a + (72 - 1)d) = 2(a + (34 - 1)d)
hence proved 72nd term is twice the 34th term.
i am done minor step jumps if you have any problem comment under the answer.

Answer 2

let first term of the AP be A , difference be D, then,
24th term = A+23Dand ..(1)
10th term= A+9D ,
72nd term= A+71D
34th term= A+33D
Now, ATQ,
24th term = 2(A+9D) , ....(2).
also , from(1),
A+23D = 2(A+9D).=
A+23D= 2A+18D.
on further solving it,
A= 5D. (2)

we have to prove: 72nd term is twice the 34th term
i.e. A+71D= 2(A+33D)
A+71D= 2A+66D.
on further solving it,
A= 5D (3)
from (2) and (3) ..we can see that the values are getting equal...
Hence , proved.