Question

If 8th term of an A.P is twice the 13th term then show that second term is twice the 10th term.

Answer 1

Answer:

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$\huge \fbox \colorbox{lime}{Solution࿐}$

$\tt a_{8} = 2(a_{13})$

$\tt = > a + 7d = 2(a + 12d)$

$\tt = > a + 7d = 2a + 24d$

$\tt = > a = - 7d$

Now,

$\tt a_{2} = 2( a_{10})$

$\tt = > a + d = 2(a + 9d)$

$\tt = > - 17d + d = 2( - 17d + 9d)$

$\tt = > - 16d = 2( - 8d)$

$\tt = > - 16d = - 16d$

$\tt = > 16d = 16d$

Therefore , L.H.S = R.H.S

Both are equal.

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Answer 2

Given: 8th term of an A.P is twice the 13th term

To Find: To show that second term is twice the 10th term.

Solution:

Formula used:

$a_{n}$ = a + ( n- 1) d

Since, $a_{8}$ = 2.$a_{13}$

     => a + (8 - 1)d = 2( a + (13 - 1) d)

     => a + 7d = 2a + 24 d

     => 2a - a = 17d -24d

      => a = -17d ---> (1)

Now,

L.H.S = $a_{2}$ = a + (2-1) d

                = a + d ---> (2)

               putting (1) in (2)

            $a_{2}$ = -17 + d = -16d

R.H.S = 2($x_{10}$)

          = 2(a + (10 - 1) d

          =2a + 18d --->(3)

           putting (1) in (3)

      = 2 x -17 + 18d

       = -34d + 16d = -16d

We can clearly see that ,

L.H.S = R.H.S

Hence, second term is twice the 10th term.