Question

for a.p. 2, 5, 8, 11,… 302, show that twice the middle term is equal to sum of first and last term.

Answer 1

Proof :

The A.P. is 2, 5, 8, 11, . . ., 302

1st term, a = 2

Common difference, d = 5 - 2 = 3

  Let, the nth term is 302

  ⇨ a + (n - 1)d = 302

  ⇨ 2 + (n - 1) * 3 = 302

  ⇨ 2 + 3n - 3 = 302

  ⇨ 3n - 1 = 302

  ⇨ 3n = 302 + 1

  ⇨ 3n = 303

  ⇨ n = 101

Thus, there are 101 terms in the A.P.

So, the middle term be the $\bold{\dfrac{101+1}{2}}$ st term,

i.e., 51st term

  The 51st term = a + (51 - 1)d

  = 2 + 50 * 3

  = 2 + 150

  = 152

⇨ 2 * 51st term = 152 * 2 = 304

Now, the sum of the 1st term and the last term

  = 2 + 302 = 304

Hence, proved.

#MarkAsBrainliest

Answer 2

$<b>$
The A.P. is 2, 5, 8, 11, . . ., 302

1st term, a = 2

Common difference, d = 5 - 2 = 3

  Let, the nth term is 302

  a + (n - 1)d = 302

 2 + (n - 1) * 3 = 302

 2 + 3n - 3 = 302

  3n - 1 = 302

  3n = 302 + 1

  3n = 303

   n = 101

Thus, there are 101 terms in the A.P.

101.+1 / 2 term,

i.e., 51st term

  The 51st term = a + (51 - 1)d

  = 2 + 50 * 3

  = 2 + 150

  = 152

⇨ 2 * 51st term = 152 * 2 = 304

Now, the sum of the 1st term and the last term

  = 2 + 302 = 304