Question

If sum of first and last term of an A.P. is 230 then find the sum of third and third last term of same A.P.

Answer 1

The first term and common difference of an A.P  is 230.

Last term of an A.P is a + (n - 1)d

First term of an A.P is a and last term is a a(subscript)n

So a + a + (n - 1)d = 230

       2a + (n - 1)d = 230 or 2a + nd - d = 230................(1)

Now the sum of third and third last term of same A.P

     a3 + a(sub)n-2 = a + (3-1)d + a + (n - 2 - 1)d

                                = a + 2d + a + nd -3d

                     = 2a + nd -d----------------(2)

From 1 and 2 we have 2a + (n-1)d will get cancelled

So a3 + a(subscript)n-2 = 230

Answer 2

Let first term is $a_1$ and last term is $a_n$

A/C to question,
$a_1+a_n=230$

we know, $a_n=a_1+(n-1)d$, d is common difference of AP.

$a_1+a_1+(n-1)d=230$

$2a_1+(n-1)d=230$......(1)

now, 3rd term,$a_3=a_1+2d$
3rd last term,$a_{n-2}=a_1+(n-2-1)d=a_1+(n-3)d$
[ 3rd last term is $a_{n-2}$]

so, $a_3+a_{n-2}=a_1+2d+a_1+(n-3)d$

= $2a_1+(n-1)d$

from equation (1),

so, 3rd term + 3rd last term = 230