Question

consider the AP 2, 5,8,11,.....302 show that twice of the middle term of the above AP is equal to the sum of its first and last term​

Answer 1

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Answer 2

For the A.P.

first term, a = 2

common difference, d = 3

last term, A = 302

let the total number of terms be 'n'

An = 302

a + (n - 1)d = 302

2 + (n - 1)(3) = 302

(n - 1)(3) = 300

n - 1 = 100

n = 101

number of terms is odd, hence position of the middle term will be

$\frac{n + 1}{2} = \frac{101 + 1}{2} = \frac{102}{2} = 51$

hence, middle term–

$a_{51} = a + 50d = 2 + 50 \times 3 = 2 + 150 \\ a_{51} = 152 \\ 2 a_{51} = 304 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (i)$

sum of first and last term–

$\: \: \: \: \: \: a + a_{101} \\ = 2 + 302 \\ = 304 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (ii)$

from (i) and (ii)

$2 a_{51} = a + a_{101}$

Hence, for the given A.P, twice the middle term is equal to the sum of the first and the last term.