Question

can anyone answer this question from class 11 ....✌️​

Answer 1

yes i can answer this question

hope it helps you

please mark as the

Brainliest

Answer 2

For the A.P.

first term = a

common difference = d

we need to prove that

$a_{k} = \frac{ a_{k - 1} \: \: \: + \: \: \: \: a_{k + 1} }{2}$

now we know that,

$a_{k} = a + (k - 1)d \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (i)$

so,

$\: \: \: \: \: \frac{ a_{k - 1} \: \: \: + \: \: \: \: a_{k + 1} }{2} \\ = \frac{a + (k - 1 - 1)d + a + (k + 1 - 1)d}{2} \\ = \frac{2a + (k - 2)d + k d }{2} \\ = \frac{2a + kd - 2d + kd}{2} = \frac{2a + 2kd - 2d}{2} \\ = \frac{2(a + kd - d)}{2} = a + kd - d \\ = a + (k - 1)d \\ = a_{k} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (from \: (i))$

Hence, any term of an A.P. is the arithmetic mean of its immediate predecessor term and its immediate successor term.