Question

solve the above question of ch- arithmetic progression ​

Answer 1

Answer:-

Let us assume that l , m , n are in AP.

So,

Using Arithmetic mean ,

→ 2m = l + n -- equation (1)

Given:

l²(m + n) , m²(n + l) , n²(l + m) are in AP.

Similarly,

2[m²(n + l)] = l²(m + n) + n²(l + m)

→ 2(m²n + lm²) = l²m + l²n + n²l + mn²

→ 2m²n - mn² + 2lm² - l²m = l²n + n²l

Taking mn and lm common in LHS we get,

→ mn(2m - n) + lm(2m - l) = l²n + n²l

Putting the values from equation (1) we get,

→ mn(l + n - n) + lm(l + n - l) = l²n + n²l

Taking "ln" common in RHS we get,

→ mn(l) + lm(n) = l²n + ln²

→ lmn + lmn = nl(l + n)

→ 2lmn = ln(l + n)

Cancelling "ln" both sides we get,

→ 2m = l + n

Hence we can say that l , m , n are in AP.