The sum of four numbers in AP is $176$ . The product of 1st and last is $1855$ . The mean of middle two is

42

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44

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Solution

The correct option is C $44$
Let the numbers in A.P. be $(a - 3 d), (a - d), (a + d), (a + 3 d)$
$∴ (a - 3 d) + (a - d) + (a + d) + (a + 3 d) = 176$
$\Rightarrow 4 a = 176$
$∴ a = 44$
$∴$ The mean of middle two is:
$= \frac{(a - d) + (a + d)}{2}$
$= a = 44$

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The sum of four numbers in AP is 176. The product of 1st and last is 1855. The mean of middle two is

The correct option is C 44Let the numbers in A.P. be (a−3d),(a−d),(a+d),(a+3d)∴(a−3d)+(a−d)+(a+d)+(a+3d)=176⇒4a=176∴a=44∴ The mean of middle two is:=(a−d)+(a+d)2=a=44

The sum of four numbers in AP is $176$ . The product of 1st and last is $1855$ . The mean of middle two is

The correct option is C $44$
Let the numbers in A.P. be $(a - 3 d), (a - d), (a + d), (a + 3 d)$
$∴ (a - 3 d) + (a - d) + (a + d) + (a + 3 d) = 176$
$\Rightarrow 4 a = 176$
$∴ a = 44$
$∴$ The mean of middle two is:
$= \frac{(a - d) + (a + d)}{2}$
$= a = 44$