The mean of five numbers in AP is $89$ . The product of first and last terms is $7885$ . The AM of first, third and fifth term is

83

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86

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89

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90

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Solution

The correct option is B $89$
Let the numbers in A.P. be $(a - 2 d), (a - d), a, (a + d), (a + 2 d)$
$∴ \frac{(a - 2 d) + (a - d) + a + (a + d) + (a + 2 d)}{5} = 89$
$\Rightarrow \frac{5 a}{5} = 89$
$∴ a = 89$
$∴$ The AM of 1st, 3rd & 5th term is:
$= \frac{(a - 2 d) + a + (a + 2 d)}{3}$
$= a = 89$

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The mean of five numbers in AP is 89. The product of first and last terms is 7885. The AM of first, third and fifth term is

The correct option is B 89Let the numbers in A.P. be (a−2d),(a−d),a,(a+d),(a+2d)∴(a−2d)+(a−d)+a+(a+d)+(a+2d)5=89⇒5a5=89∴a=89∴ The AM of 1st, 3rd & 5th term is:=(a−2d)+a+(a+2d)3=a=89

The mean of five numbers in AP is $89$ . The product of first and last terms is $7885$ . The AM of first, third and fifth term is