Question

Three number are in A.P. such that their sum is 18 and sum of their squares is 158. The greatest number among them is [UPSEAT 2004]
A) 10 B) 11 C) 12 D) None of these

Answer 1

Given:

Three number are in A.P. such that their sum is 18

The sum of their squares is 158

To find:

The greatest number among them is?

Solution:

Let's assume the 3 numbers in A.P. be "(a + d)", "a" & "(a - d)".

It is given that their sum is 18, so we can write the eq. as,

$(a + d) + a + (a - d) = 18$

$\implies a + d + a + a - d = 18$

$\implies a + a + a = 18$

$\implies 3a = 18$

$\implies a = \frac{18}{3}$

$\implies \bold{a = 6}$

Also given that the sum of their squares is 158, so we can write the eq. as,

$(a + d)^2 + a^2 + (a - d)^2 = 158$

$\implies (a^2 + 2ad +d^2) + a^2 + (a^2 -2ad + d^2) = 158$

$\implies a^2 + 2ad +d^2 + a^2 + a^2 -2ad + d^2 = 158$

$\implies a^2 +d^2 + a^2 + a^2 + d^2 = 158$

$\implies3 a^2 + 2d^2 = 158$

substituting the value of a = 6

$\implies3 (6)^2 + 2d^2 = 158$

$\implies108 + 2d^2 = 158$

$\implies 2d^2 = 158 - 108$

$\implies 2d^2 = 50$

$\implies d^2 = 25$

$\implies \bold{d =}$ ± 5

Now,

When d = +5

Then,

a + d = 6 + 5 = 11

a = 6

a - d = 6 - 5 = 1

∴ 3 numbers are → 11, 6 & 1.

and

When d = -5

Then,

a + d = 6 - 5 = 1

a = 6

a - d = 6 + 5 = 11

∴ 3 numbers are → 1, 6 & 11.

Thus, $\boxed{\bold{The \:greatest\:number\:among\:them\:is\:option (B)\rightarrow \underline{11}}}.$

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Also View:

Three numbers in A.p have sum 18 and the sum of their squares is 180.find the numbers

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Three numbers are in ap and their sum Is 18 and sum of their squares is 140 find the numbers

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Answer 2

Answer:

yes it is correct..........,...