Question

Find the three numbers in AP whose sum is 15 and sum of their squares is 107

Answer 1

Answer:

Required terms are 1 , 5 and 9 .

Step-by-step explanation:

Let the required terms of that AP be ( a + d ) , a and ( a - d ).

According to the question :

= > Sum of terms = 15

= > ( a + d ) + a + ( a - d ) = 15

= > a + d + a + a - d = 15

= > 3a = 15

= > a = 5

Also,

= > Sum of their squares = 107

= > ( a + d )^2 + a^2 + ( a - d )^2 = 107

= > ( 5 + d )^2 + 5^2 + ( 5 - d )^2 = 107 ...from above, a = 5

= > 25 + d^2 + 10d + 25 + 25 + d^2 - 10d = 107

= > 75 + 2d^2 = 107

= > 2d^2 = 107 - 75

= > 2d^2 = 32

= > d^2 = 16

= > d = ±4

Hence the required terms are :

Either : -

• a + d = 5 + 4 = 9
• a = 5
• a - d = 5 - 4 = 1

Or : -

• a + d = 5 - 4 = 1
• a = 5
• a - d = 5 - ( - 4 ) = 5 + 4 = 9

Hence the required terms are 1 , 5 and 9 .

Answer 2

$\huge\text{\underline{Answer}}$

Let the three terms in AP be $\bold{ a-d , a , a+d }$

Where a is first term and d is common difference

$\huge\text{\underline{Given}}$

The sum of three number is 15.

The sum of their square is 107.

According to question,

$\bold{a - 1 + a + a +1= 15}$

$\implies$ $\bold{3a = 15 }$

$\implies$ $\bold{a = 5 }$

$\bold{{(a - d)}^{2} + {a}^{2} + {(a + d)}^{2} = 107}$

Using,

$\boxed{{(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab }$

$\boxed{{(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab }$

$\implies$ $\bold{{a}^{2} + {d}^{2} - 2ad + {a}^{2} + {a}^{2} + {d}^{2} + 2ad = 107}$

$\implies$ $\bold{3 {a}^{2} + 2 {d}^{2} = 107 }$

Putting the value of a

$\implies$ $\bold{3 \times {5}^{2} + 2 {d}^{2} = 107 }$

$\implies$ $\bold{ 2 {d}^{2} = 107 - 75}$

$\implies$ $\bold{{d}^{2} = 16}$

$\implies$ $\bold{d = ±4 }$

Now,the three no. in AP are :-

a = 5

(a + 1)= 5+4 = 9

(a - 1) = 5 - 4 = 1

or,

a= 5

(a+1) = 5 -4 = 1

(a-1) = 5+4 = 9

$\sf{\underline{ Verification}}$

Adding this three no. we get 15

and it's given that sum of three terms in AP is 15 .

hence, verified