Question

2,4,2,?,52 find the value of ?

Answer 1

Answer:

a+b=6………………………………..(1)

a-b=4…………………………………(2)

Adding equation (1) and (2),we get,

(a+b)+(a-b)=10

=> a+b+a-b=10

=> 2a=10

=> 2a/2=10/2

=> a=5

Putting a=5 in equation (1),we get,

5+b=6

=> b=6–5

=> b=1

Now,

a^2+b^2

=(5)^2+(1)^2

=25+1

=26

2nd Method:

a+b=6

On squaring both sides,

(a+b)^2=(6)^2

=> a^2+2ab+b^2=36…………………….(1)

a-b=4

On squaring both sides,

(a-b)^2=(4)^2

=a^2–2ab+b^2=16……………..………….(2)

Adding equation (1) and (2), we get,

(a^2+2ab+b^2)+(a^2–2ab+b^2)=36+16

=> a^2+2ab+b^2+a^2–2ab+b^2=52

=> a^2+a^2+b^2+b^2+2ab-2ab=52

=> 2a^2+2b^2=52

=> 2(a^2+b^2)=52

=> a^2+b^2=52/2

=>a^2+b^2=26

Hence, a^2+b^2=26 .

Answer 2

Answer:

Step-by-step explanation: