Question

det the numbers be ( a_{1}, a_{2} Delta a_{3} ) ( a=13 quad d quad a=1 )
( 80,2 d=7-a )
( vec{y} d=7-13 quad 0 mathrm{A} )
[
a, r a_{3}-a_{2}=6
]
( 0 times e )
( a_{1}, a_{2}, a_{3} ) are in AP. ( A P ) can be
( 3-12 )
Let the common difference be ( d )
[
begin{array}{l}text { . }end{array}
]
( operatorname{seg} alpha a_{1}=a )
( operatorname{sog} a_{2}=a+d )
( a_{3}=a+2 d ). & at ( 2 d ) are in ( A cdot P . quad 7 quad 0.2 ), ( , 7,13 )
nus ( +a+2 d=21 )
feom ( left(1, frac{a+a+d}{3 a+3 d}=21right. )
[
Rightarrow quad 3(a+d)=21
]
( a+d=7 Rightarrow d=7-80 )
And ( int frac{operatorname{sen}(2)}{a}(a+2 d)-(a+d)=6 )
( Rightarrow quad a^{2}+2 a d-a-d=6 )
I ( quad a^{2}+2 a(7-a)-a-(7-a)=6 )
( Rightarrow quad a^{2}+14 a-2 a^{2}-a-7+a=6 )
( exists quad-a^{2}+14 a-7^{prime}=6 )
( 7 quad a^{2}-14 a+7+6=0 )
( Rightarrow quad a^{2}-14 a+13=0 )
( 7 a^{2}-13 a-a+13=0 )
2
( (a-13)(a-1)=0 )

# Three numbers whose sum is 21 are in AP. If the product of the first and the third numbers exceeds the second number by 6 then find the numbers. (Ans : 1, 7, 13)

Solution