Question

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)

11th - 12th Class

Maths

Solution

198

4.0 (1 ratings)

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 )