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Given a,b,c are in A.P.,b,c,d are in G.P and c,d,e are in H.P .If a=2 and e=18 , then the sum of all possible value of c is ________. |
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Answer» Given,`a,b,c` are in AP. `therefore " " b=(a+c)/(2) " " "…….(i)"` `b,c,d` are in GP. `therefore c^(2)=bd " " "…….(ii)"` and `c,,e` are in HP. `therefore d=(2ce)/(c+e) " " "…..(iii)"` Now, substituting the values of b and d in Eq. (ii), than ` c^(2)= ((a+c)/(2))((2ce)/(c+e))` `implies c(c+e)=e(a+c)` `implies c^(2)=ae " " ".......(iv)"` Given, `a=2,e=18` From Eq. (iv), `c^(2)=(2)(18)=36` ` therefore " " c= pm 6 ` From Eq. (i), `b=(2 pm 6)/(2)=4,-2` and from Eq. (ii), `d=(c^(2))/(b)=(36)/(b)=(36)/(4) " or " (36)/(-2)` `therefore " " d=9 " or " -18` Hence, `c=6,b=4,d=9 " or " c=-6,b=-2,d=-18` |
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