InterviewSolution
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InterviewSolution
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In the following A.P.'s, find the missing terms in the boxes. |
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Answer» <P> Solution :(i) `2, square , 26``RARR 2,x,26` are in A.P. (say) `rArr x=(2+26)/(2)=14` (ii)`square, 13,square,3` `rArr x,13,y,3` are in A.P. (say) `:. Y=(13+3)/(2)=8` and `(x+y)/(2)=13` `rArrx+y=26` `rArr x+8=26` `rArr x=18` `rArr x=18,y=8` (ii) `5,square,square,9(1)/(2)` `rArr 5,x,y,(19)/(2)` are in A.P.. (say) Here, a=5 and `a_(4)=(19)/(2)` `rArr 5+3D=(19)/(2)` `rArr 3d=(19)/(2)-5=(9)/(2) rArr d=(3)/(2)` `:. x=5+d=5+(3)/(2)=(13)/(2)` and `y=5+2d=5+2xx(3)/(2)=8` `:. x=(13)/(2),y=8` (iv) `-4square,square,square,square,6` `rArr -4,p,q,r,s,6` are in A.P. (say) Here, `a=-4` and `a_(6)=6` `rArr -4+5d=6` `rArr 5d=10 rArrd=2` `:. p=-4+d=-4+2=-2` `q=-4+2d=-4+2xx2=0` `r=-4+3d=-4+3xx2=2` `s=-4+4d=-4+4xx2=4` `p=-2,q=0,r=2,s=4` (v)`square,38,square,square,square,-22` `rArr p, 38,q,r,s,-22 ` are in A.P. (say) Here, `a_(2)=38` `rArr p+d=38 "" ...(1)` and `a_(6)=-22` `rArr p+5d=-22 "" ...(2) ` Subtract equation (1) from (2), we get `{:(p+5d=""-22),(underset(-)" "punderset(-)+d=""underset(-)38),(bar(""4d=-60" ")):}` `rArr"" d=-15` put d=-15 in equation (1), we get p+(-15)=38 `rArr p=53` `:. q=38+d=38+(-15)=23` `r=q+d=23+(-15)=8` `s=r+d=8+(-15)=-7` `:. p=53, q=23, r=8, s=-7` |
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