The best way to solve all the problems of arithmetic progressions of exercise 5.2 of class 10th math, How i solve easily to this question In the following APs find the missing term in the boxes.

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(i)Â For the given A.P., 2,2 , 26

The first and third term are;

aÂ = 2a_{3}Â = 26As we know, for an A.P.,

aÂ =Â_{n}a+(nÂ âˆ’1)dTherefore, putting the values here,

a_{3}Â = 2+(3-1)d26 = 2+2

d24 = 2

ddÂ = 12a_{2}Â = 2+(2-1)12= 14

Therefore, 14 is the missing term.

(ii)Â For the given A.P., , 13, ,3a_{2}Â = 13 anda_{4}Â = 3As we know, for an A.P.,

aÂ =Â_{n}a+(nâˆ’1)ÂdTherefore, putting the values here,

a_{2}Â =ÂaÂ +(2-1)d13 =Â

a+dÂ â€¦â€¦â€¦â€¦â€¦â€¦.Â(i)a_{4}Â =Âa+(4-1)d3 =Â

a+3dÂ â€¦â€¦â€¦â€¦..Â(ii)On subtracting equationÂ

(i)Â fromÂ(ii), we get,â€“ 10 = 2

ddÂ = â€“ 5From equationÂ

(i), putting the value of d,we get13 =Â

a+(-5)aÂ = 18a_{3}Â = 18+(3-1)(-5)= 18+2(-5) = 18-10 = 8

Therefore, the missing terms are 18 and 8, respectively.

(iii)For the given A.P.,aÂ= 5 anda_{4}Â = 19/2As we know, for an A.P.,

aÂ =Â_{n}a+(nâˆ’1)dTherefore, putting the values here,

a_{4}Â =Âa+(4-1)d19/2 =Â

5+3d(19/2) â€“ 5 = 3d

3d =Â 9/2

d = 3/2

a_{2}Â =Âa+(2-1)da_{2}Â =Â5+3/2a_{2}Â =Â 13/2a_{3}Â =Âa+(3-1)da_{3}Â =Â5+2Ã—3/2a_{3}Â =Â8Therefore, the missing terms are 13/2 and 8, respectively.

(iv) For the given A.P.,

aÂ = âˆ’4 anda_{6}Â = 6As we know, for an A.P.,

aÂ =Â_{n}aÂ +(nâˆ’1)ÂdTherefore, putting the values here,

a

_{6}Â = a+(6âˆ’1)d6 = âˆ’ 4+5

d10 = 5

ddÂ = 2a_{2}Â =Âa+dÂ = âˆ’ 4+2 = âˆ’2a_{3}Â =Âa+2dÂ = âˆ’ 4+2(2) = 0a_{4}Â =Âa+3dÂ = âˆ’ 4+ 3(2) = 2a_{5}Â =Â a+4dÂ = âˆ’ 4+4(2) = 4Therefore, the missing terms are âˆ’2, 0, 2, and 4 respectively.

(v)For the given A.P.,a_{2}Â = 38a_{6}Â = âˆ’22As we know, for an A.P.,

aÂ =Â_{n}a+(nÂ âˆ’1)dTherefore, putting the values here,

a_{2}Â =Âa+(2âˆ’1)d38 =Â

a+dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â(i)a_{6}Â =Âa+(6âˆ’1)dâˆ’22 =Â

a+5dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (ii)On subtracting equationÂ

(i)Â fromÂ(ii), we getâˆ’ 22 âˆ’ 38 = 4

dâˆ’60 = 4

ddÂ = âˆ’15aÂ =Âa_{2}Â âˆ’ÂdÂ = 38 âˆ’ (âˆ’15) = 53a_{3}Â =ÂaÂ+ 2dÂ= 53 + 2 (âˆ’15) = 23a_{4}Â =ÂaÂ + 3dÂ = 53 + 3 (âˆ’15) = 8a_{5}Â =ÂaÂ + 4dÂ = 53 + 4 (âˆ’15) = âˆ’7Therefore, the missing terms are 53, 23, 8, and âˆ’7, respectively.