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# In the following APs find the missing term in the boxes. Q.3 (all)

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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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1. (i)Â For the given A.P., 2,2 , 26

The first and third term are;

aÂ = 2

a3Â = 26

As we know, for an A.P.,

anÂ =Â a+(nÂ âˆ’1)d

Therefore, putting the values here,

a3Â = 2+(3-1)d

26 = 2+2d

24 = 2d

dÂ = 12

a2Â = 2+(2-1)12

= 14

Therefore, 14 is the missing term.

(ii)Â For the given A.P., , 13, ,3

a2Â = 13 and

a4Â = 3

As we know, for an A.P.,

anÂ =Â a+(nâˆ’1)Â d

Therefore, putting the values here,

a2Â =Â aÂ +(2-1)d

13 =Â a+dÂ â€¦â€¦â€¦â€¦â€¦â€¦.Â (i)

a4Â =Â a+(4-1)d

3 =Â a+3dÂ â€¦â€¦â€¦â€¦..Â (ii)

On subtracting equationÂ (i)Â fromÂ (ii), we get,

â€“ 10 = 2d

dÂ = â€“ 5

From equationÂ (i), putting the value of d,we get

13 =Â a+(-5)

aÂ = 18

a3Â = 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 and

a4Â = 19/2

As we know, for an A.P.,

anÂ =Â a+(nâˆ’1)d

Therefore, putting the values here,

a4Â =Â a+(4-1)d

19/2 =Â 5+3d

(19/2) â€“ 5 = 3d

3d =Â 9/2

d = 3/2

a2Â =Â a+(2-1)d

a2Â =Â 5+3/2

a2Â =Â 13/2

a3Â =Â a+(3-1)d

a3Â =Â 5+2Ã—3/2

a3Â =Â 8

Therefore, the missing terms are 13/2 and 8, respectively.

(iv) For the given A.P.,

aÂ = âˆ’4 and

a6Â = 6

As we know, for an A.P.,

anÂ =Â aÂ +(nâˆ’1)Â d

Therefore, putting the values here,

a6Â = a+(6âˆ’1)d

6 = âˆ’ 4+5d

10 = 5d

dÂ = 2

a2Â =Â a+dÂ = âˆ’ 4+2 = âˆ’2

a3Â =Â a+2dÂ = âˆ’ 4+2(2) = 0

a4Â =Â a+3dÂ = âˆ’ 4+ 3(2) = 2

a5Â =Â a+4dÂ = âˆ’ 4+4(2) = 4

Therefore, the missing terms are âˆ’2, 0, 2, and 4 respectively.

(v) For the given A.P.,

a2Â = 38

a6Â = âˆ’22

As we know, for an A.P.,

anÂ =Â a+(nÂ âˆ’1)d

Therefore, putting the values here,

a2Â =Â a+(2âˆ’1)d

38 =Â a+dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (i)

a6Â =Â a+(6âˆ’1)d

âˆ’22 =Â a+5dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (ii)

On subtracting equationÂ (i)Â fromÂ (ii), we get

âˆ’ 22 âˆ’ 38 = 4d

âˆ’60 = 4d

dÂ = âˆ’15

aÂ =Â a2Â âˆ’Â dÂ = 38 âˆ’ (âˆ’15) = 53

a3Â =Â aÂ + 2dÂ = 53 + 2 (âˆ’15) = 23

a4Â =Â aÂ + 3dÂ = 53 + 3 (âˆ’15) = 8

a5Â =Â aÂ + 4dÂ = 53 + 4 (âˆ’15) = âˆ’7

Therefore, the missing terms are 53, 23, 8, and âˆ’7, respectively.

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