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.

(i) For the given A.P., 2,2 , 26

The first and third term are;

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

a=_{n}a+(n−1)dTherefore, putting the values here,

a_{3}= 2+(3-1)d26 = 2+2

d24 = 2

dd= 12a_{2}= 2+(2-1)12= 14

Therefore, 14 is the missing term.

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

a=_{n}a+(n−1)dTherefore, putting the values here,

a_{2}=a+(2-1)d13 =

a+d……………….(i)a_{4}=a+(4-1)d3 =

a+3d…………..(ii)On subtracting equation

(i)from(ii), we get,– 10 = 2

dd= – 5From equation

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

a+(-5)a= 18a_{3}= 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.,a= 5 anda_{4}= 19/2As we know, for an A.P.,

a=_{n}a+(n−1)dTherefore, putting the values here,

a_{4}=a+(4-1)d19/2 =

5+3d(19/2) – 5 = 3d

3d = 9/2

d = 3/2

a_{2}=a+(2-1)da_{2}=5+3/2a_{2}= 13/2a_{3}=a+(3-1)da_{3}=5+2×3/2a_{3}=8Therefore, the missing terms are 13/2 and 8, respectively.

(iv) For the given A.P.,

a= −4 anda_{6}= 6As we know, for an A.P.,

a=_{n}a+(n−1)dTherefore, putting the values here,

a

_{6}= a+(6−1)d6 = − 4+5

d10 = 5

dd= 2a_{2}=a+d= − 4+2 = −2a_{3}=a+2d= − 4+2(2) = 0a_{4}=a+3d= − 4+ 3(2) = 2a_{5}=a+4d= − 4+4(2) = 4Therefore, the missing terms are −2, 0, 2, and 4 respectively.

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

a=_{n}a+(n−1)dTherefore, putting the values here,

a_{2}=a+(2−1)d38 =

a+d…………………….(i)a_{6}=a+(6−1)d−22 =

a+5d………………….(ii)On subtracting equation

(i)from(ii), we get− 22 − 38 = 4

d−60 = 4

dd= −15a=a_{2}−d= 38 − (−15) = 53a_{3}=a+ 2d= 53 + 2 (−15) = 23a_{4}=a+ 3d= 53 + 3 (−15) = 8a_{5}=a+ 4d= 53 + 4 (−15) = −7Therefore, the missing terms are 53, 23, 8, and −7, respectively.