Class 10

Math

All topics

Arithmetic Progressions

In the following APs, find the missing terms in the boxes :

(i) For this A.P.,

$a=2$

$a_{3}=26$

We know that, $a_{n}=a+(n−1)d$

$a_{3}=2+(3−1)d$

$26=2+2d$

$24=2d$

$d=12$

$a_{2}=2+(2−1)12$

$=14$

Therefore, $14$ is the missing term.

(ii) For this A.P.,

$a_{2}=13$ and

$a_{4}=3$

We know that, $a_{n}=a+(n−1)d$

$a_{2}=a+(2−1)d$

$13=a+d$ ... (i)

$a_{4}=a+(4−1)d$

$3=a+3d$... (ii)

On subtracting (i) from (ii), we get,

$−10=2d$

$d=−5$

From equation (i), we get,

$13=a+(−5)$

$a=18$

$a_{3}=18+(3−1)(−5)$

$=18+2(−5)=18−10=8$

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

(iii) For this A.P.,

$a_{1}=5$ and

$a_{4}=921 $

We know that, $a_{n}=a+(n−1)d$

$a_{4}=5+(4−1)d$

$921 =5+3d$

$d=23 $

$a_{2}=a+d$

$a_{2}=5+23 $

$a_{2}=213 $

$a_{3}=a_{2}+23 $

$a_{3}=8$

Therefore, the missing terms are $621 $ and $8$ respectively.

(iv) For this A.P.,

$a=−4$ and

$a_{6}=6$

We know that,

$a_{n}=a+(n−1)d$

$a_{6}=a+(6−1)d$

$6=−4+5d$

$10=5d$

$d=2$

$a_{2}=a+d=−4+2=−2$

$a_{3}=a+2d=−4+2(2)=0$

$a_{4}=a+3d=−4+3(2)=2$

$a_{5}=a+4d=−4+4(2)=4$

Therefore, the missing terms are $−2,0,2,$ and $4$ respectively.

(v) For this A.P.,

$a_{2}=38$

$a_{6}=−22$

We know that

$a_{n}=a+(n−1)d$

$a_{2}=a+(2−1)d$

$38=a+d$... (i)

$a_{6}=a+(6−1)d$

$−22=a+5d$ ... (ii)

On subtracting equation (i) from (ii), we get

$−22−38=4d$

$−60=4d$

$d=−15$

$a=a_{2}−a =38−(−15)=53$

$a_{3}=a+2d=53+2(−15)=23$

$a_{4}=a+3d=53+3(−15)=8$

$a_{5}=a+4d=53+4(−15)=−7$

Therefore, the missing terms are $53,23,8$ and $−7$ respectively.