Question

In an arithmetic progression 3rd term is 10 and 7th term is 22, find its 10" term ​

Answer 1

Answer:

t3= 10

t7= 22

we have to find t 10

a+ (3-1) d = 10

a+ 2d = 10. (1)

a+ (7-1) d = 22

a+ 6d = 22. (2)

subract above equation (1) from equation (2)

a+ 6d = 22

a + 2d = 10

4d = 12

d = 12/4

d= 3

put value d in equation (1)

a+ 2(3)= 10

a = 10+6

a = 4

to find t10

tn = a + ( n-1 ) d

t10= 4 + ( 10-1) 3

= 4 + 9×3

= 4+27

t10 = 31

Answer 2

Answer:

a+2d=10

a+6d=22

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-4d=3

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d=3

a+6=10

a=4

AP,a,a+d,a+2d

4,3,10