Question

The fifth term of an A.P is 26 and its 10th term is 51. Find the A.P​

Answer 1

Answer:

given: t5=26

a+24d=26........1

t10=51

a+9d=51.....2

subtracting 1 and 2

we get,

15d=25

d= 5/3.....

by putting the value of d in 1 we get,

a=-14.....

a.p= a,a+d, a+2d.....

=-14,-37/3,-32/3..... answer

Answer 2

Answer:

6,11,16, ............

Step-by-step explanation:

5th term = a+4d = 26 (1)

10th term = a+9d = 51 (2)

Subtracting (1) from (2), we have,

a+9d -(a+4d) = 51-26

a+9d-a-4d = 25

5d = 25

d = 25/5 = 5 (3)

Put (3) in (1),

a+4(5) = 26

a+20 = 26

a = 26-20

a = 6 (4)

Now,

1st term = a = 6

2nd term = a2 = a+d = 6+5 = 11

3rd term = a3 = a+2d = 6+2(5) = 16

So, the required A.P. becomes;

6,11,16, ...........