Question

The 4th term of an A.P. is 13 and its 10th term is 25. Find its: [3]

i. 1st term

ii. Common difference

iii. 17th term​

Answer 1

Answer:

T4 = a+3d = 13 …(1)

T10 = a+9d = 25 …(2)

Subtract (1) from (2)

a+9d-a-3d = 6d = 25–13 = 12

So d = 2.

From (1), a = 13 - 6 = 7

Answer 2

Answer:

4th term, 10th term are 13 and 25

4th term, 10th term are 13 and 25a+

4th term, 10th term are 13 and 25a+3d=13a+9d=

4th term, 10th term are 13 and 25a+3d=13a+9d=25

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−13

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=7

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=7First term

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=7First term=

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=7First term=7

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=7First term=7, common difference

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=7First term=7, common difference=

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=7First term=7, common difference=2

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=7First term=7, common difference=217th term

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=7First term=7, common difference=217th term=

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=7First term=7, common difference=217th term=a+

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=7First term=7, common difference=217th term=a+16d=

4th term, 10th term are 13 and 25a+3d=13a+9d=25(a+9d)−(a+3d)=25−136d=12d=2a=7First term=7, common difference=217th term=a+16d=39