Question

5th term of an arithmetic sequence is 39 and29th term is 56. find the 15th term​

Answer 1

Step-by-step explanation:

a5=39

a +4d =39 --------(1)

a29=56

a + 28d = 56 -------(2)

subract (1) and (2)

a +28d = 56

- a + 4d =39

- - -

solution :- 24d =17

d = 17/24

a + 4d =39

a + 4(17/24)=39

a + 17/6=39

a =39-17/6

a = 39*6/6-17/6 ......( lcm)

a = 234 /6 -17/6

a = 234-17/6

a = 217/6

a15=a +14d

=217/6 +14(17/24)

=217/6 + 7*17/12

=217*2/6*2.+ 119 /12. .....( lcm )

= 234 +119/12

= 353/12

Answer 2

Answer:

$a_{5} = 39 \\ a_{29} = 56$

= a + 4d = 39 a + 28 d = 56

= a = 39 - 4d a = 56-28d

i.e. 39-4d = 56-28d

24d = 17

d= 17/24

Now,

$a = 39 - 4 \times \frac{17}{24}$

a = 217/6

Now 15 term is

a+14d = 217/6 + 14 * 17/24

217/6 + 7*17/12

217/6 + 119/12

46.08 is the answer