Question

Five numbers are in AP and their sum is 50.Find its 3rd number.

Answer 1

Sn=50
n=5
Sn=n/2[2a+(n-1)d]
50=n/2[2a+4d]
50×2=2a+4d
100=2(a+2d)
100/2=a+2d
50=a+2d
so, T3=a+2d
=50
hope it's helpful for you

Answer 2

Answer:

10

Step-by-step explanation:

Let , t1 = a

t1 + t2 + t3 + t4 + t5

Therefore , n = 5

And Sn = 50 c.d. = d [d not equals to 0]

Therefore ,

Sn = n/2 [2 × t1 + (n - 1) × d]

=> 50 = 5/2 [2a + (5 - 1)d]

=> 100 = 5 (2a + 4d)

=> 100 = 10 (a + 2d)

=> 10 = a + 2d

Now , t3 = a + 2d

= 10

HOPE MY ANSWER WILL HELP YOU !

THANK YOU.