Five numbers are in AP and their sum is 50.Find its 3rd number.
Answers
Answered by
4
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
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
Zico26:
How can the 3rd no be 50
there r total 5 nos
i think other terms are in minus also
Sum is 50
not possible
no answer is right other terms in minuse
calculate that T3=50 S5=50 then find a and d then u realise
Answered by
5
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.
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