Question

If 51+53+55+....+tn= 5151,then find the value of tn.​

Answer 1

Answer:

number of terms are 51

Step-by-step explanation:

this sequence is an AP

here common difference is constant

here a=51 ,d=53-51=2

s=n/2[2a+(n-1)d]

5151=n/2[2(51)+(n-1)2]

10302=n[102+2n-2]

10302=n[100+2n]

10302=2n(50+n)

5151=n(50+n)

5151=50n+n2

n2+50n-5151=0

n2+101n-51n-5151=0

n(n+101)-51(n+101)=0

(n+101)(n-51)=0

(n+101)=0 or (n-51)=0

n=-101 or n=51

negative value is not possible

therefore n=51

Answer 2

Answer:

tn = 151

Step-by-step explanation:

t1 = 51 c.d. = d = 53 - 51 = 2

Now ,

Sn = n/2 [ 2 t1 + (n - 1) d ]

=> 5151 = n/2 [ 102 + (n - 1) 2 ]

=> 5151 = n ( 51 + n - 1 )

=> n^2 + 50 n - 5151 = 0

=> n^2 + ( 101 - 51 ) n - 5151 = 0

=> n^2 + 101 n - 51 n - 5151 =0

=> (n + 101) (n - 51) = 0

Either , | Or ,

n = - 101 | n = 51

Since , ' n ' can't be (-)ve integer .

Therefore , n not equals to - 101.

Therefore , n = 51

So , tn = t1 + (n - 1) d

tn = 51 + (51 - 1) 2

tn = 51 + 100

tn = 151

HOPE THIS WILL HELP YOU !

THANK YOU.