Question

how many terms of this sequence lies between 100 and 550( sequence= 11, 19,27,35​

Answer 1

Step-by-step explanation:

The A.P is 11,19,27,35....

Here a = 11 , d = t2-t1 = 19-11 =8 , t100 = ?

Therefore, tn =a + ( n-1 ) d

t100 = 11 + ( 100-1 ) 8

t100 = 11 + 99 × 8

t100 = 11 +792

t100 = 803

Now, a=11 , d = t2-t1= 19-11 = 8 , t550 = ?

tn = a + ( n-1 ) d

t550 = 11 + ( 550-1 ) 8

t550 = 11 + 549 × 8

t550 = 11 + 4392

t550 = 4403

Now, t550 - t100

4403 - 803

= 3600

Therefore, 3600 terms of this sequence lies between 100 and 550.

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