Question

In the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 ....... , where n consecutive terms have the value n, the 150th term is

Answer 1

To find the N th term the formula used is n(n+1)/2 since it is a series.

Let X be the term required

So X = 150

X = n(n+1)/2
150 = n(n+1/2

150*2 = n(n+1)
300 = n² +n
n² + n - 300 = 0

Using quadratic formula we get n = −1±√1201 / 2

It comes to be around 17 approx

So 150 th term might be 17

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Answer 2

Answer: 17

Step-by-step explanation:

Given sequence

$(1),(2,2),(3,3,3),(4,4,4,4).....((n-1),(n-1)....(n-1),(n,n....n)$ .....1

So total no of terms upto n is $n\frac{(n+1)}{2}$

Because in 1st group-- 1 term , 2nd group --2 so on. In n groups n(n+1)/2

As we assume 150th term as "n" so from 1st eq

$n\frac{(n-1)}{2} < 150 ≤ n\frac{(n+1)}{2}$

$n(n-1) < 300 ≤ n(n+1)$

So finding 2 consecutive integers whose product is nearby 300.

By hit and trial,

We got $n=17$

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