# Ex.5.2 Q6 Arithmetic Progressions Solution - NCERT Maths Class 10

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## Question

Check Whether \(-150\) is a term of the AP \( \,11,\,8,\,5,\,2\,\dots\)

Video Solution

Arithmetic Progressions

Ex 5.2 | Question 6

## Text Solution

**What is Known:?**

The AP

**What is Unknown?**

Whether \(-150\) is a term of AP.

**Reasoning:**

\({a_n} = a + \left( {n - 1} \right)d\)

Where \({a_n}\) is the \(n\rm{th}\) term, \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.

**Steps:**

\(11,\,8,\,5,\,2 \) are in AP

First term \(a = 11\)

Common difference \(d = 8--11 = - 3\)

\[\begin{align}{a_n} &= a + (n - 1)d\\a + (n - 1)d &= - 150\\11 + (n - 1)( - 3) &= - 150\\n - 1 &= \frac{{161}}{3}\\n &= \frac{{164}}{3}\end{align}\]

\(n = \frac{{164}}{3}\) which is a fraction. Given number is not a term of AP \(11,\, 8,\, 5,\, 2\,\dots\)\(n\) should be positive integer and not a fraction.