# Ex.5.2 Q5 Arithmetic Progressions Solution - NCERT Maths Class 10

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## Question

Find the number of terms in each of the following APs:

i) \begin{align} 7,13,19,……,205\end{align}

ii) \begin{align}18,15 \frac{1}{2}, 13, \ldots . .,-47\end{align}

Video Solution
Arithmetic Progressions
Ex 5.2 | Question 5

## Text Solution

What is Known?

The AP

What is Unknown?

No. of terms of the AP.

Reasoning:

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

i) \begin{align} 7,13,19,……,205\end{align}

Steps:

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

Where \begin{align}{a_n}\end{align} is the $$n^{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

\begin{align}{a}&=7 \\ {d}&=13-7=6 \\ {a_{n}}&=205 \\ {a_{n}} &=a+(n-1) d\\ {a+(n-1) d}&=205 \\ {7+(n-1) 6}&=205 \\ {(n-1) 6}&=205-7 \\ {(n-1) 6}&=198 \\ {n-1}&=33 \\ {n}&=33 + 1 \\ {n}&=34 \end{align}

Number of terms in the given AP is $$34.$$

ii) \begin{align}18,15 \frac{1}{2}, 13, \ldots . .,-47\end{align}

Steps:

\begin{align}d &= 15\frac{1}{2} - 18\\ &= \frac{{31}}{2} - 18\\&= {-\frac{{5}}{2}} \\{a_n} &= a + (n - 1)d\\a + (n - 1)d &= - 47\\18 + (n - 1)({-\frac{{5}}{2}}) &= - 47\\n - 1 &= 26\\n &= 26 + 1\\ &= 27 \end{align}

The number of terms in the given AP is $$27.$$