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 \(27.\)