Ex.5.3 Q5 Arithmetic Progressions Solution - NCERT Maths Class 10

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Question

The first term of an AP is \(5,\) the last term is \(45\) and the sum is \(400.\) Find the number of terms and the common difference.

 Video Solution
Arithmetic Progressions
Ex 5.3 | Question 5

Text Solution

 

What is Known?

\(a,{\rm{ }}l\), and \({S_n}\)

What is Unknown?

\(n\) and \(l\)

Reasoning:

Sum of the first \(n\) terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) or \({S_n} = \frac{n}{2}\left[ {a + l} \right]\), and \(nth\) term of an AP is \(\,{a_n} = a + \left( {n - 1} \right)d\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms and \(l\) is the last term.

Steps:

Given,

  • First term, \(a = 5\)
  • Last term, \(l = 45\)
  • Sum up to \(n\rm{th}\) terms, \({S_n} = 400\)

We know that sum of \(n\) terms of AP

\[\begin{align}{S_n} &= \frac{n}{2}\left( {a + l} \right)\\400& = \frac{n}{2}\left( {5 + 45} \right)\\400&= \frac{n}{2} \times 50\\n &= 16\end{align}\]

\[\begin{align}l &= {a_n} = a + \left( {n - 1} \right)d\\45 &= 5 + \left( {16 - 1} \right)d\\40 &= 15d\\d &= \frac{{40}}{{15}}\\d& = \frac{8}{3}\end{align}\]