Ex.5.3 Q8 Arithmetic Progressions Solution - NCERT Maths Class 10

Go back to  'Ex.5.3'

Question

Find the sum of first \(51\) terms of an AP whose second and third terms are \(14\) and \(18\) respectively.

 Video Solution
Arithmetic Progressions
Ex 5.3 | Question 8

Text Solution

What is Known?

\({a_2}\) and \({a_3}\)

What is Unknown?

\({S_{51}}\)

Reasoning:

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

  • 2nd term, \({a_2} = 14\)
  • 3rd term, \({a_3} = 18\)
  • Common difference, \(d = {a_3} - {a_2} = 18 - 14 = 4\)

We know that \(n\rm{th}\) term of AP, \({a_n} = a + \left( {n - 1} \right)d\)

\[\begin{align}{a_2} &= a + d\\14 &= a + 4\\a& = 10\end{align}\]

Sum of \(n\) terms of AP series,

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{51}} &= \frac{{51}}{2}\left[ {2 \times 10 + \left( {51 - 1} \right)4} \right]\\ &= \frac{{51}}{2}\left[ {20 + 50 \times 4} \right]\\ &= \frac{{51}}{2} \times 220\\ &= 51 \times 110\\ &= 5610\end{align}\]