Ex.5.3 Q9 Arithmetic Progressions Solution - NCERT Maths Class 10

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Question

If the sum of first \(7\) terms of an AP is \(49\) and that of \(17\) terms is \(289,\) find the sum of first \(n\) terms.

 Video Solution
Arithmetic Progressions
Ex 5.3 | Question 9

Text Solution

What is Known?

\({S_7}\) and \({S_{17}}\)

What is Unknown?

\({S_n}\)

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 \(n\rm{th}\) 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,

  • Sum of first 7 terms, \({S_7} = 49\)
  • Sum of first 17 terms, \({S_{17}} = 289\)

We know that sum of \(n\) term of AP is,

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_7} &= \frac{7}{2}\left[ {2a + \left( {7 - 1} \right)d} \right]\\49 &= \frac{7}{2}\left[ {2a + 6d} \right]\\a + 3d &= 7\qquad \qquad ....\left( \rm{i} \right)\\\\{S_{17}}&= \frac{{17}}{2}\left[ {2a + \left( {17 - 1} \right)d} \right]\\289 &= \frac{{17}}{2}\left[ {2a + 16d} \right]\\a + 8d &= 17\qquad \qquad ....\left( {\rm{ii}} \right)\end{align}\]

Subtracting equation (i) from equation (ii),

\[\begin{align}5d &= 10\\d &= 2\end{align}\]

 From equation (i),

\[\begin{align}7& = a + 3 \times 2\\7 &= a + 6\\a&= 1\end{align}\]

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\ &= \frac{n}{2}\left[ {2 \times 1 + \left( {n - 1} \right)2} \right]\\ &= \frac{n}{2}\left[ {2 + 2n - 2} \right]\\ &= \frac{n}{2} \times 2n\\& = {n^2}\end{align}\]