# 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}

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