# Ex.5.4 Q2 Arithmetic progressions Solutions - NCERT Maths Class 10

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## Question

The sum of the third and the seventh terms of an A.P is $$6$$ and their product is $$8.$$ Find the sum of first sixteen terms of the A.P.

Video Solution
Arithmetic Progressions
Ex 5.4 | Question 2

## Text Solution

What is Known?

$${a_3} + {a_7} = 6$$ and $${a_3} \times {a_7} = 8$$

What is Unknown?

Sum of first $$16$$ terms, $${S_{16}}$$

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.

Steps:

Given:

$${a_3} + {a_7} = 6 \qquad \dots {\rm{ Equation}}\left( 1 \right)$$

$${a_3} \times {a_7} = 8 \qquad\dots {\rm{Equation }}\left( 2 \right)$$

We know that $$n^\rm{th}$$ term of AP Series,

$$\,{a_n} = a + \left( {n - 1} \right)d$$

Third term, $$\,{a_3} = a + \left( {3 - 1} \right)d$$

$$\,{a_3} = a + 2d \qquad\dots {\rm{Equation }}\left( 3 \right)$$

Seventh term, $$\,{a_7} = a + \left( {7 - 1} \right)d$$

$$\,{a_7} = a + 6d \quad \dots {\rm{ Equation}}\left( 4 \right)$$

Using Equation (3) and Equation (4) in Equation (1)

\begin{align}\left( {a + 2d} \right) + \left( {a + 6d} \right) &= 6\\2a + 8d &= 6\\a + 4d &= 3\\a &= 3 - 4d \qquad \dots \rm{Equation}\left( 5 \right)\end{align}

Using Equation (3) and Equation (4) in Equation (2)

$\left( {a + 2d} \right) \times \left( {a + 6d} \right) = 8$

Substituting the value of Equation (5) in above,

\begin{align}\left( {3 - 4d + 2d} \right) \times \left( {3 - 4d + 6d} \right)&= 8\\ \left( {3 - 2d} \right) \times \left( {3 + 2d} \right) &= 8\\{\left( 3 \right)^2} - {\left( {2d} \right)^2} &= 8\\ 9 - 4{d^2} &= 8\\4{d^2} &= 1\\{d^2} &= \frac{1}{4}\\d &= \pm \frac{1}{2}\\d& = \frac{1}{2}\;{\rm{ or }}\;\left( { - \frac{1}{2}} \right)\end{align}

Hence by substituting both the values of $$d,$$

When $$d = \frac{1}{2}$$

\begin{align}a &= 3 - 4d\\ &= 3 - 4 \times \frac{1}{2}\\& = 3 - 2\\ &= 1\end{align}

When $$d = - \frac{1}{2}$$

\begin{align}a &= 3 - 4d\\ &= 3 - 4 \times \left( { - \frac{1}{2}} \right)\\ &= 3 + 2\\& = 5\end{align}

We know that Sum of $$n^\rm{th}$$ term of AP Series, $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

When $$a = 1$$ and d = \frac{1}{2}\) \[\begin{align}{S_{16}} &= \frac{{16}}{2}\left[ {2 \times 1 + \left( {16 - 1} \right) \times \frac{1}{2}} \right]\\ &= 8\left[ {2 + \frac{{15}}{2}} \right]\\ &= 8 \times \frac{{19}}{2}\\ &= 76\end{align}

When $$a = 5$$ and $$d = - \frac{1}{2}$$

\begin{align}{S_{16}} &= \frac{{16}}{2}\left[ {2 \times 5 + \left( {16 - 1} \right) \times \left( { - \frac{1}{2}} \right)} \right]\\ &= 8\left[ {10 - \frac{{15}}{2}} \right]\\ &= 8 \times \frac{5}{2}\\& = 20\end{align}

Therefore, $${S_{16}} = 20,76$$

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