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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