
How many terms of the A.P. 3, 5, 7, 9…. must be added to get the sum of 120?
Answer
528.3k+ views
Hint:
Find the first term and common difference of the given A.P. Then substitute these values in the formula to find the sum of n terms and thus find the number of terms, n.
Complete step-by-step answer:
A.P. represents arithmetic progression. It is a sequence of numbers such that the difference between the consecutive terms is constant. Difference here means the second minus the first term. Let us consider ‘d’ as the common difference of the arithmetic progress and d is a natural number.
We have been given the A.P.- 3,5,7,9….
where first term, a = 3
common difference, d = 5 – 3 = 2.
The sum of first n terms of an A.P. can be found using the formula,
\[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]\]
where n is the number of terms, which we need to find.
We have been given, \[{{S}_{n}}=120\].
\[\therefore \]Put \[{{S}_{n}}=120\], a = 3 and d = 2.Find the value of n.
\[\begin{align}
& 120=\dfrac{n}{2}\left[ 2\times 3+(n-1)2 \right] \\
& 240=n\left[ 6+2n-2 \right] \\
\end{align}\]
\[\begin{align}
& 240=n\left[ 4+2n \right] \\
& \therefore 240=4n+2{{n}^{2}} \\
& \Rightarrow 2{{n}^{2}}+4n-240=0 \\
\end{align}\]
Divide the entire equation by 2.
\[\therefore {{n}^{2}}+2n-120=0\]
The above expression is similar to the quadratic equation \[a{{x}^{2}}+bx+c=0\].
Thus comparing both equations, we get,
a = 1, b = 2 and c = -120.
Put the values in quadratic formula, \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
\[n=\dfrac{-2\pm \sqrt{4-4\times 1\times (-120)}}{2\times 1}=\dfrac{-2\pm
\sqrt{4+480}}{2}=\dfrac{-2\pm 22}{2}\]
\[n=\dfrac{-2+22}{2}\] and \[n=\dfrac{-2-22}{2}=\dfrac{-24}{2}=-12\]
\[\therefore n=\dfrac{20}{2}=10\].
The number of terms cannot be negative, so don’t take = -12.
\[\therefore n=10\]
Thus we got the number of terms as 10.
The number of terms can be added = 10 – 4 = 6 terms.
Note: Arithmetic progression is one of the very important portions to solve the sequence and series. Thus, it’s important that you remember the basic formulae to find the sum and also to get the number of terms in the series.
Find the first term and common difference of the given A.P. Then substitute these values in the formula to find the sum of n terms and thus find the number of terms, n.
Complete step-by-step answer:
A.P. represents arithmetic progression. It is a sequence of numbers such that the difference between the consecutive terms is constant. Difference here means the second minus the first term. Let us consider ‘d’ as the common difference of the arithmetic progress and d is a natural number.
We have been given the A.P.- 3,5,7,9….
where first term, a = 3
common difference, d = 5 – 3 = 2.
The sum of first n terms of an A.P. can be found using the formula,
\[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]\]
where n is the number of terms, which we need to find.
We have been given, \[{{S}_{n}}=120\].
\[\therefore \]Put \[{{S}_{n}}=120\], a = 3 and d = 2.Find the value of n.
\[\begin{align}
& 120=\dfrac{n}{2}\left[ 2\times 3+(n-1)2 \right] \\
& 240=n\left[ 6+2n-2 \right] \\
\end{align}\]
\[\begin{align}
& 240=n\left[ 4+2n \right] \\
& \therefore 240=4n+2{{n}^{2}} \\
& \Rightarrow 2{{n}^{2}}+4n-240=0 \\
\end{align}\]
Divide the entire equation by 2.
\[\therefore {{n}^{2}}+2n-120=0\]
The above expression is similar to the quadratic equation \[a{{x}^{2}}+bx+c=0\].
Thus comparing both equations, we get,
a = 1, b = 2 and c = -120.
Put the values in quadratic formula, \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
\[n=\dfrac{-2\pm \sqrt{4-4\times 1\times (-120)}}{2\times 1}=\dfrac{-2\pm
\sqrt{4+480}}{2}=\dfrac{-2\pm 22}{2}\]
\[n=\dfrac{-2+22}{2}\] and \[n=\dfrac{-2-22}{2}=\dfrac{-24}{2}=-12\]
\[\therefore n=\dfrac{20}{2}=10\].
The number of terms cannot be negative, so don’t take = -12.
\[\therefore n=10\]
Thus we got the number of terms as 10.
The number of terms can be added = 10 – 4 = 6 terms.
Note: Arithmetic progression is one of the very important portions to solve the sequence and series. Thus, it’s important that you remember the basic formulae to find the sum and also to get the number of terms in the series.
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