# How many terms of the series 12, 16, 20….. must be taken to make the sum equal to 208?

Last updated date: 20th Mar 2023

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Answer

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Hint: First of all, take the total number of terms as n. We can see that the given series is in A.P. So, find its first term a and the common difference d and use the formula of sum of n terms that is \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] to get the value of n.

Complete step-by-step answer:

We are given the terms 12, 16, 20, 24…… We have to find the number of terms that must be taken to make the sum of the terms equal to 208.

Let us consider the series given in the question,

12, 16, 20, 24…..

We can see that in the above series, each term is greater than the previous term by 4. Also, we know that the general terms of the arithmetic progression are as follows:

\[a,a+d,a+2d,a+3d.....\]

Here, the first term is ‘a’ and the common difference is ‘d’.

Similarly, in the given series, the first term is 12 and the common difference is 4. So, this series forms an A.P.

Let us consider that there are n terms in this series. We know that sum of n terms of A.P. \[=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] where a, d and n are the first term, common difference, and the number of terms respectively.

So, we get the sum of total n terms of the given series in A.P \[=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]

By substituting a = 12 and d = 4, we get,

Sum of total n terms of the given series \[=\dfrac{n}{2}\left[ 2\left( 12 \right)+\left( n-1 \right)4 \right]\]

\[=\dfrac{n}{2}\left[ 24+\left( n-1 \right)4 \right]\]

Now, we are given that this sum is equal to 208, so we get,

\[\dfrac{n}{2}\left[ 24+\left( n-1 \right)4 \right]=208\]

By cross multiplying the above equation, we get,

\[n\left( 24+\left( n-1 \right)4 \right)=208\times 2\]

Or, \[24n+4n\left( n-1 \right)=416\]

By dividing 2 on both sides of the above equation, we get,

\[12n+2n\left( n-1 \right)=208\]

By simplifying the above equation, we get,

\[2{{n}^{2}}-2n+12n-208=0\]

Or, \[2{{n}^{2}}+10n-208=0\]

We can write 10n = 26n – 16n, so we get,

\[2{{n}^{2}}+26n-16n-208=0\]

Or, \[2n\left( n+13 \right)-16\left( n+13 \right)=0\]

By taking (n + 13) common, we get,

\[\left( n+13 \right)\left( 2n-16 \right)=0\]

So, we get n = – 13 and \[n=\dfrac{16}{2}=8\]

Since we know that the number of terms could be negative, so we get the number of terms as 8.

Note: Here, students must note that a number of terms are always positive. Also, students are advised to properly substitute the value of a and d to find the value of n and recheck each step twice. Students can also cross-check their answer by substituting the value of n in \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] and checking if LHS = RHS or not.

Complete step-by-step answer:

We are given the terms 12, 16, 20, 24…… We have to find the number of terms that must be taken to make the sum of the terms equal to 208.

Let us consider the series given in the question,

12, 16, 20, 24…..

We can see that in the above series, each term is greater than the previous term by 4. Also, we know that the general terms of the arithmetic progression are as follows:

\[a,a+d,a+2d,a+3d.....\]

Here, the first term is ‘a’ and the common difference is ‘d’.

Similarly, in the given series, the first term is 12 and the common difference is 4. So, this series forms an A.P.

Let us consider that there are n terms in this series. We know that sum of n terms of A.P. \[=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] where a, d and n are the first term, common difference, and the number of terms respectively.

So, we get the sum of total n terms of the given series in A.P \[=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]

By substituting a = 12 and d = 4, we get,

Sum of total n terms of the given series \[=\dfrac{n}{2}\left[ 2\left( 12 \right)+\left( n-1 \right)4 \right]\]

\[=\dfrac{n}{2}\left[ 24+\left( n-1 \right)4 \right]\]

Now, we are given that this sum is equal to 208, so we get,

\[\dfrac{n}{2}\left[ 24+\left( n-1 \right)4 \right]=208\]

By cross multiplying the above equation, we get,

\[n\left( 24+\left( n-1 \right)4 \right)=208\times 2\]

Or, \[24n+4n\left( n-1 \right)=416\]

By dividing 2 on both sides of the above equation, we get,

\[12n+2n\left( n-1 \right)=208\]

By simplifying the above equation, we get,

\[2{{n}^{2}}-2n+12n-208=0\]

Or, \[2{{n}^{2}}+10n-208=0\]

We can write 10n = 26n – 16n, so we get,

\[2{{n}^{2}}+26n-16n-208=0\]

Or, \[2n\left( n+13 \right)-16\left( n+13 \right)=0\]

By taking (n + 13) common, we get,

\[\left( n+13 \right)\left( 2n-16 \right)=0\]

So, we get n = – 13 and \[n=\dfrac{16}{2}=8\]

Since we know that the number of terms could be negative, so we get the number of terms as 8.

Note: Here, students must note that a number of terms are always positive. Also, students are advised to properly substitute the value of a and d to find the value of n and recheck each step twice. Students can also cross-check their answer by substituting the value of n in \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] and checking if LHS = RHS or not.

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