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

Answer

Verified

378.9k+ views

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.

Recently Updated Pages

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

State Gay Lusaaccs law of gaseous volume class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

What is pollution? How many types of pollution? Define it

Change the following sentences into negative and interrogative class 10 english CBSE

Which is the tallest animal on the earth A Giraffes class 9 social science CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE