# The sum of some terms of G.P is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.

Last updated date: 18th Mar 2023

•

Total views: 306.3k

•

Views today: 3.85k

Answer

Verified

306.3k+ views

Hint: First of all, use the formula for sum of n terms of G.P that is \[{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]. Here put \[{{S}_{n}}=315,\text{ }a=5\] and \[r=2\] to get the value of n. Then, find the last term by using the formula \[{{a}_{n}}=a{{r}^{n-1}}\] by putting the value of a, r and n.

We are given that the sum of some terms of G.P is 315 whose first term is 5 and common ratio is 2.

Here, we have to find the last term and the number of terms in the given G.P.

Before proceeding with the question, we must know that G.P or geometrical progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non – zero number called as common ratio.

Now, we know that the general form of geometric sequence is

\[a,a{{r}^{1}},a{{r}^{2}},a{{r}^{3}},a{{r}^{4}}.....\]

where a is the first term and r is the common ratio of G.P.

As we are given that the first term is 5. Therefore, let us consider a = 5.

Also, as we are given that common ratio is 2, therefore, let us consider r = 2.

Let us consider that there are total n terms in G.P.

Let us take the sum of these n terms of the G.P \[={{S}_{n}}\].

As we are given that the sum of the terms of this G.P is 315, therefore we get \[{{S}_{n}}=315\].

Now, we know that the sum of n terms of G.P is \[{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\].

where a is the first term and r is the common ratio of G.P.

By putting the values of \[{{S}_{n}}\], a and r in the above expression, we get

\[315=\dfrac{5\left( 1-{{2}^{n}} \right)}{\left( 1-2 \right)}\]

By cross multiplying and simplifying the above equation, we get

\[\Rightarrow \left( 315 \right)\left( -1 \right)=5\left( 1-{{\left( 2 \right)}^{n}} \right)\]

\[\Rightarrow -315=5-5{{\left( 2 \right)}^{n}}\]

By taking the term containing “n” on one side and the constant term on the other side, we get,

\[\Rightarrow 5.{{\left( 2 \right)}^{n}}=5+315\]

\[5.{{\left( 2 \right)}^{n}}=320\]

By dividing 5 on both sides, we get

\[\Rightarrow {{\left( 2 \right)}^{n}}=\dfrac{320}{5}\]

\[\Rightarrow {{\left( 2 \right)}^{n}}=64\]

We can write \[64={{2}^{6}}\]. Therefore, we get,

\[\Rightarrow {{2}^{n}}={{2}^{6}}\]

We know that when we have \[{{a}^{p}}={{a}^{q}}\], p and q will be equal for all values of \[a\ne 1,-1,0\]. Using this in the above equation, we get, n = 6.

Therefore, we have found that the number of terms in the given G.P is 6.

Now, we know that the last term of given G.P is nothing but the sixth term of this G.P as this contains 6 terms only.

Also, we know that the general term of G.P is given by \[{{a}_{n}}=a{{r}^{n-1}}\].

By putting the values of a = 5, r = 2 and n = 6, we get

\[{{a}_{6}}=5.{{\left( 2 \right)}^{6-1}}\]

\[\Rightarrow {{a}_{6}}=5.{{\left( 2 \right)}^{5}}\]

\[\Rightarrow {{a}_{6}}=5.\left( 32 \right)\]

\[\Rightarrow {{a}_{6}}=160\]

Therefore, we have found that the sixth or last term of this G.P is equal to 160.

Note: Here, students can cross – check the answer by manually writing the 6 terms of the G.P as follows:

We know that general form of geometric sequence is

\[a,a{{r}^{2}},a{{r}^{3}}.....a{{r}^{n-1}}\]

Here, we have found that the number of terms is 6. Therefore, our geometric sequence is \[a,a{{r}^{2}},a{{r}^{3}},a{{r}^{4}},a{{r}^{5}}\].

As we are given that a = 5 and r = 2, therefore we get G.P as

\[5,5.\left( 2 \right),5.{{\left( 2 \right)}^{2}},5{{\left( 2 \right)}^{3}}.5{{\left( 2 \right)}^{4}},5{{\left( 2 \right)}^{5}}\]

Or, \[5,\text{ }10,\text{ }20,\text{ }40,\text{ }80,\text{ }160\]

Therefore, we get the last term as 160 and sum of terms as 5 + 10 + 20 + 40 + 80 + 160 = 315 which is the same as given in the question.

Hence, our answer is correct.

We are given that the sum of some terms of G.P is 315 whose first term is 5 and common ratio is 2.

Here, we have to find the last term and the number of terms in the given G.P.

Before proceeding with the question, we must know that G.P or geometrical progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non – zero number called as common ratio.

Now, we know that the general form of geometric sequence is

\[a,a{{r}^{1}},a{{r}^{2}},a{{r}^{3}},a{{r}^{4}}.....\]

where a is the first term and r is the common ratio of G.P.

As we are given that the first term is 5. Therefore, let us consider a = 5.

Also, as we are given that common ratio is 2, therefore, let us consider r = 2.

Let us consider that there are total n terms in G.P.

Let us take the sum of these n terms of the G.P \[={{S}_{n}}\].

As we are given that the sum of the terms of this G.P is 315, therefore we get \[{{S}_{n}}=315\].

Now, we know that the sum of n terms of G.P is \[{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\].

where a is the first term and r is the common ratio of G.P.

By putting the values of \[{{S}_{n}}\], a and r in the above expression, we get

\[315=\dfrac{5\left( 1-{{2}^{n}} \right)}{\left( 1-2 \right)}\]

By cross multiplying and simplifying the above equation, we get

\[\Rightarrow \left( 315 \right)\left( -1 \right)=5\left( 1-{{\left( 2 \right)}^{n}} \right)\]

\[\Rightarrow -315=5-5{{\left( 2 \right)}^{n}}\]

By taking the term containing “n” on one side and the constant term on the other side, we get,

\[\Rightarrow 5.{{\left( 2 \right)}^{n}}=5+315\]

\[5.{{\left( 2 \right)}^{n}}=320\]

By dividing 5 on both sides, we get

\[\Rightarrow {{\left( 2 \right)}^{n}}=\dfrac{320}{5}\]

\[\Rightarrow {{\left( 2 \right)}^{n}}=64\]

We can write \[64={{2}^{6}}\]. Therefore, we get,

\[\Rightarrow {{2}^{n}}={{2}^{6}}\]

We know that when we have \[{{a}^{p}}={{a}^{q}}\], p and q will be equal for all values of \[a\ne 1,-1,0\]. Using this in the above equation, we get, n = 6.

Therefore, we have found that the number of terms in the given G.P is 6.

Now, we know that the last term of given G.P is nothing but the sixth term of this G.P as this contains 6 terms only.

Also, we know that the general term of G.P is given by \[{{a}_{n}}=a{{r}^{n-1}}\].

By putting the values of a = 5, r = 2 and n = 6, we get

\[{{a}_{6}}=5.{{\left( 2 \right)}^{6-1}}\]

\[\Rightarrow {{a}_{6}}=5.{{\left( 2 \right)}^{5}}\]

\[\Rightarrow {{a}_{6}}=5.\left( 32 \right)\]

\[\Rightarrow {{a}_{6}}=160\]

Therefore, we have found that the sixth or last term of this G.P is equal to 160.

Note: Here, students can cross – check the answer by manually writing the 6 terms of the G.P as follows:

We know that general form of geometric sequence is

\[a,a{{r}^{2}},a{{r}^{3}}.....a{{r}^{n-1}}\]

Here, we have found that the number of terms is 6. Therefore, our geometric sequence is \[a,a{{r}^{2}},a{{r}^{3}},a{{r}^{4}},a{{r}^{5}}\].

As we are given that a = 5 and r = 2, therefore we get G.P as

\[5,5.\left( 2 \right),5.{{\left( 2 \right)}^{2}},5{{\left( 2 \right)}^{3}}.5{{\left( 2 \right)}^{4}},5{{\left( 2 \right)}^{5}}\]

Or, \[5,\text{ }10,\text{ }20,\text{ }40,\text{ }80,\text{ }160\]

Therefore, we get the last term as 160 and sum of terms as 5 + 10 + 20 + 40 + 80 + 160 = 315 which is the same as given in the question.

Hence, our answer is correct.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE