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# 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. Verified
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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.