Question

The sum of a GP with common ratio 3 is 364 and last term is 243, then the number of terms is
(a) 6
(b) 5
(c) 4
(d) 10

Answer 1

Hint:By using the formula of ${{n}^{th}}$ term of geometric progression. Find a relation between first term and last term. By that you get the first term in terms of n as you know ${{n}^{th}}$ term and common ratio. By using this value of a in sum of n terms formula the whole equation converts into terms of n, which you can solve easily and find the value of n as required in the question. The ${{n}^{th}}$ term t of a geometrical progression with first term ‘a’ and common ratio ‘r’. we write it in the form of:
$T_{nth}=a{{r}^{n-1}}$

Complete step-by-step answer:
If the first term is ‘a’ and the common ratio is ‘r’ for a geometric progression.
Given in question that common ratio value is 3. Given in question that ${{n}^{th}}$ term of this sequence is 243. We know that ${{n}^{th}}$ term formula is given by $T_{nth}=a{{r}^{n-1}}$.
By substituting value of $T_{nth}$ , r we get,
$243=a{{3}^{n-1}}$
By simplifying we get the value of ‘a’ as given below.
$a=\dfrac{343\times 3}{{{r}^{n}}}=\dfrac{729}{{{3}^{n}}}$
Let the sum of n terms of the GP be represented as S:
By general knowledge of progression, we know formula of S:
$S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$
By substituting ‘a’ as the value obtained in the equation, and substitute the value of S as given in question i.e 364. We get
$364=\dfrac{\dfrac{729}{{{3}^{n}}}\left( {{3}^{n}}-1 \right)}{3-1}$
By cross multiplying the equation terms we get,
$364\times 2=729-\dfrac{729}{{{3}^{n}}}$
By simplifying the equation ore, we get
\[{{3}^{n}}={{3}^{6}}\]
By comparing the both sides of equation, we can say \[n=6\]
So, satisfying all given conditions, we need 6 terms.
Option (a) is correct.

Note: Be careful while taking the sum of terms, the whole equation is ${{3}^{n}}$ . So, try to convert remaining terms in ${{3}^{n}}$.Students should remember the general term of G.P i.e $T_{nth}=a{{r}^{n-1}}$ and sum of n terms $S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ as they given n finite terms we have to apply this formula,don't get confused with sum of infinite series of G.P formula i.e ${\dfrac{a}{1-r}}$.