Question

In a geometric progression with a common ratio ‘q’, the sum of the first 109 terms exceeds the sum of the first 100 terms by 12. If the sum of the first nine terms of the progression is $\dfrac{\lambda }{{{q}^{100}}}$, then find the value of $\lambda$?
(a) 10
(b) 14
(c) 12
(d) 22

Answer 1

We start solving the problem by assigning the variable for the first term of the geometric progression. We then recall the definition of sum of first n terms of geometric progression as ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}$. We use this definition for the first condition that the sum of the first 109 terms exceeds the sum of the first 100 terms by 12 and gets a result. We then use the definition sum of geometric progression for the first 9 terms and compare results to get the required value of $\lambda$.

Complete step-by-step solution:
According to the problem, we have given a geometric progression with common ratio ‘q’ and the sum of the first 109 terms exceeds the sum of the first 100 terms by 12. We need to value of $\lambda$ if it is given that the sum of the first nine terms of the progression is $\dfrac{\lambda }{{{q}^{100}}}$.
Let us assume the first terms of the given geometric progression be ‘p’.
We know that sum of first n terms in a geometric progression is defined as ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}$, where ‘a’ is the first term of geometric progression and ‘r’ is the common ratio of that geometric progression.
According to the problem, the sum of the first 109 terms exceeds the sum of the first 100 terms by 12.
So, we have ${{S}_{109}}={{S}_{100}}+12$.
$\Rightarrow \dfrac{p\left( {{q}^{109}}-1 \right)}{\left( q-1 \right)}=\dfrac{p\left( {{q}^{100}}-1 \right)}{\left( q-1 \right)}+12$.
$\Rightarrow \dfrac{p\left( {{q}^{109}}-1 \right)}{\left( q-1 \right)}-\dfrac{p\left( {{q}^{100}}-1 \right)}{\left( q-1 \right)}=12$.
$\Rightarrow \dfrac{p\left( {{q}^{109}}-1-\left( {{q}^{100}}-1 \right) \right)}{\left( q-1 \right)}=12$.
$\Rightarrow \dfrac{p\left( {{q}^{109}}-1-{{q}^{100}}+1 \right)}{\left( q-1 \right)}=12$.
$\Rightarrow \dfrac{p\left( {{q}^{109}}-{{q}^{100}} \right)}{\left( q-1 \right)}=12$.
$\Rightarrow \dfrac{p\left( {{q}^{100}} \right)\left( {{q}^{9}}-1 \right)}{\left( q-1 \right)}=12$.
$\Rightarrow \dfrac{p\left( {{q}^{9}}-1 \right)}{\left( q-1 \right)}=\dfrac{12}{{{q}^{100}}}$ ------(1).
According to the problem, we have given that the sum of the first nine terms of the progression is $\dfrac{\lambda }{{{q}^{100}}}$.
So, we have $\dfrac{p\left( {{q}^{9}}-1 \right)}{\left( q-1 \right)}=\dfrac{\lambda }{{{q}^{100}}}$ -----(2).
Comparing equations (1) and (2), we get $\lambda =12$.
$\therefore$ The value of $\lambda$ is 12.
The correct option for the given problem is (c).

Note: We should confuse with the power of common ratio present in the definition of the sum of first n terms in geometric progression. We can also find the relation between the first term and common ratio which is to be substituted in finding the sum of the first 9 terms but it is a long process and needs precise calculation. Similarly, we can also expect this type of problem in arithmetic and harmonic progression.