Question

Sachin invested in a national savings scheme. In the first year he invested $Rs.5000$, in the second year $Rs.7000$, in the third year $Rs.9000$ and so on. Find the total amount that he invested in $12$ years.

Answer 1

Since the amount is increased sequentially every year, we can use the concept of Arithmetic sequences. The fixed amount which is increasing will be the common difference. Then using the equation for the sum of terms of an arithmetic sequence we can find the total amount invested in $12$ years.

Useful formula:
For an arithmetic sequence with first term $a$ and common difference $d$, the ${n^{th}}$ term is
${x_n} = a + (n - 1)d$.
Sum of first $n$ terms is ${S_n} = \dfrac{1}{2}n({x_n} + a)$

Complete step by step solution:
Given that in the first year Sachin invested $Rs.5000$, in the second year $Rs.7000$, in the third year $Rs.9000$ and so on.
So we can see that every year the amount increased by $7000 - 5000 = 9000 - 7000 = 2000$.
We are asked to find the total amount invested in $12$ years.
Here we can use the concept of Arithmetic Sequences.
For an arithmetic sequence with first term $a$ and common difference $d$, the ${n^{th}}$ term is
${x_n} = a + (n - 1)d$.
Here $a = 5000,d = 2000$
Put $n = 12$, we get the amount deposited by Sachin in the ${12^{th}}$ year.
$\Rightarrow {x_{12}} = a + (12 - 1)d = a + 11d$
Substituting for $a,d$ we have,
$\Rightarrow {x_{12}} = 5000 + (11 \times 2000) = 5000 + 22000 = 27000$
To find the total amount invested in $12$ years, we use the sum formula.
Sum of first $n$ terms is ${S_n} = \dfrac{1}{2}n({x_n} + a)$
Substituting for $n,a,{x_n}$ we get,
$\Rightarrow {S_{12}} = \dfrac{1}{2}12(27000 + 5000)$
Simplifying we get,
$\Rightarrow {S_{12}} = 6 \times 32000 = 192000$
Therefore the total amount invested in $12$ years by Sachin is $Rs.19200$.

$\therefore$ The answer is $Rs.192000$.

Note: We can find the sum of amount invested in $12$ years using another equation.
${S_n} = 2a + (n - 1)d$, where $a,d$ are the first term and common difference respectively.
So direct substitution will lead to an answer. However, if we want to find the amount deposited in the ${12^{th}}$ year as well, we use the first method.