Simple Arithmatic progrssion problem

Question

A man saves Rs $32$ during first year ,$36$ in the next year $40$ in $3$rd year .if he continue his savings in this sequence,in how many years he saves $2000$ Rs.

Rs=currency

Answer 1

Notice that the above numbers form a sequence in AP $32,36,40,....$

So, take the first term as $a=32$ and the common difference is $d=36-32=4$

Use the formula $S_n=\dfrac n2[2a+(n-1)d]$

$2000=\dfrac n2[2(32)+(n-2)4]$

and continue from here to find $n$

Answer 2

We note: each term of the sequence $32, 36, 40 ...$ has a common difference of $4$.

The $n$-th term of a sequence which is arithmetic is given by $a_n = a_1 + (n-1)d$ where $d$ is the common difference and $a_1$ the first term (here, $32$).

The sum of the terms $a_1 + a_2 + ... + a_n$ is also given by $S_n = \frac{n}{2}(a_1 + a_n)$.

Substitution of our definition of $a_n$, $a_1 = 32$, $S_n = 2000$, and $d = 4$ should give you an equation you can solve for $n$.