Question

How many terms of the AP 21, 18, 15 … must be added to get the sum 0?

Answer 1

Hint: Here, we have to use the concept of the Arithmetic progression (A.P.) as the series given is in AP form. Arithmetic progression (A.P.) is the sequence of numbers such that the difference between the consecutive numbers remains constant. We will use the formula for the sum of $$n$$ number of terms in an AP and equate it to zero to find out the number of terms which should be added to get the sum of 0.

Formula used: We will use the formula of Sum of $$n$$ number of terms in an $$\mathrm{A}\mathrm{P}={\displaystyle \frac{n}{2}}\left[2a+(n-1)d\right]$$.

Complete step-by-step answer:
Series of numbers given is 21, 18, 15 …
Here in this series of AP first term is 21.
We can calculate the common difference by subtracting either 21 from 18 or 18 from 15.
$$d=18-21=15-18=-3$$
So, the common difference is $$-3$$.
Therefore, $$a=21$$and $$d=-3$$.
Now it is given that the sum of the given AP series is 0. So, we have to equate the formula of the sum of n number of terms in an AP equals to 0 and then by solving we will get the value of n (number of terms).
Therefore, Sum of n number of terms in an AP $$={\displaystyle \frac{n}{2}}\left[2a+(n-1)d\right]=0$$
Take the term $${\displaystyle \frac{\mathrm{n}}{2}}$$on the other side then the equation becomes
$$\Rightarrow 2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}=0$$
Now, we have to putting the value of a and d in the above equation, we get
$$\Rightarrow 2(21)+(n-1)(-3)=0$$
Multiplying the terms, we get
$$\Rightarrow 42-3n+3=0$$
Now by simplifying the above equation we get
$$\Rightarrow 3n=45$$
$$\Rightarrow n={\displaystyle \frac{45}{3}}=15$$
Hence, 15 terms of the given AP series is added to get the sum zero.

Note: We always have to keep in mind the basic A.P. and G.P series
A.P. series is $$a,a+d,a+2d,a+3d,............$$
$${\mathrm{n}}^{th}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}=a+\left(n-1\right)d$$
Where a is the first term and d is the common difference
G.P. series is $$a,ar,a{r}^2,a{r}^3,a{r}^4,.........$$
$${\mathrm{n}}^{\mathrm{t}\mathrm{h}}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}=a{r}^{n-1}$$
Where, a is the first term of the G.P. and r is the common ratio.