Question

(i) How many terms of the sequence $18,16,14,...$ should be taken so that their sum is zero?
(ii) How many terms are there in the A.P. whose first and fifth terms are $-14$ and $2$ respectively and the sum of the terms is 40?
(iii) How many terms of the A.P., $9,17,25,...$ must be taken so that their sum is $636$?
(iv) How many terms of the A.P., $63,60,57,...$ must be taken so that their sum is $693$?
(v) How many terms of the A.P., $27,24,\mathrm{21...}$ should be taken so that their sum is zero?
(vi) How many terms of the A.P., $45,39,\mathrm{33...}$must be taken so that their sum is $180$?

Answer 1

In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the number of terms.

(i) A.P. is $18,16,14,...$

So here, let us find the number of terms whose sum is $0$. For that, we will use the formula,

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

Where; $a=$ first term for the given A.P.

$d=$ common difference of the given A.P.

$n=$ number of terms

The first term \((a) = 18\)

The sum of $n$ terms $S_n=0$

Common difference of the A.P. , $d=a_2-a_1$

$=16-18=-2$

$\Rightarrow d=-2$

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

$\frac{n}{2}\left(2\right(18)+(n-1\left)\right(-2)=0$

$\Rightarrow n(38-2n)=0$

$\Rightarrow 38-2n=0$ [Since the number of terms cannot be zero, $n\ne 0$]

$\Rightarrow 38=2n$

$\Rightarrow n=19$

(ii)

Given, $a=-14,a_5=2$ and sum of $n$-terms $S_n=40$

$a_5=2$

$\Rightarrow a+4d=2$

$\Rightarrow -14+4d=2$

$\Rightarrow 4d=16$

$\Rightarrow d=4$

Further, let us find the number of terms whose sum is $40$. For that, we will use the formula,

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

$40=\frac{n}{2}\left(2\right(-14)+(n-1\left)4\right)$

$\Rightarrow 80=n(-28+4n-4)$

$\Rightarrow 80=n(4n-32)$

$\Rightarrow 4n^2-32n-80=0$

$\Rightarrow n^2-8n-10=0$

$\Rightarrow n^2-10n+2n-10=0$

$\Rightarrow n(n-10)+2(n-10)=0$

$\Rightarrow (n-10)(n+2)=0$

$\Rightarrow n=10$ or $n=-2$

Since the number of terms cannot be negative. Therefore, the number of terms $n$ is $10$

(iii)

A.P. is $9,17,25,...$

So here, let us find the number of terms whose sum is $636$. For that, we will use the formula,

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

The first term $\left(a\right)=9$

The sum of $n$ terms $S_n=636$

Common difference of the A.P. ., $d=17-9=8$

So, on substituting the values in the formula for the sum of $n$ terms of an A.P., we get,

$636=\frac{n}{2}\left(2\right(9)+(n-1\left)8\right)$

$\Rightarrow 636=\frac{n}{2}\left(1\right)+8n-8)$

$\Rightarrow 636=\frac{n}{2}(10+8n)$

$\Rightarrow 636=5n+4n^2$

$\Rightarrow 4n^2+5n-636=0$

$\Rightarrow 4n^2-48n+53n-636=0$

$\Rightarrow 4n(n-12)+53(n-12)=0$

$\Rightarrow (4n+53)(n-12)=0$

$\Rightarrow 4n=-53$ or $n=12$

Since, the number of terms cannot be negative. Therefore, the number of terms $n$ is $12$

(iv) A.P. is $63,60,57,...$

So here, let us find the number of terms whose sum is $693$. For that, we will use the formula,

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

The first term $\left(a\right)=63$

The sum of $n$ terms $S_n=693$

Common difference of the A.P. , $d=60-63=-3$

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

$693=\frac{n}{2}\left(2\right(63)+(n-1\left)\right(-3\left)\right)$

$\Rightarrow 693=\frac{n}{2}(129-3n)$

$\Rightarrow 1386=129n-3n^2$

$\Rightarrow 3n^2-129n+1386=0$

$\Rightarrow n^2-43n+462=0$

$\Rightarrow n^2-22n-21n+462=0$

$\Rightarrow n(n-22)-21(n-22)=0$

$\Rightarrow (n-22)(n-21)=0$

$\Rightarrow n=22,n=21$

So, the sum of $22$ as well as $21$ terms is $693$.

Therefore, the number of terms $n$ is $21$ or $22$

(v)
The given AP is $27,24,21,...$
First term of the AP, $a=27$
Common difference $d=24-27=-3$
Let the sum of the first $x$ terms of the AP be $0$.
Sum of first $x$ terms $=\frac{x}{2}\left(2\right(27)+(x-1\left)\right(-3\left)\right)$

$\Rightarrow \frac{x}{2}\left(2\right(27)+(x-1\left)\right(-3\left)\right)=0$

$\Rightarrow 54-3x+3=0$

$\Rightarrow 57-3x=0$

$\Rightarrow 3x=57$

$\Rightarrow x=19$
Thus, the sum of the first $19$ terms of the AP is $0$.

(vi) Given AP is $45,39,\mathrm{33...}$

So here, let us find the number of terms whose sum is $180$. For that, we will use the formula,

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

The first term $\left(a\right)=45$

The sum of $n$ terms $\left(Sn\right)=180$

Common difference of the A.P. , $d=39-45=-6$

$180=\frac{n}{2}(2a+(n-1\left)d\right)$

$\Rightarrow 180=\frac{n}{2}\left(2\right(45)+(n-1\left)\right(-6\left)\right)$

$\Rightarrow 360=n(90-6n+6)$

$\Rightarrow 360=n(96-6n)$

$\Rightarrow 360=96n-6n^2$

$\Rightarrow 6n^2-96n+360=0$

$\Rightarrow n^2-16n+60=0$

$\Rightarrow n^2-10n-6n+60=0$

$\Rightarrow n(n-10)-6(n-10)=0$

$\Rightarrow (n-10)(n-6)=0$

$\Rightarrow n=10,n=6$

So, the sum of $10$ as well as $6$ terms is $180$.

Therefore, the number of terms $n$ is $6$ or $10$.