Question

How many terms of the AP: $- 15, - 13, - 11,.....,$ are needed to make the sum $- 55$ ?

Answer 1

The given problem requires us to find the sum of an arithmetic progression. The first few terns and the last term of the series is given to us in the question. For finding out the sum of an arithmetic progression, we need to know the first term, the common difference and the number of terms in the arithmetic progression. We can find out the common difference of an arithmetic progression by knowing the difference of any two consecutive terms of the series.

Complete step by step answer:
So, we have the arithmetic progression as $- 15, - 13, - 11,.....,$ The difference of any two consecutive terms of the given series is constant. So, the given sequence is an arithmetic progression. Now, we have to find the sum of this arithmetic progression. Here, first term $= a = - 15$. Now, we can find the common difference of the arithmetic progression by subtracting any two consecutive terms in the series.
So, common ratio $= d = - 13 - \left( { - 15} \right) = - 13 + 15 = 2$
So, $d = 2$ .

Now, we know the first term and the common difference between the terms of the given AP. So, we will find the sum of n terms of the provided arithmetic progression and equate it to $- 55$ to find the value of n, the number of terms. Now, we can find the sum of the given arithmetic progression using the formula,
$S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Hence, the sum of AP $= S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Substituting the values of a and d in the formula and equating it to $- 55$, we get,
$\Rightarrow S = \dfrac{n}{2}\left[ {2\left( { - 15} \right) + \left( {n - 1} \right)\left( 2 \right)} \right] = - 55$
Opening the brackets and cancelling the common factors in numerator and denominator, we get,
$\Rightarrow \dfrac{n}{2}\left[ { - 30 + 2n - 2} \right] = - 55$

Taking $2$ common from the bracket, we get,
$\Rightarrow \dfrac{{2n}}{2}\left[ {n - 16} \right] = - 55$
Cancelling the common factors in numerator and denominator, we get,
$\Rightarrow n\left( {n - 16} \right) = - 55$
Multiplying the terms by opening the brackets, we get,
$\Rightarrow {n^2} - 16n + 55 = 0$
Solving the quadratic equation by splitting the middle term, we get,
$\Rightarrow {n^2} - 11n - 5n + 55 = 0$
We split the middle term $- 16n$ into two terms $- 11n$ and $- 5n$ since the product of these two terms, $55{n^2}$ is equal to the product of the constant term and coefficient of ${x^2}$ and sum of these terms gives us the original middle term, $- 16n$.

Taking n common from first two terms and $- 5$ from last two terms, we get,
$\Rightarrow n\left( {n - 11} \right) - 5\left( {n - 11} \right) = 0$
$\Rightarrow \left( {n - 11} \right)\left( {n - 5} \right) = 0$
Now, if the product of two terms is zero. So, we know that either of the two terms must be zero. Hence, we have,
$\Rightarrow \left( {n - 11} \right) = 0$
$\Rightarrow \left( {n - 5} \right) = 0$
Finding the value of n, we get,
$\Rightarrow n = 11$ or $\Rightarrow n = 5$
So, the possible values of $n$ are: $5$ and $11$.

Hence, the sum of first $5$ and 11 terms of the arithmetic progression $- 15, - 13, - 11,.....,$ is $- 55$.

Note: Arithmetic progression is a series where any two consecutive terms have the same difference between them. The common difference of an arithmetic series can be calculated by subtraction of any two consecutive terms of the series. The sum of n terms of an arithmetic progression can be calculated if we know the first term, the number of terms and difference of the arithmetic series as: $S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.