Question

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

Answer 1

Answer:

$\large\boxed{\sf{5\;\;and\;\;11\;\;terms}}$

Step-by-step explanation:

Given an AP such that,

• -15, -13, -11, ..........

Here, we have ,

• First term, a = -15
• Common difference, d = - 13 - ( -15 )= 2

To find the number of terms needed to make the sum -55,

We know that, sum of n terms of an AP is,

• $\large\red{S_{n} = \dfrac{n}{2}\left[2a+(n-1)d\right]}$

Let the number of terms required is n.

Therefore, we will get,

$= > S_{n} = \dfrac{n}{2} (2 \times ( - 15) +2 (n - 1)) \\ \\ = > \frac{n}{2} ( - 30 + 2n - 2) = - 55 \\ \\ = > n(2n - 32) = - 55 \times 2 \\ \\ = > 2 {n}^{2} - 32n = - 110 \\ \\ = > 2( {n}^{2} - 16n + 55) = 0 \\ \\ = > {n}^{2} - 16n + 55 = 0 \\ \\ = > {n}^{2} - 11n - 5n + 55 = 0 \\ \\ = > n(n - 11) - 5(n - 11) = 0 \\ \\ = > (n - 11)(n - 5) = 0$

Now, we have two cases.

CASE I:

When n - 11 = 0

=> n = 11

CASE II:

When n - 5 = 0

=> n = 5

Hence, the sum of 5 and 11 terms of the given AP will be equal to -55.