**IBDP** **Mathematics**** Question Analysis Topic: Mathematics - Arithmetic Progression**

**Exam Question:**

The 14th term of an A.P. is twice its 8th term. If its 6th term is −8, then find the sum of its first 20 terms.

**Answer:**

For IBDP Mathematics, you should know:

First, let's define the terms. Let **a **be the **first term **and **d **be the **common difference**.

Then, we have:

a_{14} = a + (14−1)d = a + 13d

a_{8} = a + (8−1)d = a + 7d

a_{6} = a + (6−1)d = a + 5d

**We also know that given ****a**_{14}** is twice of ****a**_{8}**:**

∴ a + 13d = 2(a+7d)

⇒ a + 13d = 2a + 14d

⇒ 2a + 14d − a − 13d = 0

⇒ a + d = 0

⇒ a = −d .....(1)

a_{6} = a + (6−1)d = a + 5d

∴ a + 5d = −8 ......(2)

**Then, substitute value ****a ****= −****d**** in equation (2). From this, we get:**

−d + 5d = −8

⇒ 4d = −8

⇒ d = −2

**Inversely, substituing the value of ****d ****=−****2**** in equation (1) we get**

a = 2.

Then S_{20} = (20/2)[4 + (10 − 1)(−2)] = 10(4 − 38) = −340.

Work hard for your IBDP Mathematics examination!

End of analysis. Great!