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:
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First, let's define the terms. Let a be the first term and d be the common difference.
Then, we have:
a14 = a + (14−1)d = a + 13d
a8 = a + (8−1)d = a + 7d
a6 = a + (6−1)d = a + 5d
We also know that given a14 is twice of a8:
∴ a + 13d = 2(a+7d)
⇒ a + 13d = 2a + 14d
⇒ 2a + 14d − a − 13d = 0
⇒ a + d = 0
⇒ a = −d .....(1)
a6 = 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 S20 = (20/2)[4 + (10 − 1)(−2)] = 10(4 − 38) = −340.
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