Arithmetic Progressions
The 14th term of an AP is twice its 8th term. If its 6th terms is -8, then find the sum of its first 20 terms.
Let a and d be the first term and the common difference of the AP, respectively.
∴ nth term of the AP, an = a + (n-1)d
So,
a14 = a+ (14-1)d = a + 13d
a8 = a + (8 - 1)d = a + 7d
a6 = a+ (6 - 1)d = a + 5d
According to the question,
a14 = 2a8
⇒ a + 13d = 2 (a + 7d)
⇒ a + d = 0 .... (i)
Also,
a6 = a + 5d = - 8 ... (ii)
solving (i) and (ii), we get
a = 2 and d = -2
∴ S20 = 20/2 [2 x 2 + (20 - 1)(-2)]
= - 340
Hence, the sum of the first 20 terms is -340.
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Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a – 10, d = 10
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = –2, d = 0
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = 4, d = – 3
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = – 1.25, d = – 0.25
For the following APs, write the first term and the common difference
3, 1, – 1, – 3, . . .
For the following APs, write the first term and the common difference
– 5, – 1, 3, 7, . . .
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