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Question
The 5th term and the 9th term of an Arithmetic Progression are 4 and – 12 respectively.
Find:
- the first term
- common difference
- sum of 16 terms of the AP.
Solution
Let a and d be the first term and common difference of A.P.
Then by Tn = a + (n – 1)d
Given T5 = 4
`\implies` a + 4d = 4 ...(1)
And T9 = – 12
`\implies` a + 8d = – 12 ...(2)
Solving equations (1) and (2), we get
– 4d = 16
`\implies` d = – 4
Put this value in equation (1)
a + 4 × (– 4) = 4
a – 16 = 4
a = 20
∴ a. First term a = 20
b. Common difference d = – 4
c. Sum of n terms = `n/2 [2a + (n - 1)d`
∴ Sum of 16 terms = `16/2 [2 xx 20 + 15 xx (-4)]`
= 8 [40 – 60]
= 8 × (– 20)
= – 160
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