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Question
Find the sum of first 17 terms of an AP whose 4th and 9th terms are –15 and –30 respectively.
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Solution
Let the first term, common difference and the number of terms in an AP are a, d and n, respectively.
We know that, the nth term of an AP,
Tn = a + (n – 1)d ...(i)
∴ 4th term of an AP,
T4 = a + (4 – 1)d = –15 ...[Given]
⇒ a + 3d = –15 ...(ii)
And 9th term of an AP,
T9 = a + (9 – 1)d = –30 ...[Given]
⇒ a + 8d = –30 ...(iii)
Now, subtract equation (ii) from equation (iii), we get
a + 8d = –30
a + 3d = –15
– – +
5d = –15
⇒ d = –3
Put the value of d in equation (ii), we get
a + 3(–3) = –15
⇒ a – 9 = –15
⇒ a = –15 + 9 = – 6
∵ Sum of first n terms of an AP,
Sn = `n/2[2a + (n - 1)d]`
∴ Sum of first 17 terms of an AP,
S17 = `17/2 [2 xx (-6) + (17 - 1)(-3)]`
= `17/2 [-12 + (16)(-3)]`
= `17/2(-12 - 48)`
= `17/2 xx (-60)`
= 17 × (–30)
= –510
Hence, the required sum of first 17 terms of an AP is –510.
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