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प्रश्न
Find the sum:
1 + (–2) + (–5) + (–8) + ... + (–236)
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उत्तर
Here, first term (a) = 1
And common difference (d) = (–2) – 1 = –3
∵ Sum of n terms of an AP,
Sn = `n/2[2a + (n - 1)d]`
⇒ Sn = `n/2[2 xx 1 + (n - 1) xx (-3)]`
⇒ Sn = `n/2(2 - 3n + 3)`
⇒ Sn = `n/2(5 - 3n)` ...(i)
We know that, if the last term (l) of an AP is known, then
l = a + (n – 1)d
⇒ –236 = 1 + (n – 1)(–3) ...[∵ l = –236, given]
⇒ –237 = – (n – 1) × 3
⇒ n – 1 = 79
⇒ n = 80
Now, put the value of n in equation (i), we get
Sn = `80/2[5 - 3 xx 80]`
= 40(5 – 240)
= 40 × (–235)
= –9400
Hence, the required sum is –9400.
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