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प्रश्न
Which term of the AP: –2, –7, –12,... will be –77? Find the sum of this AP upto the term –77.
उत्तर
Given, AP: –2, –7, –12,...
Let the nth term of an AP is –77
Then, first term (a) = –2 and
Common difference (d) = –7 – (–2)
= –7 + 2
= –5
∵ nth term of an AP, Tn = a + (n – 1)d
⇒ –77 = –2 + (n – 1)(–5)
⇒ –75 = –(n – 1) × 5
⇒ (n – 1) = 15
⇒ n = 16
So, the 16th term of the given AP will be –77
Now, the sum of n terms of an AP is
Sn = `n/2[2a + (n - 1)d]`
So, sum of 16 terms i.e., upto the term –77
S16 = `16/2 [2 xx (-2) + (n - 1)(-5)]`
= 8[–4 + (16 – 1)(–5)]
= 8(–4 – 75)
= 8 × (–79)
= –632
Hence, the sum of this AP upto the term –77 is –632.
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What is the sum of the first 30 natural numbers ?
There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.
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Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
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Now,
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