The sum of 11 terms of an AP whose middle term s 30 is

(1) 320 (2)330 (3)340 (4)350

n=11 (this is the total no of terms)

(a1 + a11)/2 = 30 (sum of first and eventh term)

which gives a1 + a11 = 60 (first and 11th term)

now sum of AP = n x (a1 + a11) /2

= 11 x 60 /2

= 11 x 30

= 330

Hope this helps. Thumbs up if you comprehendo ! xD 

  • 30
a6 = a + 5d (a6 is the middlemost term)?
30 = a + 5d (1)?

S11 = [11 (2a + 10d)] ? 2?
S11 = [11 (a + 5d)]?
S11 = [11 (30)] From (1)?
hence?
S11 = 330

So answer will be second (2)
  • 33
Find H.P whose 3rd and 14th term are respectively 6/7 and 1/3
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Find the GP in which 5th term is 27/16 and 9th term is 1/3
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  • 52
hope it will help you : )
  • -2
11*30=330
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There numbers whose sum is 18 are in A.P.;if 2,4,11 are added to them respectively,the resulting numbers are in G.P. Determine the numbers.
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330 is the answer

  • 0
330 is the correct answer.
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Answer =11?30=330
Option:-(2)
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A11=an+(11-1)d
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