Login

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
View Full Answer
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
  • 5
Find the GP in which 5th term is 27/16 and 9th term is 1/3
  • 1
6 sol:

  • -2
2(i) in the clicked image

  • -2
Please find your solution below

  • 52
hope it will help you : )
  • -2
11*30=330
  • 0
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.
  • 0
330 is the answer

  • 0
330 is the correct answer.
  • 0
Answer =11?30=330
Option:-(2)
  • 0
A11=an+(11-1)d
  • 0
find an explicit formula for the Arithmetic sequence - 2 - 14 - 26 - 38
  • 0
Ok mam
  • 0
Ysjskanba
  • 0
Tuwowkebsgssm sbsnsmsbsb.shsms
Ssbsksnsns
S
Bssjjemsmememmekdmd.d.d
D.d
D
D
D
Dremmejejejsjjsjsie s
Snsjheuebnxjdnd.dndjsnnsns.d
D.jdjejejenejeedddsd
Da
D. Red Jjsjenr
  • 0
How old are margie and tommy
  • 0
What are you looking for?