If 12th term of an A.P. is 6, then the sum of first 23 terms​

2 thoughts on “If 12th term of an A.P. is 6, then the sum of first 23 terms​”

1. Answer:

   a+11d=6.—-1

   multiplying 1 with 2 we get

   2a+22d=12

   sum of first 23 terms will be

   11.5(a+a+22d)

   11.5((2a+22d)

   11.5×12=138.

   boom.its solved.

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2. Step-by-step explanation:

   12

   th

   term of the A.P is −13

   sum of firts four terms is 24

   Let, the 1st term of the A.P is a.

   and common difference of the A.P is d.

   We know, the formula for nth of an A.P. is ,

   t

   n

   =a+(n−1)d

   and, the formula for the sum of n−terms of an A.P. is,

   S

   n

   =

   2

   n

   [2a+(n−1)d]

   By the question,

   t

   12

   ⇒a+(12−1)d=−13

   ⇒a+11d=−13

   ⇒a=−13−11d——(i)

   S

   4

   →

   2

   4

   [2a+(4−1)d]=24

   ⇒2a+3d=12

   ⇒a=

   2

   12−3d

   ——(ii)

   comparing (i) and (ii) we get,

   −13−11d=

   2

   12−3d

   ⇒−22d+3d=26+12

   ⇒−19d=38 ⇒d=−2

   Now, from (i) we get,

   a=−13+22

   ⇒a=9

   Hence, sum of the first 10 terms, is,

   S

   10

   =

   2

   10

   [18+(10−1).(−2)]

   =5[18−18]=0

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