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