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. Reply

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 Reply

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.

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