Question

If 12th term of an AP is -13 and the sum of its 4 terms is 24, what is the sum of its first 10 terms?​

Answer 1

Answer:

let the 4 terms be (a - 3d), (a - d), (a + d) and (a + 3d).

a/q => a - 3d + a - d + a + d + a + 3d = 24

=> 4a = 24

=> a = 24/4 = 6

Now 12th term =-13

=> a+11d = -13

=> 6+11d = -13

=> 11d = -13-6

=> d = -19/11

Step-by-step explanation:

$sum \: of \: n \: terms \: = \frac{n}{2}(2a + (n - 1)d) \\ n = 10 , \: a = 6 \: ,d = - \frac{19}{11} \\ = > \: sum \: of \: 10 \: terms \: = \frac{10}{2} (2 \times 6 + (10 - 1) \frac{ - 19}{11} ) \\ = 5(12 + \frac{ - 9 \times 19}{11} ) \\ = 5 (\frac{132 - 171}{11} ) \\ = \frac{ - 39 \times 5}{11} \\ = - \frac{195}{11}$