One of the most important and exam oriented question from Chapter name- Arithmetic Progression
Class 10th
Chapter number- 9
Exercise :- 9.6
This type of question has been asked in previous years exams.
In this question we have been given that 12th term of an A.P. is –13 and the sum of the first four terms is 24
Now we have to find the sum of first 10 terms
CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 17
We are given,
12th term of the A.P., (a12) = a + 11d = −13 …. (1)
Sum of first four terms = S4 = 4[2a + (4 − 1)d]/2 = 24
=> 24 = 4[2a + 3d]/2
=> 2a + 3d = 12 ….. (2)
On multiplying eq(1) by 2 and subtracting eq(2) from it we get,
=> (2a+3d) − 2(a+11d) = 12 − 2(−13)
=> 2a + 3d − 2a − 22d = 38
=> −19d = 38
=> d = −2
On putting the value of d in eq(1), we get,
=> a + 11(−2) = −13
=> a = −13 + 22
=> a = 9
Now, sum of first 10 terms is given by,
S10 = 10[2(9) + (10 − 1)(−2)]/2
= 10(18 − 18)/2 = 0
Hence, the sum of first 10 terms of the given A.P. is 0.