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 in an A.P. second and third terms are 14 and 18 respectively.
Now we have to find the sum of the first 51 terms
CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 21
Given A.P. has,
Second term, a2 = a + d = 14 …..(1)
Third term, a3 = a + 2d = 18
Hence, common difference(d) = a3 − a2 = 18 − 14 = 4.
On putting d = 4 in (1), we get,
=> a + 4 = 14
=> a = 10
We know sum of n terms of an A.P. is given by Sn = n[2a + (n − 1)d] / 2.
Here a = 10, d = 4 and n = 51. So we get,
S51 = 51[2(10) + (51 − 1)(4)]/2
= 51[20 + 200]/2
= 51[110] = 5610
Hence, the sum of first 51 terms for the given A.P. is 5610.