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 an arithmetic progression. And we have to find out how many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693.
CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 10(iv)
Given A.P. has first term(a) = 63,
Common difference(d) = 60 – 63 = -3 and sum(Sn) = 636.
We know sum of n terms of an A.P. is given by, Sn = n[2a + (n – 1)d] / 2.
=> 693 = n[2(63) + (n − 1)(−3)]/2
=> 693 = n[126+(−3n + 3)]/2
=> 693 = n[129 − 3n]/2
=> 129n − 3n2 = 1386
=> 3n2 − 129n + 1386 = 0
=> n2 − 43n + 462 = 0
=> n2 − 22n − 21n + 462 = 0
=> n(n − 22) − 21(n − 22) = 0
=> (n −22) (n −21) = 0
=> n = 22 or n = 21
So, the value of n can be both 21 as well as 22 because,
a22 = a + 21d = 63 + 21(− 3) = 63 − 63 = 0
Sum remains the same even if we add the 22nd term because its value is 0.
Hence, the number of terms (n) is 21 or 22.