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 the sum of first m terms of an A.P. is 4m2 – m.
Also. It is given that its nth term is 107,
Now we have to find the value of n. Also, the 21st term of this A.P.
CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 44
Given Sm = 4m2 – m,
On putting m = 1, we get the first term(a), S1 = a = 4(1)2 – 1 = 3.
On putting m = 2 gives S2 = a + a + d = 4(2)2 – 2 = 14
=> d = 14 – 2a
=> d = 14 – 6 = 8
The nth term of the A.P., an = a + (n – 1)d = 107
=> 3 + (n – 1)8 = 107
=> 8(n – 1) = 104
=> n – 1 = 13
=> n = 14
21st term of the A.P., a21 = a + 20d
= 3 + 20(8)
= 163
Hence, the value of n is 14 and 21st term of the A.P. is 163.