This is the basic and conceptual question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise 9.6
In this question we have been given that the sum of the first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is -30 and common difference is 8.
Now we have to Find the value of n.
CBSE DHANPAT RAI publications
Class:- 10th
Solutions of CBSE Mathematics
Question 73
Let us assume S1 be the sum of the first n terms of an A.P.
whose first term is 8 and the common difference is 20.
Hence, S1 = n[2(8) + (n – 1)20] / 2
= n[16 + 20n – 20] / 2
= n[10n – 2]
Let us assume S2 be the sum of first 2n terms of another A.P.
whose first term is -30 and common difference is 8.
Hence, S2 = 2n[2(–30) + (2n – 1)8] / 2
= n[–60 + 16n – 8]
= n[16n – 68]
According to the question, we have
=> S1 = S2
=> n[10n – 2] = n[16n – 68]
=> 10n – 2 = 16n – 68
=> 6n = 66
=> n = 11
Hence, the value of n is 11.