This is an arithmetic progression based 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. is 4n – n2,

We have to. Find out what is the first term.

Also we have to find the sum of first two terms the second term and Similarly find the 3rd, 10th and nth terms.

CBSE DHANPAT RAI PUBLICATIONS

Understanding CBSE Mathematics

Class :- 10th

Question no 45

Given S

_{n}= 4n – n^{2},On putting n = 1, we get the first term(a), S

_{1}= a = 4(1) – 1^{2}= 3.On putting n = 2 gives S

_{2}= a + a + d = 4(2) – 2^{2}= 4=> d = 4 – 2a

=> d = 4 – 6 = –2

Second term(a

_{2}) = a + d = 3 + (–2) = 1By using the formula of nth term of an A.P.

a

_{n}= a + (n – 1)dSo,

= 3 + (n – 1)(–2)

= 5 – 2n

Hence, third term(a

_{3}) = 5 – 2(3) = –1And tenth term(a

_{10}) = 5 – 2(10) = –15Hence, the first term is 3, sum of first two terms is 4, second term is 1and 3rd, 10th and nth terms are –1, –15 and 5 – 2n respectively.