Question :

Q. 1 If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the 10th and the nth terms.

Q. 2 Check whether – 150 is a term of the AP: 11, 8, 5, 2 . . .

Maths

Class 10

##### Views

633

Anaisha Qayanat

Answer 1 :  Given that, Sn = 4n − n2
First term will be, a = S1 = 4(1) − (1)2 = 4 − 1 = 3
Sum of first two terms = S= 4(2) − (2)2 = 8 − 4 = 4
Second term, a2 = S2 − S1 = 4 − 3 = 1
Common difference, d = a2 − a = 1 − 3 = −2

Now we find the nth term is using formula
an = a + (n − 1)d  = 3 + (n − 1) (−2) = 3 − 2n + 2 = 5 − 2n

Therefore, a3 = 5 − 2(3) = 5 − 6 = −1
a10 = 5 − 2(10) = 5 − 20 = −15

Hence, the sum of first two terms is 4. The second term is 1.

The 3rd, 10th, and nth terms are −1, −15, and 5 − 2n respectively.

For the given, A.P. 11, 8, 5, 2, …, Here First term is a = 11, common difference is d = a2 − a1 = 8 − 11 = −3

Let −150 be the nth term of this A.P. Here we know, for an A.P, an = a + (n − 1) d, substitute the value from previous steps, we get

–150 = 11 + (n – 1)(–3)
–150 = 11 – 3n + 3
–164 = –3n
n = 164/3

Clearly, n is not an integer but a fraction, therefore, –150 is not a term of this A.P.

## Advantages Of NCERT, CBSE & State Boards Solutions For All Subjects

• All the NCERT Solutions have been prepared by academic experts having 10+ years of teaching experience.
• They have prepared all the solutions in simple and easy language so that each and every student can understand the concepts easily.
• All the solutions have been explained step to step-wise in details with better explanations.
• Students can also use these question and answers for your assignments and in homework help.
• All the solutions have been explained in detail and the answers have been compiled in a step-wise manner.
• All the questions and answers are commonly prepared according to the Latest Syllabus of Board Education and Guidelines.
• Students can know about the various types of questions asked in the exams with the help of these solutions.