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 . . .

Mathematics

Class 11

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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 = S2 = 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.

Also Check : The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.

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