10th Grade > Mathematics
ARITHMETIC PROGRESSIONS MCQs
Total Questions : 107
| Page 5 of 11 pages
Answer: Option C. -> 65
Answer: Option B. -> 156
Question 43.
Soma purchases Nation Saving Certificates every year whose value exceeds the previous year’s purchase by Rs. 400. After 8 years, she find that she has purchased certificates whose total face value is Rs. 48000. what is the face value of the certificates purchased by her in the first year ?
Answer: Option D. -> Rs. 4600
Answer: Option B. -> 1330
Answer: Option C. -> 14280
Answer: Option C. -> 1540
Answer: Option C. -> 89
:
C
The nth term of an AP of first term a and common difference d is
an=a+(n−1)d.⇒a7=a+6d=34−−−(1)a13=a+12d=64−−−(2)(2)−(1)⇒6d=30⇒d=5Substituting d = 5 in (1), we geta=4.∴a18=a+17d=4+(17)5=89
:
C
The nth term of an AP of first term a and common difference d is
an=a+(n−1)d.⇒a7=a+6d=34−−−(1)a13=a+12d=64−−−(2)(2)−(1)⇒6d=30⇒d=5Substituting d = 5 in (1), we geta=4.∴a18=a+17d=4+(17)5=89
:
Here, S14= 1050, n = 14, a = 10.
As Sn= (n2) [2a + (n-1) d]
So,
1050=(142) [20 + 13d]
⇒1050=140+91d
⇒910=91d
⇒d=10
Therefore, a20=10+(20–1)×10=200
i.e. 20th term is 200.
Answer: Option B. -> 4n+1
:
B
Given, Sr=(2r2+3r), the sum upto r terms of an AP.
We know that the nth term of AP is
Tn=Sn−Sn−1.
⇒Sn=2n2+3n and
Sn−1=2(n−1)2+3(n−1).
∴Tn=Sn−Sn−1
=2n2+3n−[2(n2−2n+1)+3n−3]
=2n2+3n−2n2+4n−2−3n+3=4n+1
:
B
Given, Sr=(2r2+3r), the sum upto r terms of an AP.
We know that the nth term of AP is
Tn=Sn−Sn−1.
⇒Sn=2n2+3n and
Sn−1=2(n−1)2+3(n−1).
∴Tn=Sn−Sn−1
=2n2+3n−[2(n2−2n+1)+3n−3]
=2n2+3n−2n2+4n−2−3n+3=4n+1
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