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Arithmetic Progressions Practice 2: Sum of Terms

Section A: Sum Formula Concepts

Question 1mcq
Why can the sum of an AP be found by pairing terms from the beginning and the end?
A.Each pair has the same sum.
B.Each pair has the same product.
C.The common difference becomes zero.
D.The first term must equal the last term.
Question 2mcq
For 2+5+8+11+142+5+8+11+14, which pairing best explains the AP sum idea?
A.2+52+5, 8+118+11
B.2+142+14, 5+115+11, and middle term 88
C.2×142\times14, 5×115\times11
D.14214-2, 11511-5
Question 3mcq
Find the sum of the first 1010 terms of the AP 3,7,11,3, 7, 11, \ldots.
A.180180
B.190190
C.200200
D.210210
Question 4mcq
Find S20S_{20} for the AP with a=5a=5 and d=3d=3.
A.620620
B.640640
C.670670
D.700700
Question 5mcq
A student calculates S12S_{12} for 4,10,16,4,10,16,\ldots as 122[2(4)+12(6)]\frac{12}{2}[2(4)+12(6)]. What is the error?
A.They used nn instead of n1n-1.
B.They used the wrong first term.
C.They should divide by 1212.
D.They changed the common difference sign.

Section B: Calculate and Check Sums

Question 6mcq
Find the sum of 1515 terms of an AP whose first term is 66 and last term is 4848.
A.405405
B.410410
C.420420
D.430430
Question 7mcq
An AP has 2020 terms. The first term is 33 and the last term is 7979. What is the sum?
A.780780
B.800800
C.820820
D.840840
Question 8mcq
The first nn natural numbers form the AP 1,2,3,,n1,2,3,\ldots,n. What is the sum of the first 5050 natural numbers?
A.12251225
B.12501250
C.12751275
D.25002500
Question 9mcq
The sum of nn terms is found using Sn=n2(a+l)S_n=\frac{n}{2}(a+l). Which information is enough to use this directly?
A.a,d,na,d,n only
B.a,l,na,l,n
C.d,ld,l only
D.a,da,d only
Question 10mcq
Assertion: In an AP, the average of the first and last terms equals the average of all terms. Reason: Opposite terms pair to the same sum. Which is correct?
A.Both are true and the reason explains the assertion.
B.Both are true but the reason does not explain it.
C.Assertion is true but reason is false.
D.Assertion is false but reason is true.