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September 9, 2023
An Arithmetic Progression is Represented in the form a, (a + d), (a + 2d), (a + 3d), …where a = the first term, and d = the common difference.
n is number of Terms
General form, Tn = a + (n-1)d
Where Tn is nth term of an Arithmetic Progression
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Question 1 : Find 10th term in the series 1, 3, 5, 7, …
Options A 20 B 19 C 15 D 21
Solution: We know that,
tn = a + (n – 1)dwhere tn = nth term,
a= the first term ,
d= common difference,
n = number of terms in the sequence
In the given series,
a (first term) = 1
d (common difference) = 2 (3 – 1, 5 – 3)
Therefore, 10th term = t10 = a + (n-1) d
t10 = 1 + (10 – 1) 2
t10 = 1 + 18
t10 = 19
Correct option: B
Question 2 : Find last term in the series if there are 8 term in this series 13 , 17 , 21 ,25….
Options A 33 B 41 C 37 D 39
tn = a + (n – 1)d where tn = nth term,
a = the first term ,
d = common difference,
a (first term) = 13
d (common difference) = 4(17 – 13, 21 – 17)
Therefore, 8th term = t8 = a + (n-1) d
t8 = 13 + (8 – 1) 4
t8 = 13 + 28
t8 = 41
Question 1 : Find the number of terms in the series 7, 11, 15, . . .71
OptionsA 12B 25C 22D 17
Solution: We know that, n= [ \frac{(l-a)}{d} ]+ 1
where n = number of terms,
a= the first term,
l = last term,
d= common difference
a (first term) = 7
l (last term) = 71
d (common difference) = 11 – 7 = 4
n= [\frac{(71-7)}{4}]+ 1
n = \frac{64}{4} + 1
n = 16 + 1
n = 17
Correct option: D
Question 2 : Find the number of terms if First term = 22 ,Last term = 50 and common difference is 4
Options:
A 10 B 9 C 8 D 7
a = the first term,
d = common difference
a (first term) = 22
l (last term) = 50
d (common difference) = 4
n = [\frac{(50-22)}{4}]+ 1
n = \frac{28}{4} +1
n = 7+ 1
n = 8
Correct option: C
Question 1. Find the sum of the series 1, 3, 5, 7…. 201
Options A 12101 B 25201 C 22101 D 10201
Sn = \frac{n}{2}[2a + (n − 1) d]
OR
\frac{n}{2} (a+l)where,
l = tn = nth term = a + (n-1)d
a = 1, d = 2, and l = 201
Since we know that, l = a + (n – 1) d
201 = 1 + (n – 1) 2
201 = 1 + 2n -2
202 = 2n
n = 101
Sn = \frac{ n}{2}(a+l)
Sn = \frac{101}{2} (1+201)
Sn = 50.5 (1 + 201)
Sn = 50.5 x 202
Sn = 10201
Question 2. Find the sum of the Arithmetic series if First term of this series is 45 , common difference is 5 and number of terms in this series is 8. Options A 500 B 300 C 400 D 200
\frac{n}{2} (a+l) where, a = the first term,
a = 45, d = 5, and n = 8
Sn = \frac{8}{2}[2\times 45 + (8 − 1) \times 5]
Sn = 4 (90 + 35)
Sn = 4 x 125
Sn = 500
Correct option: A
Question 1.Find the arithmetic mean of first five prime numbers.
A 6.6 B 3.6 C 5.6 D 7.6
Solution: We know that
b = \frac{1}{2} (a + c)
Here, five prime numbers are 2, 3, 5, 7 and 11
Therefore, their arithmetic mean (AM) = \frac{(2+3+5+7+11)}{5} = 5.6
Question 2.Find Second Number if arithmetic mean of two numbers is 24 and First number is 10.
A 32 B 38 C 34 D 36
Let second Number is b
Therefore, their arithmetic mean (AM) = \frac{(10 + b)}{2} = 24
b = 38
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plz check your type 3 question 1 answer and option don’t match
Really sorry for the silly mistake Sunishtha, we’ll fix it up