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The 11th and 13th terms of an AP are 35 and 41 respectively, its common difference is
11th term is 35 so a+10d = 35 => equation 1
13th term is 41 so a+12d = 41 => equation 2
solve both equations by elimination method
we get common difference as 3
An AP consists of 31 terms. If its 16th term is m, then sum of all the terms of this AP is
S_{31} = 31/2(2a + 30d)
a_{16} = a + 15d = m
⇒ S_{31} = (31/2) x 2(a + 15d) ⇒ S_{31} = 31 m
Which term of the AP: 18, 23, 28, 33,... is 98?
98=a+(n1)d
98=18+(n1)5
80=(n1)5
16=n1
17=n
this explains that 98 is 17th term of the AP
A= 15
d = a_{2}  a_{1}
d = 25 / 2  15
d = ( 30  25 / 2 )
d =  5/ 2
16th term of an AP = a + 15 d
15 +( 15 × ( 5 / 2 ))
15 + (75 / 2 )
=  45 / 2
The list of numbers 10, 6, 2, 2,... is
a_{1}=10
a_{2}=6
so the common difference
d=(6(10)) = 4
The sum of first 16 terms of the AP : 10, 6, 2,... is
The formula for the sum of first n terms of a AP is n/2(2a+(n1)d)
here n=16 a=10 d=610=26=4
so sum of first 16 terms=16/2(2·10+(161)·4)
=8(20+15·4)
=8(2060)
=8(40)
=320
First five multiple of 3 are:
3, 6, 9, 12, 15
Here first term, a = 3
common difference, d = 6 – 3 = 3
Number of term, n = 5
⇒ s_{5} = 9 × 5 = 45
Two APs have same common difference. The first term of one of these is 1 and that of the other is  8. Then the difference between their 4th term is
Correct Answer : c
Explanation : ad_{1} = 1
ad_{2} = 8
4th term of ad_{1} is
a_{4} = a + 3d
a_{4} = 1 + 3d.....(1)
Similarly ad_{2} is
a_{4} = 8 + 3d.......(2)
Subtracting (2) from (1), we get
1 + 3d (8 + 3d)
=> 7
The 21st term of the AP whose first two terms are 3 and 4, is
a_{1}=3,
a_{2}=4,
d=4(3)=7,
a_{21}=?,
a_{21}=a+(n1)d
=3+(211)7
= 3+(20)(7)
= 3+140
=137
If 18, a, b, 3 are in AP, then a + b is equal to
We know
AM is ( a + b)/2
(a+b) = [ {18+(3) }/2] X 2 = 15
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