Sequences and Series Test - 3

 By formula  
Sn = n/2(a+l)

Even natural numbers which are divisible by 5 are 10,20.30...
Difference between the consecutive terms is same  
∴They form an AP whose first term (a) = 10,
common on difference (d) = 10,
and number of terms (n) = 100
 

∴The sum of first hundred even natural numbers which are divisible by 5  is 50500.

10 x 10th term = 15 x 15th term
let a is the first term and d is the common difference .
10(a+9d) = 15(a+14d)
5a + 120d = 0
a + 24d = 0
now , 25th term = a + (25-1)d
= a+24d = 0
hence 25th term = 0

The series is 1² , 2² , 3² , 4² , 5² , 6² , 7² , ...
Hence, next term = 8²  = 64

a = 62  d = 57 - 62 = -5
tn = a + (n-1)d
= 62 + (n-1)(-5)
= 62 - 5n + 5
= 67 - 5n
From the options, we take '14 '
= 67 - 5(14)
= 67 - 70
= -3 (The first negative term will be at the 14th term)

The series is 24,20,16...in AP
so a = 1st term=24 &common difference = 20-24 = -4
sum = 72 is of say n terms
then 72 = n/2[2*24+(n-1)*-4] = n/2[48-4n+4] = n/2[52-4n]
or 144 = 52n-4n²

or 4n² -52n+144 = 0
or n² -13n+36 = 0
or n² -9n-4n+36 = 0
or n(n-9)-4(n-9) = 0
or (n-9)(n-4) = 0
or n = 9 or 4

Given  x  ²  - 3x = 0  
Factor x out in the expression on the left. 
x (x - 3) = 0  
For the product x (x - 3) to be equal to zero we nedd to have  
x = 0 or x - 3 = 0  
Solve the above simple equations to obtain the solutions. 
x = 0  
or  
x = 3  

Let the three numbers be a-d, a, a+d
ATQ,
a - d + a + a + 2d = 45
3a = 45
a = 15
Middle term is 15

The general form of an AP is  a,  a+d, a+2d,.....
where a is the first term and d is the common difference  
If we double the terms,  the new sequence would be
A , a+2d, a+4d,......
We can observe that this sequence is also an AP
First term is a
Common difference is 2d  nth term= 2a+(n-1)2d         
= 2[a+(n-1)d]